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a) \(\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(\Leftrightarrow[\left(x-1\right)\left(x+7\right)][\left(x-2\right)\left(x+8\right)]+8\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)\)+8
Đặt \(x^2+6x-7=a\)
\(\Rightarrow a\left(a-9\right)+8\)\(\)
\(\Leftrightarrow a^2-a-8a+8\)
\(\Leftrightarrow a\left(a-1\right)-8\left(a-1\right)\)
\(\Leftrightarrow\left(a-1\right)\left(a-8\right)\)
\(\Leftrightarrow\left(x^2+6x-8\right)\left(x^2+6x-15\right)\)
Chúc bạn học tốt !!
a) Ta có : a2x + a2y - 7x - 7y
= a2(x + y) - (7x + 7y)
= a2(x + y) - 7(x + y)
= (x + y)(a2 - 7)
b) Ta có : x3 + y(1 - 3x2) + x(3x2 - 1) - y3
= x3 - y(3x2 - 1) + x(3x2 - 1) - y3
= x3 - y3 + [x(3x2 - 1) - y(3x2 - 1)]
= x3 - y3 - (3x2 - 1)(x - y)
= (x - y)(x2 + xy + y2) - (3x2 - 1)(x - y)
= (x - y)[(x2 + xy + y2) - (3x2 - 1)]
= (x - y)(x2 + xy + y2 - 3x2 + 1)
= (x - y)(-2x2 + xy + y2 + 1)
bài 2:a. \(5x.\left(y^2-2yz+z^2\right)\)
\(=5x.\left(y-z\right)^2\) .......k bít dc chưa
b.\(\left(x^2y-x\right)+\left(xy^2-y\right)\)
\(=x.\left(xy-1\right)+y.\left(xy-1\right)\)
\(=\left(xy-1\right).\left(x+y\right)\)
\(b,x^2+4x+3=x^2+3x+x+3.\)
\(=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)
\(c,16x-5x^2-3=x-5x^2+15x-3\)
\(=x\left(1-5x\right)+3\left(5x-1\right)\)
\(=\left(x+3\right)\left(1-5x\right)\)
\(d,x^4+4=x^4+4x^2+4-4x^2=\left(x+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
a) x5 + x + 1
= x5 - x4 + x4 - x3 + x3 - x2 + x2 + x + 1
= ( x5 + x4 + x3) - ( x4 + x3 + x2) + (x2 + x +1)
= x3( x2 + x + 1) - x2( x2 + x + 1) + (x2 + x +1)
= ( x2 + x +1).( x3 - x2 + 1)
b) ( x2 + x)2 -2(x2 + x) -15
=( x2 + x)2 -2(x2 + x).1 + 1- 16
=( x2 + x - 1)2 - 42
=( x2 + x - 1 - 4).( x2 + x - 1+ 4)
=(x2 + x - 5).( x2 + x + 3)
c) x4 + 5x3 + 10x - 4
= (x2)2 - 22 + 5x.( x2 + 2)
=( x2 -2).(x2 + 2) + 5x.( x2 + 2)
= ( x2 + 2).(x2 -2 + 5x)
d) x8 + x7 + 1
= x8 + x7 + x6 - x6 + 1
= x6 ( x2 + x + 1) - ( x6 - 1)
= x6( x2 + x + 1) - ( x3 - 1).(x3 + 1)
= x6( x2 + x + 1) - ( x- 1).( x2 + x + 1).(x3 + 1)
= ( x2 + x + 1).[ x6 -( x- 1).(x3 + 1)]
= ( x2 + x + 1).( x6 - x4 + x3 - x +1)
cau b)
\(B=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-16=\left(x^2+x-1\right)^2-4^2\)\(B=\left(x^2+x-3\right)\left(x^2+x+1\right)\)
\(B=\left[\left(x+\dfrac{1}{2}\right)^2-\left(\sqrt{\dfrac{13}{4}}\right)^2\right]\left(x^2+x+1\right)\)
\(B=\left(x+\dfrac{1-\sqrt{13}}{2}\right)\left(x+\dfrac{1+\sqrt{13}}{2}\right)\left(x^2+x+1\right)\)
b,\(^{x^6-x^4+4x^3+2x^2}\)
\(x^6+4x^3+4-x^4+2x^2-4\)
\(\left(x^3+2\right)^2-\left(x^2-2\right)^2\)
\(\left(x^3-x^2+4\right)\cdot\left(x^3+x^2\right)\)
c \(a^2\cdot\left(x+y\right)+b^2\cdot\left(x+y\right)-2ab\cdot\left(x+y\right)\)
\(\left(x+y\right)\cdot\left(a^2+b^2-2ab\right)\)
\(\left(x+y\right)\cdot\left(a-b\right)^2\)
xin lỗi vì ko có thời gian nên phần d bn tự làm nha