\(a,\left(2x-5\right)^2=\left(x-2\right)^2\)

\(\Rightarrow\left(2x-5\right)^2-\left(x-2\right)^2=0\)

\(\Rightarrow\left(2x-5-x+2\right)\left(2x-5+x-2\right)=0\)

\(\Rightarrow\left(x-3\right)\left(3x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\3x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)

\(b,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+1\)

\(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10=-4x^2+1\)

\(\Rightarrow-10x-4x^2-12=-4x^2+1\)

\(\Rightarrow-10x-4x^2-12+4x^2-1=0\)

\(\Rightarrow-10x-13=0\)

\(\Rightarrow x=-\dfrac{13}{10}\)

\(b,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+1\) \(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10+4x^2-1=0\)

\(\Rightarrow-10x-9=0\)

\(\Rightarrow x=-\dfrac{9}{10}\)

a) -105 - 5.x = (-5)^2

=>-105 - 5.x = 25

=> 5.x = -105 - 25

=> 5.x = -130

=> x = -130: 5

=> x = -26

c) 400 -4.|5-x| = (-6)^2

=>400- 4.|5-x| = 36

=> 4.|5-x| = 400-36

=> 4.|5-x| = 364

=> |5-x| = 364:4

=> |5-x| = 91

=> \(\left[\begin{matrix}5-x=91\\5-x=-91\end{matrix}\right.\)

=> \(\left[\begin{matrix}x=5-91\\x=5-\left(-91\right)\end{matrix}\right.\)

^{=> \(\left[\begin{matrix}x=-86\\x=96\end{matrix}\right.\)}

d) 2.|x-1| -5 = -15

=>2.|x-1| = -15+5

=>2.|x-1| = 10

=> |x-1| = 10:2

=> |x-1| = 5

=> \(\left[\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)

=>

Bài 1:

\(B=\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)\left(x+2\right)+\left(x-1\right)\)

\(=\left(x-1\right)\left[\left(x-2\right)\left(x-3\right)+x+2+1\right]\)

\(=\left(x-1\right)\left(x^2-5x+6+x+3\right)\)

\(=\left(x-1\right)\left(x^2-4x+9\right)\)

\(=\left(5-1\right)\cdot\left(5^2-4\cdot5+9\right)\)

\(=4\cdot\left(25-20+9\right)=4\cdot14=56\)

5.

(x^2 -1)(x^2 +9) <0

(x+3)(x+1)(x-1)(x-3)<0

x \(\in\)(-3;-1)U(1;3)

b) x² - 5x - 3 = 0

=>

(x+1)(2-x)-(3x+5)(x+2)=-4x²+2

(x+1)(2-x)-(3x+5)(x+2)+4x²+2=0

2x-x²+2-x-3x²-6x-5x-10+4x²+2=0

(-x²-3x²+4x²) + (2x-x-6x-5x) + (2-10+2)=0

0 + -10x + (-6)

-10x-6=0

-10x=0+6

-10x=6

x=\(\dfrac{-6}{10}\) = \(\dfrac{-3}{5}\)

vậy x=\(\dfrac{-3}{5}\)

\(a,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+2\)

\(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10+4x^2-2=0\)

\(\Rightarrow-10x-10=0\)

\(\Rightarrow-10\left(x+1\right)=0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

\(b,x^2-5x-3=0\) câu này đề sai

Mk làm câu tương tự , bn xem lại đề r tự lm nhé

\(x^2-4x+3=0\)

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

\(\Rightarrow\left(x^2-3x\right)-\left(x-3\right)=0\)

\(\Rightarrow x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Đáp án D

Ta có