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a)
\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=-\dfrac{1}{4}-y\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-\dfrac{1}{3}+x=-\dfrac{1}{4}-y\\\dfrac{1}{2}-\dfrac{1}{3}+x=\dfrac{1}{4}+y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-\dfrac{5}{12}\\x-y=\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
ta thấy : \(\left|x-y\right|\ge0\\ \left|y+\dfrac{9}{25}\right|\ge0\)\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
đẳng thửc xảy ra khi : \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow x=y=-\dfrac{9}{25}\)
vậy \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)
c) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
ta thấy \(\left(\dfrac{1}{2}x-5\right)^{20}\:và\:\left(y^2-\dfrac{1}{4}\right)^{10}\) là các lũy thừa có số mũ chẵn
\(\Rightarrow\:\)\(\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\ \left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy cặp số x,y cần tìm là \(\left(10;\dfrac{1}{2}\right)\:hoặc\:\left(10;-\dfrac{1}{2}\right)\)
d)
\(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=x\left(vì\:x\ge0\right)\\ \Leftrightarrow x\left(x^2-\dfrac{9}{4}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy x cần tìm là \(-\dfrac{3}{2};0;\dfrac{3}{2}\)
e)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
ta thấy: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
vậy cặp số cần tìm là \(0;\dfrac{1}{10}\)
a) \(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=4+1\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(x+20\right)^{100}+\left|y+4\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\) \(\forall x,y\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+20=0\\y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-20\\y=0-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{-20;-4\right\}.\)
Chúc bạn học tốt!
b)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[1-\left(x-1\right)^4\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^4=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\Rightarrow x=2\end{matrix}\right.\)
Vậy ...
Làm cho hết chớ .-.
a, \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)\(\Rightarrow x=\frac{5}{6}\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^4=1\end{cases}}\)
Giải: \(\left(x-1\right)^4=1\)\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
c, Vì \(\left(x+20\right)^{100}\ge0\)\(\forall x\inℝ\); \(\left|y+4\right|\ge0\)\(\forall y\inℝ\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)\(\forall x,y\inℝ\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)
d, \(2^{x-1}=16\)\(\Rightarrow2^{x-1}=2^4\)=> x - 1 = 4 => x = 5
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
3: |2x-1|=|x+1|
=>2x-1=x+1 hoặc 2x-1=-x-1
=>x=2 hoặc 3x=0
=>x=2 hoặc x=0
4: \(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{5}=0\\y-\sqrt{3}=0\\x-y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{5}\\y=\sqrt{3}\\z=x-y=-\sqrt{5}-\sqrt{3}\end{matrix}\right.\)
1/ \(\left(x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{25}\\x+1=-\sqrt{25}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
Vậy...
2/ \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^{x+4}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^{x+4}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy...
3/ Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)
Mà \(\left(x+20\right)^{100}+\left|y+4\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+20=0\\y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)
Vậy..
1) (x - 1)2 = 25
(x - 1)2 = 52
=> x - 1 = 5
x = 5 + 1
x = 6