Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(8x^2+30x+7\)
\(=8x^2+28x+2x+7\)
\(=4x\left(2x+7\right)+\left(2x+7\right)\)
\(=\left(2x+7\right)\left(4x+1\right)\)
b) Ta có: \(4x^3-12x^2+9x\)
\(=x\left(4x^2-12x+9\right)\)
\(=x\left(2x-3\right)^2\)
c) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=\left(x+2\right)\cdot3x\)
d) Ta có: \(ab+c^2-ac-bc\)
\(=\left(ab-bc\right)+\left(c^2-ac\right)\)
\(=b\left(a-c\right)+c\left(c-a\right)\)
\(=b\left(a-c\right)-c\left(a-c\right)\)
\(=\left(a-c\right)\left(b-c\right)\)
e) Ta có: \(4x^2-y^2+1-4x\)
\(=\left(4x^2-4x+1\right)-y^2\)
\(=\left(2x-1\right)^2-y^2\)
\(=\left(2x-1-y\right)\left(2x-1+y\right)\)
f) Ta có: \(6x^2-7x-20\)
\(=6x^2-15x+8x-20\)
\(=3x\left(2x-5\right)+4\left(2x-5\right)\)
\(=\left(2x-5\right)\left(3x+4\right)\)
\(4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\), \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=\left(x+2\right)3x\)
\(ab+c^2-ac-bc=ab-ac-bc+c^2=a\left(b-c\right)-c\left(b-c\right)=\left(b-c\right)\left(a-c\right)\)
\(4x^2-y^2+1-4x=4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-y-1\right)\left(2x+y-1\right)\)
\(6x^2-7x-20=6x^2-15x+8x-20=3x\left(2x-5\right)+4\left(2x-5\right)=\left(2x-5\right)\left(3x+4\right)\)
\(8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(4x+1\right)\left(2x+7\right)\)
\(2x-1^3+8\)
\(=2x-9\)
\(=\left(\sqrt{2x}\right)^2-3^2\)
\(=\left(\sqrt{2x}-3\right)\left(\sqrt{2x}+3\right)\)
_________
\(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)
\(=\left(2x-1\right)^3\)
_______________
\(8x^3-12x^2+6x-2\)
\(=8x^3-12x^2+6x-1-1\)
\(=\left(2x-1\right)^3-1\)
\(=\left(2x-1-1\right)\left(4x^2-4x+1+2x-1+1\right)\)
\(=\left(2x-2\right)\left(4x^2-2x+1\right)\)
\(=2\left(x-1\right)\left(4x^2-2x+1\right)\)
________
\(9x^3-12x^2+6x-1\)
\(=x^3+8x^3-12x^2+6x-1\)
\(=x^3+\left(2x-1\right)^3\)
\(=\left(x+2x-1\right)\left(x^2-2x^2-x+4x^2-4x+1\right)\)
\(=\left(3x-1\right)\left(3x^2-5x+1\right)\)
b: 8x^3-12x^2+6x-1
=(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3
=(2x-1)^3
c: =(8x^3-12x^2+6x-1)-1
=(2x-1)^3-1
=(2x-1-1)[(2x-1)^2+2x-1+1]
=2(x-1)(4x^2-4x+1+2x)
=2(x-1)(4x^2-2x+1)
\(A=\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\times\frac{x^2-36}{12x^2+12}\)
\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\times\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)
\(A=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}\times\frac{1}{12\left(x^2+1\right)}\)
\(A=\frac{12\left(x^2+1\right)}{x}\times\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)
a) Ta có:(6x+1)^2 +(6x-1)^2 +2(6x+1)(6x-1) =[(6x+1)+(6x-1)]^2 =(12x)^2=(12^2)(x^2)=144.x^2
b) Ta có:(6x+1)^2 +(6x-1)^2 -(12x+2)(6x-1)=(6x+1)^2 +(6x-1)^2 -2(6x+1)(6x-1)=[(6x+1)-(6x-1)]^2=2^2=4
c) Ta có:(ac+bd)^2 +(ad-bc)^2=(ac)^2 +2(ac)(bd) +(bd^2) +(ad)^2 -2(ad)(bc) +(bc)^2
=a^2.c^2 +2abcd +b^2 d^2 +a^2.d^2 -2abcd +b^2.c^2=a^2.c^2 +b^2.d^2 +a^2.d^2 +b^2.c^2
=(a^2 +b^2)(c^2 +d^2)
d) Ta có:(ac-bd)(ac+bd)=(ac)^2 -(bd)^2=a^2.c^2 -b^2.d^2