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Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
1) \(B=-7x^2+9\)
Do \(x^2\ge0\forall x\Rightarrow-7x^2\le0\forall x\)
\(\Rightarrow B=-7x^2+9\le9\)
\(maxB=9\Leftrightarrow x=0\)
2) \(C=2-\left(3x-4\right)^4\)
Do \(\left(3x-4\right)^4\ge0\forall x\Rightarrow-\left(3x-4\right)^4\le0\forall x\)
\(\Rightarrow C=2-\left(3x-4\right)^4\le2\)
\(maxC=2\Leftrightarrow x=\dfrac{4}{3}\)
3) \(D=\dfrac{1}{2}x^2+3\)
Do \(\dfrac{1}{2}x^2\ge0\forall x\Rightarrow D=\dfrac{1}{2}x^2+3\ge3\)
\(minD=3\Leftrightarrow x=0\)
4) \(E=\dfrac{2016}{2-x^2+3}=\dfrac{2016}{-x^2+5}\)
Do \(x^2\ge0\forall x\Rightarrow-x^2+5\le5\forall x\)
\(\Rightarrow E=\dfrac{2016}{-x^2+5}\ge\dfrac{2016}{5}\)
\(minE=\dfrac{2016}{5}\Leftrightarrow x=0\)
\(B=-7x^2+9\)
Vì \(-7x^2\le0\forall x\)
\(\Rightarrow-7x^2+9\le9\forall x\)
\(\Rightarrow B_{max}=9\Leftrightarrow-7x^2=0\Leftrightarrow x=0\)
\(C=2-\left(3x-4\right)^4\)
Vì \(-\left(3x-4\right)^4\le0\forall x\)
\(\Rightarrow-\left(3x-4\right)^4+2\le2\forall x\)
\(\Rightarrow C_{max}=2\Leftrightarrow-\left(3x-4\right)^4=0\Leftrightarrow x=\dfrac{4}{3}\)
Nếu tìm GTLN thì câu \(d\) là \(D=-\dfrac{1}{2}x^2+3\)
Vì \(-\dfrac{1}{2}x^2\le0\forall x\)
\(\Rightarrow-\dfrac{1}{2}x^2+3\le3\forall x\)
\(\Rightarrow D_{max}=3\Leftrightarrow-\dfrac{1}{2}x^2=0\Leftrightarrow x=0\)
\(E=\dfrac{2016}{2-x^2+3}=\dfrac{2016}{5-x^2}\)
Vì \(x^2\ge0\forall x\)
\(\Rightarrow5-x^2\le5\forall x\)
\(\Rightarrow E_{min}=5\Leftrightarrow x=\dfrac{2016}{5}\)
a) 7/x-1=x+1/9
7/x-x-1=1/9
7/x-x=1/9+1
7/x-x=10/9
còn lại thì bạn tự tính nhé
câu b làm tương tự
Lời giải:
a.
PT $\Leftrightarrow -5x^2+15x-5+x+5x^2=x-2$
$\Leftrightarrow 16x-5=x-2$
$\Leftrightarrow 15x=3$
$\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}$
b.
PT $\Leftrightarrow -4x^2+20x+7x^2-28x-3x^2=12$
$\Leftrightarrow -8x=12$
$\Leftrightarrow x=\frac{-3}{2}$
\(\left(-\dfrac{3}{4}x+1\right)\div\dfrac{2}{3}=1\)
\(-\dfrac{3}{4}x+1=1\times\dfrac{2}{3}\)
\(-\dfrac{3}{4}x+1=\dfrac{2}{3}\)
\(-\dfrac{3}{4}x=\dfrac{2}{3}-1\)
\(-\dfrac{3}{4}x=-\dfrac{1}{3}\)
\(x=-\dfrac{1}{3}\div\left(-\dfrac{3}{4}\right)\)
\(x=\dfrac{4}{9}\)
x+3=6
x=6-3
x=3
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
\(\left(\frac{1}{4}\right)^3\cdot4^3=\left(\frac{1}{4}\cdot4\right)^3=1^3=1\)
\(\frac{1000^4}{250^4}=4^4=256\)
\(2^2\cdot9\cdot\frac{1}{54}\cdot\left(\frac{4}{9}\right)^2=2^2\cdot3^2\cdot2\cdot3^3\cdot\left(\frac{4}{9}\right)^2=\left[\left(2\cdot3\cdot\frac{4}{9}\right)^2\right]\cdot2\cdot3^3=\frac{64}{9}\cdot2\cdot27=384\)
2. a) 2x = 9 => x không thỏa mãn
b) x2 = 9 => x = \(\pm\)3
c) (x + 1)2 = 4 => (x + 1)2 = \(\pm\)22
=> \(\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
Bài 1 :
\(a,\left(\frac{1}{4}\right)^3.4^3\)
\(=\frac{1}{4^3}.4^3\)
\(=1\)
\(b,\frac{1000^4}{250^4}=\frac{\left(250.4\right)^4}{250^4}=\frac{250^4.4^4}{250^4}=4^4=256\)
\(d,2^2.9.\frac{1}{54}.\left(\frac{4}{9}\right)^2\)
\(=36.\frac{1}{54}.\frac{4^2}{9^2}\)
\(=\frac{18.2.16}{18.3.81}\)
\(=\frac{32}{243}\)
Bài 2 :
\(a,2^x=9\)
\(\Rightarrow\)x không thỏa mãn
\(b,x^2=9\)
\(\Rightarrow x^2=3^2\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(c,\left(x+1\right)^2=4\)
\(\Rightarrow\left(x+1\right)^2=2^2\)
\(\Rightarrow\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
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