Ta có 

x*(x-3)+5*(x-3)=0

=>(x-3)(x+5)=0

=> x-3=0 hoặc x+5=0

=> x=3 hoặc x=-5

Ta có

(x+3)*(x55) là tương tự trên

Ta có

7*(x-3)-4(x-3)=0

=>(7-4)(x-3)=0

=>x=3

KL

Vì (3.x-5)-(2.x-7)=0 nên 3.x-5=2.x-7( vì 2 số bằng nhau trừ cho nhau bằng 0)

Có 3.x-5=2.x-7 ( bên dưới mình áp dụng quy tắc chuyển vế nhé)

3.x-2.x=5-7

x=-2

Tick cho mình nhé. Chắc chắn đúng

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Leftrightarrow x^2-5x-3x+15+1=0\)

\(\Leftrightarrow x^2-8x+16=0\)

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

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\)

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Rightarrow x^2-3x-5x+15+1=0\)

\(\Rightarrow x^2-8x+16=0\)

\(\Rightarrow x^2-2x.4+4^2=0\)

\(\Rightarrow\left(x-4\right)^2=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

\(\left(x-3\right)\left(x-5\right)+1=0\)

\(\Leftrightarrow x^2-5x-3x+15+1=0\)

\(\Leftrightarrow x^2-8x+16=0\)

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

\(\Leftrightarrow x=4\)

 20 . 2^x + 1 = 10.4^2 + 1 

 20 . 2^x + 1 = 10 . 16 + 1

20 . 2^x + 1 = 161

20 . 2^x = 161 - 1

20 . 2^x = 160

2^x = 8

2^x = 2^3

=> x = 3

Ban co giai dc phan 2 ko ?

( 4 - x : 2 )^3 - 1 = 2 . ( 2^3 - 5 : 2^0 )

( 4 - x : 2 )^3 - 1 = 2 . ( 8 - 5 : 1 )

( 4 - x : 2 )^3 - 1 = 2 . 3

( 4 - x : 2 )^3 - 1 = 6

( 4 - x : 2 )^3  = 7

=> ko tìm đc x

\(\dfrac{5}{6}-\dfrac{1}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{2}{5}x=0\Rightarrow\dfrac{1}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{2}{5}x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{2}{5}x=\dfrac{5}{6}\Rightarrow\dfrac{1}{2}x-\dfrac{2}{5}x=\dfrac{5}{6}+\dfrac{1}{6}=1\)

\(\Rightarrow x\left(\dfrac{1}{2}-\dfrac{2}{5}\right)=1\Rightarrow\dfrac{1}{10}x=1\Rightarrow x=1:\dfrac{1}{10}=10\)

Vậy x = 10

( x-3 ) ( x-5 ) < 0

\(\Rightarrow\)\(\left\{{}\begin{matrix}x-3< 0\Rightarrow x< 3\\x-5>0\Rightarrow x>5\end{matrix}\right.\)

\(\Rightarrow\) 5 < x < 3

\(\Rightarrow\) x \(\in\) {\(\varnothing\)}

\(\Rightarrow\)\(\left\{{}\begin{matrix}x-3>0\Rightarrow x>3\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\)

\(\Rightarrow\)3 < x < 5

\(\Rightarrow\) x = 4