a) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)

\(\Leftrightarrow\left(2.x-1\right)^8-\left(2.x-1\right)^6=0\)

\(\Leftrightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=1\end{matrix}\right.\)

Vậy : \(x\in\left\{\frac{1}{2},1\right\}\)

b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)

\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy : \(x\in\left\{0,1,2\right\}\)

Chúc học tốt nhé !!

a) \(\left(x-1\right)^3=8=2^3\)

\(x-1=2\)

\(x=2+1=3\)

b) \(7^{2x-6}=49=7^2\)

\(2x-6=2\)

\(2x=6+2=8\)

\(x=8:2=4\)

c) \(\left(2x-14\right)^7=128=2^7\)

\(2x-14=2\)

\(2x=14+2=16\)

\(x=16:2=8\)

d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)

\(x=5\)

e) \(3\cdot\left(x+2\right):7\cdot4=120\)

\(x+2=120:3\cdot7:4\)

\(x+2=70\)

\(x=70-2=68\)

Lời giải:

a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$

$\Rightarrow x=3$

b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

c. $(2x-14)^7=128=2^7$

$\Rightarrow 2x-14=2$

$\Rightarrow 2x=16$

$\Rightarrow x=18$

d.

$x^4.x^5=5^3.5^6$

$x^9=5^9$

$\Rightarrow x=5$

e. 

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$

bó tay

-100 taij x=0

câu a) bỏ 1 số 6 đi nha

Ta có (x - 2)² - (x - 3) (x + 3) = 6 

=> (x - 2)² - (x² - 3²) = 6

=> x² - 4x + 2² - x² + 3² = 6

=> x² - x² - 4x + 4 + 9 = 6

=> - 4x + 13 = 6 

=> -4x = -7

=> x = \(\frac{-7}{-4}=\frac{7}{4}\)

a) \(\Leftrightarrow\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9=6\)

\(\Leftrightarrow13-4x=6\)

\(\Leftrightarrow4x=7\)

\(\Leftrightarrow x=\frac{7}{4}\)

b) \(4(x−3)^2−(2x−1)(2x+1)=10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)−4x^2+1=10\)

\(\Leftrightarrow4x^2-24x+36−4x^2+1=10\)

\(\Leftrightarrow37-24x=10\)

\(\Leftrightarrow24x=27\)

\(\Leftrightarrow x=\frac{9}{8}\)

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