\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

Sao đề là phân tích mà lại "= 0" vậy bạn?

\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)

a: \(5x\left(2x+3\right)+6x+9\)

\(=5x\left(2x+3\right)+\left(6x+9\right)\)

\(=5x\left(2x+3\right)+3\left(2x+3\right)\)

\(=\left(2x+3\right)\left(5x+3\right)\)

b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)

\(=\left(x+4\right)\left(3x+48+5\right)\)

=(x+4)(3x+53)

Ta có:

(x + 2)(x + 3)(x + 4)(x + 5) - 24

= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24

= (x² + 5x + 2x + 10)(x² + 4x + 3x + 12) - 24

= (x² + 7x + 10)(x² + 7x + 12) - 24

Đặt x² + 7x + 10 = k 

=> k(k + 2) - 24 = k² + 2k - 24 = k² + 6x - 4x - 24 

                            = k(k + 6)  - 4(k  + 6)

                          = (k - 4)(k + 6)

=> (x + 2)(x + 3)(x + 4)(x + 5) - 24

= (x² + 7x + 10 - 4)(x² + 7x + 10 + 6)

= (x² + 7x + 6)(x² + 7x + 16)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)

Đặt \(x^2+7x+11=t\)thay vào (1) ta được:
\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24\)

\(=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:

\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left[x\left(x+1\right)+6\left(x+1\right)\right]\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

\(\left(x+2\right).\left(x+5\right).\left(x+3\right).\left(x+4\right)-24\)

\(\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Dat \(x^2+7x+10=y\)

ta co :\(y\left(y+2\right)-24=y^2+2y-24\)

\(y^2+6y-4y-24=\left(y^2+6y\right)-\left(4y+24\right)\)

\(y\left(y+6\right)-4\left(y+6\right)=\left(y-4\right)\left(y+6\right)\)

tra lai bien ta co

\(\left(x^2+7x+10-4\right)\left(x^2+7x+12+6\right)\)

\(\left(x^2+7x+6\right)\left(x^2+7x+18\right)=\left(x^2+6x+x+6\right)\left(x^2+7x+18\right)\)

\(\left(x+1\right)\left(x+6\right)\left(x^2+7x+18\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Let \(t=x^2+7x+10\) we have:

\(=t\left(t+2\right)-24=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

(x - 5)² - 4(x - 3)² + 2(2x - 1)(x - 5) + (2x - 1)²

= [(x - 5)² + 2(2x - 1)(x - 5) + (2x - 1)²) - [2(x - 3)]²

= (x - 5 + 2x - 1)² - (2x - 6)²

= (3x - 6)² - (2x - 6)²

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

( x - 5 )² - 4( x - 3 )² + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )²

= [ ( x - 5 )² + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )² ] - 2²( x - 3 )²

= ( x - 5 + 2x - 1 )² - ( 2x - 6 )²

= ( 3x - 6 )² - ( 2x - 6 )²

= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )

= x( 5x - 12 )

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

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

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

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

\(=\left[\left(3x-6\right)-\left(2x-6\right)\right].\left[\left(3x-6\right)+\left(2x-6\right)\right]\)

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

\(=\left(5x-12\right)x\)

B

Lời giải:

a. 

$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

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

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

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