giống cái kia thôi bn

Mik làm rồi mà

Mà cái bn Nguyễn Duy Đạt gì đó làm thiếu 1 trường hợp

Mà bn vẫn kik hở

Sao zzzzz??????

chắc nhầm ! Xin lỗi ! Xin lỗi !

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

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

''...'' vế sau á :), ko chắc 

.............................................................................................................................................................................................????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????.....................................................................................................................................................................?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

bó tay

-100 taij x=0

\(VT=\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=\left|4\right|=4\)

\(VP=\frac{8}{3\left(x+1\right)^2+2}\le\frac{8}{2}=4\)

\(VT\ge VP\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(2x+3\right)\left(1-2x\right)\ge0\left(1\right)\\\left(x+1\right)^2=0\left(2\right)\end{cases}}\)

\(\left(2\right)\)\(\Leftrightarrow\)\(x=-1\) ( thỏa mãn\(\left(1\right)\) ) 

... 

Ta có: \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

=>  \(\left(2x-1\right)\in\left\{1;-1;0\right\}\)

* Nếu 2x - 1 = 1

=> 2x          = 2

=>   x          = 2 : 2 = 1

* Nếu 2x - 1 = -1

=> 2x          = (-1) +  1

=> 2x           = 0

=>  x            = 0 : 2 = 0

* Nếu 2x - 1 = 0

=> 2x          =  0 + 1 

=> 2x          = 1

=>  x           = 1 : 2

=> x            = 1/2

Vậy x = { 1; 0 ; 1/2 } thì \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

CHÚC BẠN HỌC TỐT

( 2x - 1 )⁶ = ( 2x - 1 )⁸

( 2x - 1 )⁸ - ( 2x - 1 )⁶ = 0

( 2x - 1 )⁶ . ( ( 2x - 1 )² - 1 ) ) = 0

Vậy ( 2x - 1 )⁶ = 0 hoặc ( 2x - 1 )² - 1 = 0

2x - 1 = 0 hoặc \(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}}\)

x=1/2 hoặc \(\orbr{\begin{cases}x=1\\x=0\end{cases}}\)

Vậy x \(\in\){ 1/2; 0 ;1 }

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. 

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

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

$x+2=210:3=70$

$x=70-2=68$

\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)

\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)

\(b,\Rightarrow\left[{}\begin{matrix}6-2x-x-3=8\left(x\le-3\right)\\6-2x+x+3=8\left(-3\le x\le3\right)\\2x-6+x+3=8\left(x>3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\left(ktm\right)\\x=1\left(tm\right)\\x=\dfrac{11}{3}\left(tm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{11}{3}\end{matrix}\right.\)