Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(1-2x\right)^2=\left(3x-2\right)^2\)
\(=\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
\(\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
\(\left(3-5x\right)\left(x-1\right)=0\)
\(\Rightarrow3-5x=0\) \(x-1=0\)
\(\Rightarrow x=\frac{3}{5}\) or \(x=1\)
b)\(\left(x-2\right)^3+\left(5-2x\right)^3\)
=\(\left(x-2+5-2x\right)\left(\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right)\)
\(\left(3-x\right)\left(x^2-4x+4-5x+2x^2+10-4x+25-20x+4x^2\right)\)
(\(\left(3-x\right)\left(7x^2-33x+39\right)\)
..............
Bài 9:
a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)
\(=4x^2+4xy+y^2-4x^2+y^2-xy-y^2\)
\(=3xy-y^2\)
\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)
b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{31}{2}\)
Bài 7:
a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)
b) \(93\cdot107=100^2-7^2=10000-49=9951\)
c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)
d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)
e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-18^8+1=1\)
f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)
\(a,=\left(73-27\right)\left(73^2+73\cdot27+27^2\right)=46\cdot8029=369334\\ b,=\left(63-27\right)\left(63+27\right)+\left(72-18\right)\left(72+18\right)\\ =100\cdot36+54\cdot100=100\left(36+54\right)=100\cdot90=9000\\ c,=\left(105-5\right)^3=100^3=1000000\)
a: \(=x^2+2x-8-x^2-2x-1=-9\)
b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
h) \(y\left(y-x\right)^3-x\left(x-y\right)^2+xy\left(x-y\right)=y\left(y-x\right)^3-x\left(y-x\right)^2-xy\left(y-x\right)=\left(y-x\right)\left[y\left(y-x\right)^2-x-xy\right]=\left(y-x\right)\left[y\left(y^2-2xy+x^2\right)-x-xy\right]=\left(y-x\right)\left(y^3-2xy^2+x^2y-x-xy\right)\)
i) \(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2=10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(a-2b\right)^2=\left(a-2b\right)^2\left(10x^2-x^2-2\right)=\left(a-2b\right)^2\left(9x^2-2\right)\)
Bài 1:
\(a,=6x^2+19x-7-6x^3-4x^2+7x=-6x^3+2x^2+26x-7\\ b,B=26\cdot\left(63^2+63\cdot37+37^2\right):26+63\cdot37\\ =63^2+63\cdot37+37^2+63\cdot37\\ =\left(63+37\right)^2=100^2=10000\)
Bài 2:
\(a,=x\left(y^2-25\right)=x\left(y-5\right)\left(y+5\right)\\ b,=\left(x-y\right)\left(x+2\right)\\ c,=\left(x-3\right)\left(x^2-4\right)=\left(x-2\right)\left(x-3\right)\left(x+2\right)\)
g: \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(5x+3\right)\)
f: \(4x^2\left(x+1\right)+2x^2\left(x+1\right)\)
\(=6x^2\left(x+1\right)\)
(3x + 1)2 - (3x - 1)2
= (3x + 1 + 3x - 1)(3x + 1 - 3x + 1)
= 6x.2
= 12x
các bạn giúp mih nốt mấy bài kia với mih đang cần gấp