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1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
Gửi Thắng Nguyễn: Mình không biết tại sao lại ko phân tích được?
\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left[\left(a-b\right)+\left(a+c\right)\right]\)
\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left(a-b\right)+bc\left(a+c\right)\)
\(=\left(a-b\right)\left(ab+bc\right)+\left(a+c\right)\left(bc-ac\right)\)
\(=b\left(a-b\right)\left(a+c\right)-c\left(a+c\right)\left(a-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+c\right)\)
a) 2x + 2y - x2 - xy
= 2(x + y) + x(x + y)
= (x + y) (x + 2)
mk ko bít phân tích đúng ko đúng thì t i c k nhé!! 245433463463564564574675687687856856846865855476457
a)\(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
b)\(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)
\(=\left(x+3\right)\left[\left(x+3\right)-\left(2x-5\right)\right]\)
\(=\left(x+3\right)\left(8-x\right)\)
c)\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)
\(=\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x-2\right)^2\)
\(=\left(3x+2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]+\left(3x-2\right)\left[\left(3x-2\right)-\left(3x+2\right)\right]\)
\(=4\left(3x+2\right)-4\left(3x-2\right)\)
\(=4\left(3x+2-3x+2\right)\)
=4.4=16
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
Bài 1 : Ta có : x3 + 2x2 + x
= x3 + x2 + x2 + x
= x2(x + 1) + x(x + 1)
= (x2 + x)(x + 1)
= x(x + 1)2
Bài : 2 :
a) Ta có : \(\frac{2}{3}x\left(x^2-4\right)=0\)
\(\Rightarrow\frac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
=> x = 0
x - 2 = 0
x + 2 = 0
=> x = 0
x = 2
x = -2
\(4\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-6\right)-3x^2\)
\(=4\left(x-2\right)\left(x-6\right)\left(x-3\right)\left(x-4\right)-3x^2\)
\(=4\left(x^2-8x+12\right)\left(x^2-7x+12\right)-3x^2\)
Đặt: \(x^2-8x+12=y\)
\(\Leftrightarrow4y\left(y-x\right)-3x^2\)
\(=4y^2-4yx-3x^2\)
\(=\left(2y+x\right)\left(2y-3x\right)\)
\(=\left(2x^2-15x+24\right)\left(2x^2-19x+24\right)\)