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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: \(\left(xy+ab\right)^2+\left(bx-ay\right)^2\)
\(=x^2y^2+a^2b^2+x^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(b^2+y^2\right)\left(x^2+a^2\right)\)
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 )
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
=))
ảnh giỏi lắm đừng coi thường à. olm tới 11000 điểm mà chỉ 1 tuần dc 1000 đỉm á
các bạn không trả lời cho bạn à
Bạn cx có làm đc bài đó đâu Despicable