\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

bài 1:

a) x(x-2)-5y-(x-2)=(x-5y)(x-2)

b) =(2x-3-4x)(2x-3+4x)=(-2x-3)(6x-3)

bài 2 bạn tự luyện nhé

Lời giải:
$(4x-2)^2-16x+16x=(4x-2)^2$

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$(3x-4)(3x+4)-(20x^2y-15xy^2):(5xy)$

$=(3x-4)(3x+4)-(4x-3y)$ không phân tích được thành nhân tử.

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$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$

$=3(x-2)(x^2+2x+4)-240x$

$=3(x^3-2^3)-240x=3x^3-240x-24$

$=3(x^3-80x-8)$

\(3x^2y-5xy^2=xy\left(3x-5y\right)\).

1D  2C

Câu 1: D

Câu 2: C

Làm tính nhân

(4x³+3xy²-2y³).(3x²-5xy-6y²)

=12x⁵+12y⁵-20x⁴y-36x²y³-8xy⁴

Phân tích đa thức thành nhân tử

10x³+5x²y-10x²y-10xy²+5y³

=10x³-5x²y-10xy²+5y³

=5(2x³-x²y-2xy²+y³-)

5 x² - 5xy+2x−2y = 2x ( x - y) + 2( x - y) = 2( x + 1)( x - y)

a: =(x^2-x+1)(x^2+x+1)

b: =x^2-6xy+9y^2=(x-3y)^2

c: =5x(x^2-2xy+y^2)

=5x(x-y)^2

d: =(x-3)^2

e: =(2y-z)(4x+7y)

a)HĐT:(x^2+1-x)(x^2+1+x)

b)=x^2-2.x.3y+(3y)^2

c)=5x(x^2-2xy+y^2)

=5x(x-y)^2

d)x^2-2.3.x+3^2

=(x-3)^2

e)(2y-z)+7y(2y-z)

=(2y-z)(1+7y)

Answer:

Câu đầu bạn xem lại.

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

\(=\left(3x\right)^2+2.2x.4+4^2+\left(4x\right)^2-2.4x.3+3^2+2^2-\left(5x\right)^2\)

\(=9x^2+24x+16+16x^2-24x+9+4-25x^2\)

\(=\left(9x^2+16x^2-25x^2\right)+\left(24x-24x\right)+\left(16+9+4\right)\)

\(=29\)

\(\left(5x+y\right).\left(25x^2-5xy+y^2\right)-\left(5x-y\right).\left(25x^2+5xy+y^2\right)\)

\(=\left(5x+y\right).[\left(5x\right)^2-5x.y+y^2]-\left(5x-y\right).[\left(5x\right)^2+5x.y+y^2]\)

\(=\left(5x\right)^3+y^3-[\left(5x\right)^3-y^3]\)

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

\(=2y^3\)