(8x-3)(3x+2)-(4x+7)(x+4) = (2x+1)(5x-1)-33

(24x²-9x+16x-6)-(4x²+7x+16x+28) = (10x²+5x-2x-1)-33

24x²+7x-6-4x²-23x-28 = 10x²+3x-1-33

20x²-16x-34 = 10x²+3x-34

<=> 20x²-16x = 10x²+3x

2x²-19x=0

2x(x-19)=0

=>\(\left[{}\begin{matrix}2x=0\Rightarrow x=0\\x-19=0\Rightarrow x=19\end{matrix}\right.\)

Không chắc lắm :)

\(\Leftrightarrow4\left(x^2+x-2\right)-\left(4x^2+11x-3\right)=2x-2\)

\(\Leftrightarrow4x^2+4x-8-4x^2-11x+3=2x-2\)

=>-7x-5=2x-2

=>-9x=3

hay x=-1/3

\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)

\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

Đặt \(p=x^2-4,5x-8\)ta có :

\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)

\(A=p^2-\left(2,5x\right)^2+4x^2\)

\(A=p^2-6,25x^2+4x^2\)

\(A=p^2-2,25x^2\)

\(A=p^2-\left(1,5x\right)^2\)

\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)

Thay \(p=x^2-4,5x-8\)vào A ta có :

\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)

\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)

\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(=\left(x+1\right)\left(x-8\right)\left(x-4\right)\left(x+2\right)+4x^2\)

\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

  Đặt \(x^2-2x-8=t\)

  Ta có : \(\left(t-5x\right)t+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)

    Học tốt ~~

a,       -5x^2 - x + 6 = 0

<=>    -5x^2 - x -5 -1 =0

<=>    (-5x^2-5)-(x+1)=0

<=>    -5(x^2-1)-(x+1)=0

<=>   -5(x-1)(x+1)-(x+1)=0

<=>    (x+1)(-5x+5-1)=0

<=>    (x+1)(-5x+4)=o

<=>    x+1=0  hoặc -5x+4=0

*) x+1=0<=>x=-1

*) -5x+4=0<=> x=4/5

b)      4x^2+x-5= 0

<=>  4x^2 - 4 +x-1=0

<=>  4(x^2-1)+(x-1)=0

<=> 4(x-1)(x+1)+(x-1)=0

<=>(x-1)(4x+4+1)=0

<=>(x-1)(4x+5)=0

<=>  x-1=0 hoặc 4x+5=0 

+) x-1=0<=> x=1

+)4x+5=0<=>x=-5/4

...cách làm .. ptđt thành nhân tử

  .giải .... >>

  a . \(-5x^2-x+6=0\)

 <=> (-5x^2 +5x ) - ( 6x-6)=0 <=> -5x(x-5) -6 (x-1) =0 <=> (x-1)(-5x-6/5)=0 <=> x=1 hoặc x =-6/5

b.<=> 4x^2-4x +5x-5 = 0 <=.> 4x(x-1) +5 (x-1) <=> (x-1)(4x+5)=0 <=> x= 1 hoặc x= -5/4

...... chuẩn ko cần chỉnh .. check liền tay ..The end•••

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

2) \(5x^2-5y^2-10x+10y=5\left(x^2-y^2\right)-10\left(x-y\right)\)

            \(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x+5y-10\right)\)

3) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)

4) \(ax-bx-a^2+2ab-b^2=x\left(a-b\right)-\left(a^2-2ab+b^2\right)\)

                            \(=x\left(a-b\right)-\left(a-b\right)^2=\left(a-b\right)\left(x-a+b\right)\)

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

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

6) \(x^2+4x-y^2+4=x^2+4x+4-y^2=\left(x+2\right)^2-y^2\)

                                   \(=\left(x+2-y\right)\left(x+2+y\right)\)

Phân tích các đa thức sau thành nhân tử :

1) x^3 + x^2y - 4x - 4y

2) x^3 - 3x^2 +1 - 3x

3) 3x^2 - 6xy + 3y^2 - 12z^2

4) x^2 - 2x - 15

5) 2x^2 +3x - 5

6) 2x^2 - 18

7) x^2 - 7xy + 10y^2

8) x^3 - 2x^2 + x - xy^2

Làm nhanh giúp mình với nhé .....mình đang cần gấp[[[[

\(x=\frac{1}{2}\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\frac{1}{2}.\left(\sqrt{2}-1\right)\)

\(\Rightarrow2x=\sqrt{2}-1\Rightarrow2x+1=\sqrt{2}\)

\(\Rightarrow4x^2+4x+1=2\Rightarrow4x^2+4x-1=0\)

\(B=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1-1\right]^{2018}+2018\)

\(=\left(-1\right)^{2018}+2018=2019\)