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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 \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
b) | 3x - 4 | + | 5y + 5 | = 0
Ta có \(\hept{\begin{cases}\left|3x-4\right|\ge0\\\left|5y+5\right|\ge0\end{cases}\forall xy}\)
\(\Leftrightarrow\left|3x-4\right|+\left|5y+5\right|\ge0\forall xy\)
Do đó để tổng | 3x - 4 | + | 5y + 5 | = 0 thì \(\hept{\begin{cases}\left|3x-4\right|=0\\\left|5y+5\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x-4=0\\5y+5=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x=4\\5y=-5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=-1\end{cases}}\)
Vậy \(x=\frac{4}{3}\) và y= - 1
c) | x + 3 | + | x + 1 | = 3x (*1)
Ta có \(\hept{\begin{cases}\left|x+3\right|\ge0\\\left|x+1\right|\ge0\end{cases}\forall x}\)
\(\Leftrightarrow\) | x + 3 | + | x + 1 | \(\ge0\forall\)x
\(\Leftrightarrow3x\ge0\forall x\)
\(\Leftrightarrow x\ge0\)
\(\Leftrightarrow x+3>x+1>x\ge0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=x+3\\\left|x+1\right|=x+1\end{cases}}\)
\(\Leftrightarrow\left|x+3\right|+\left|x+1\right|=x+3+x+1\)
\(\Leftrightarrow\left|x+3\right|+\left|x+1\right|=2x+4\) (*2)
Từ (*1) và (*2) <=> 2x + 4 = 3x
\(\Leftrightarrow4=3x-2x\)
\(\Leftrightarrow x=4\)
Vậy x = 4
Câu a t đang nghi sai đề
Lát t lm đc thì lm sau nhé
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
hình như mk thấy có phần tương tự trong sbt oán 7 ở phần nào đó thì phải . Bạn về nhà tìm thử xem sau đó mở đáp án ở sau mà coi
Lí luận chung cho cả 3 câu :
Vì GTTĐ luôn lớn hơn hoặc bằng 0
a) \(\Rightarrow\hept{\begin{cases}x+\frac{3}{7}=0\\y-\frac{4}{9}=0\\z+\frac{5}{11}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-3}{7}\\y=\frac{4}{9}\\z=\frac{-5}{11}\end{cases}}}\)
b)\(\Rightarrow\hept{\begin{cases}x-\frac{2}{5}=0\\x+y-\frac{1}{2}=0\\y-z+\frac{3}{5}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{1}{10}\\z=\frac{7}{10}\end{cases}}}\)
c)\(\Rightarrow\hept{\begin{cases}x+y-2,8=0\\y+z+4=0\\z+x-1,4=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=2,8\\y+z=-4\\z+x=1,4\end{cases}}}\)
\(\Rightarrow x+y+y+z+z+x=2,8-4+1,4\)
\(\Rightarrow2\left(x+y+z\right)=0,2\)
\(\Rightarrow x+y+z=0,1\)
Từ đây tìm đc x, y, z
Ta có : \(\left(3x-\frac{y}{5}\right)^2\ge0;\left(2y+\frac{3}{7}\right)^2\ge0\)
\(=>\left(3x-\frac{y}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2\ge0\)
Mà \(\left(3x-\frac{y}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2=0\)nên dấu "=" xảy ra
\(< =>\hept{\begin{cases}3x-\frac{y}{5}=0\\2y+\frac{3}{7}=0\end{cases}}< =>\hept{\begin{cases}3x-\frac{y}{5}=0\\y=-\frac{3}{14}\end{cases}}\)
\(< =>\hept{\begin{cases}x=-\frac{1}{70}\\y=-\frac{3}{14}\end{cases}}\)
Ta có : \(\left(x+y-\frac{1}{4}\right)^2\ge0;\left(x-y+\frac{1}{5}\right)^2\ge0\)
Cộng theo vế ta được : \(\left(x+y-\frac{1}{4}\right)^2+\left(x-y+\frac{1}{5}\right)^2\ge0\)
Mà \(\left(x+y-\frac{1}{4}\right)^2+\left(x-y+\frac{1}{5}\right)^2=0\)nên dấu "=" xảy ra
\(< =>\hept{\begin{cases}y+x=\frac{1}{4}\\y-x=\frac{1}{5}\end{cases}}< =>\hept{\begin{cases}y=\frac{9}{40}\\x=\frac{1}{40}\end{cases}}\)
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
=>\(3x-\frac{1}{2}=0;\frac{1}{2}y+\frac{3}{5}=0\left(\left|3x-\frac{1}{2}\right|;\left|\frac{1}{2}y+\frac{3}{5}\right|\ge0\right)\)
=>\(x=\frac{1}{6};y=\frac{-6}{5}\)
b)\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
Ta lại có:
\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\)
=>\(\frac{3}{2}x+\frac{1}{9}=0;\frac{1}{5}y-\frac{1}{2}=0\Rightarrow x=-\frac{2}{27};y=\frac{5}{2}\)
b, \(\Leftrightarrow x\left(x-3\right)+\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\2x+1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\2x=-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{-1}{2}\end{array}\right.\)
a) |x-y|+|x-9|=0
=>
b) |x2-3x|+|(x+1).(x-3)|=0
xét x2-3x|=0
=> x2-3x=0
x(x-3)=0
=>x=0 hoặc x-3=0
=> x=3
|(x+1)(x-3)|=0
=> (x+1)(x-3)=0
th1 x=0
(0+1).(0-3)=0
-1.(-3)=0(loại)
th2 x=3
(3+1)(3-3)=0
4.0=0 (lấy)
=> x=0