Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)

\(=x^2y^2+a^2b^2+x^2b^2+a^2y^2\)

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

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

(ab-xy)² - (bx- ay)²

= [ (ab- xy) - (bx - ay) ].[ (ab-xy) - (bx- ay) ]

a)x²-xy-x+y

=(x²-x)-(xy-y)

=x(x-1)-y(x-1)

=(x-1)(x-y)

b) xy+4-x²+2y

=(4-x²)+(xy+2y)

=(2-x)(x+2)+y(x+2)

=(x+2)(2-x+y)

c) xy+y-2(x+1)

=y(x+1)-2(x+1)

=(x+1)(y-2)

d) 5(x-y)+ax-ay

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

=(x-y)(5+a)

#H

Trả lời:

a, x² - xy - x + y

= ( x² - xy ) - ( x - y )

= x ( x - y ) - ( x - y )

= ( x - y ) ( x - 1 )

b, xy + 4 - x² + 2y 

= ( xy + 2y ) - ( x² - 4 )

= y ( x + 2 ) - ( x - 2 ) ( x + 2 )

= ( x + 2 ) ( y - x + 2 )

c, xy + y - 2 ( x + 1 ) 

= y ( x + 1 ) - 2 ( x + 1 )

= ( x + 1 ) ( y - 2 )

d, 5 ( x - y ) + ax - ay

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

= ( 5 + a ) ( x - y )

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

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

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

d)\(5\left(x-y\right)+ax-ay=5\left(x-y\right)+\left(ax-ay\right)=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)

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

                                         \(=\left(x-b\right)\left(x-a\right)\)

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

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

f)\(ax^2+ay-bx^2-by=a\left(x^2+y\right)-b\left(x^2+y\right)\)

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

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Mấy bài này khá đơn giản .

Bạn chỉ cần áp dụng hằng đẳng thức \(x^2-y^2=\left(x+y\right)\left(x-y\right)\) là được nhs =))

a)

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

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

\(=\left[xy+4-2\left(x+y\right)\right]\left[xy+4+2\left(x+y\right)\right]\)

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

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

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

b)

\(\left(ab-xy\right)^2-\left(bx-ay\right)^2\)

\(=\left(ab-xy-bx+ay\right)\left(ab-xy+bx-ay\right)\)

\(=\left[a\left(b+y\right)-x\left(b+y\right)\right]\left[a\left(b-y\right)+x\left(b-y\right)\right]\)

\(=\left(b+y\right)\left(a-x\right)\left(a+x\right)\left(b-y\right)\)

c)

\(=\left(x^2+8x-34+3x^2-8x-2\right)\left(x^2+8x-34-3x^2+8x+2\right)\)

\(=\left(4x^2-36\right)\left(-2x^2+16x-32\right)\)

\(=\left(2x-6\right)\left(2x+6\right)\left(-2\right)\left(x^2-8x+16\right)\)

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

Bạn liểm tra lại nhs

Mk lm hay nhấm lắm

=))