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6x2+7x-3
=6x2-2x+9x-3
=(6x2-2x)+(9x-3)
=2x(3x-1)+3(3x-1)
=(3x-1)(2x+3)
a) \(x^2-2xy+y^2-xz+yz\)
= \(\left(x-y\right)^2-z\left(x-y\right)\)
= \(\left(x-y\right)\left(x-y-z\right)\)
b) \(x^3+9x^2-4x-36\)
= \(x^3-2x^2+11x^2-22x+18x-36\)
= \(x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\)
= \(\left(x-2\right)\left(x^2+11x+18\right)\)
= \(\left(x-2\right)\left(x^2+2x+9x+18\right)\)
= \(\left(x-2\right)\left(x+2\right)\left(x+9\right)\)
Chuc ban hoc tot
x^8+x+1= (x^8 - x^5) + (x^5 - x^2) + (x^2+x+1)
= x^5.(x^3-1) + x^2.(x^3-1) + (x^2+x+1)
= x^5.(x-1).(x^2+x+1) + x^2.(x-1).(x^2+x+1) + (x^2+x+1)
=(x^2+x+1).[x^5.(x-1)+x^2.(x-1)+1]
b) \(\dfrac{2}{3}x^2y-2xy^2+4xy=2xy\left(\dfrac{1}{3}x-y+2\right)\)
c) \(4x^2-2x-3y-9y^2\)
\(=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-1\right)\)
Ta có: \(A=2\left(x+y\right)^4-3\left(x+y\right)^2-5\)
\(=2\left(x+y\right)^4-5\left(x+y\right)^2+2\left(x+y\right)^2-5\)
\(=\left(x+y\right)^2\left[2\left(x+y\right)^2-5\right]+\left[2\left(x+y\right)^2-5\right]\)
\(=\left[2\left(x+y\right)^2-5\right]\left[\left(x+y\right)^2+1\right]\)
78: \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)-2b\left(a-c\right)\)
\(=b\left(2a-2c\right)-2b\left(a-c\right)\)
=0
x3 - 5x2 + x - 5 = (x3 - 5x2) + (x - 5) = x2(x - 5) + (x - 5) = (x - 5)(x2 + 1)
cho mk hỏi rút gọn bt nay nhé A= 1phần x+1 trừ x-1phân x cộng x+2 phần x^2 +x