\((2x-7)^3=8(7-2x)^2\)

⇔ \((2x-7)^3=8(2x-7)^2\)       (*)

\(TH1: (2x-7)^2=0\)

Khi đó: \(2x-7=0\)  ⇔ \(x=\dfrac{7}{2} \)

\(TH2:\left(2x-7\right)^2\ne0\) 

Khi đó: (*) ⇔ \(2x-7=8\) (chia 2 vế cho \((2x-7)^2\))

⇔ \(x=\dfrac{15}{2} \) 

Vậy \(x=\dfrac{15}{2}\); \(x=\dfrac{7}{2}\)

`(2x+5)(2x-7)-(2x-3)^2=36`

`<=>4x^2-14x+10x-35-(4x^2-12x+9)=36`

`<=>4x^2-4x-35-4x^2+12x-9=36`

`<=>8x-44=36`

`<=>8x=80`

`<=>x=10`

Vậy `S={10}`

Ta có: \(\left(2x+5\right)\left(2x-7\right)-\left(2x-3\right)^2=36\)

\(\Leftrightarrow4x^2-14x+10x-35-\left(4x^2-12x+9\right)=36\)

\(\Leftrightarrow4x^2-4x-35-4x^2+12x-9=36\)

\(\Leftrightarrow8x-44=36\)

\(\Leftrightarrow8x=80\)

hay x=10

Vậy: S={10}

\(\Leftrightarrow\left(2x-3\right)^7\left(2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)

a ) \(\left|2x-3\right|=\left|x+5\right|\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=x+5\\2x-3=-x+5\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-x=5+3\\2x+x=5+3\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\3x=8\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=\frac{8}{3}\left(loại\right)\end{array}\right.\)

Vậy ...........

b ) \(\left|x\right|+2x=7\)

\(\Rightarrow\left[\begin{array}{nghiempt}-x+2x=7\\x+2x=7\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=7\\3x=7\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=7\left(loại\right)\\x=\frac{7}{3}\end{array}\right.\)

Vậy .........................

c ) \(\left|x\right|+2x=7\) ( giống câu b ) 

a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.

a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 ) 

<=> 2x - 3 - x + 5 = x + 7 - x - 2

<=> x = 3

b)(7x-5)-(6x+4)=(2x+3)-(2x+1)

<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1

<=> x = 11

c)(9x-3)-(8x+5)=(3x+2)

<=> 9x - 3 - 8x - 5 = 3x + 2

<=> -2x = 10

<=> x = -5

d)(x+7)-(2x+3)=(3x+5)-(2x+4)

<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4

<=> -2x = -3

<=> x = 3/2

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Mài ngu

\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{7^x.\left(7^2+7+1\right)}{57}=7^x\)

\(\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}=\frac{5^{2x}\left(1+5+5^3\right)}{131}=\frac{25^x.131}{131}=25^x\)

\(\Rightarrow7^x=25^x\Rightarrow x=0\)

ai tích mình mình tích lại cho

ai tích mình mình tích lại cho