a) (x-1)*(x+2)-(x-3)*(-x+4)=19

\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)

\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)

\(\Leftrightarrow2x^2-3x+7-19=0\)

\(\Leftrightarrow2x^2-3x-12=0\)

Đề sai??

b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)

\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)

\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)

\(\Leftrightarrow-30x^2+7x-4+17=0\)

\(\Leftrightarrow-30x^2+7x+13=0\)

???

c) (x+1)*(x+1)-(x-1)*(x-1)=9

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

\(\Leftrightarrow\left(x+1+x-1\right)\left(x+1-x+1\right)=9\)

\(\Leftrightarrow2x.2=9\)

\(\Leftrightarrow x=\frac{9}{4}\)

b1 A=10000 B=23386

b2 a,x=5 

b,x=4 x=2 

có lẽ bn nên tự lm thì hơn

a: \(x^2-10x=-25\)

\(\Leftrightarrow x-5=0\)

hay x=5

b: \(x^2-6x+8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(A=\frac{3\left(x^2+6x+13\right)+4}{x^2+6x+13}=3+\frac{4}{x^2+6x+13}=3+\frac{4}{\left(x+3\right)^2+4}\le3+1=4\)

Dấu "=" xảy ra khi: \(x+3=0\Rightarrow x=-3\)

Vậy GTLN của A là 4 khi x = -3

a) (x + 3)² - (x - 2)² = 2x

=> (x + 3 - x + 2)(x + 3 + x - 2) = 2x

=> 5(2x + 1) = 2x

=> 10x + 5 = 2x

=> 10x - 2x = -5

=> 8x = -5

=> x = -5/8

b) 7x(x - 2) = x - 2

=> 7x(x - 2) - (x - 2) = 0

=> (7x - 1)(x - 2) = 0

=> \(\orbr{\begin{cases}7x-1=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{7}\\x=2\end{cases}}\)

c) 8x³ - 12x² + 6x - 1 = 0

=> (2x - 1)³ = 0

=> 2x - 1 = 0

=> 2x = 1

=> x = 1/2