Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(3x\left(2x-4\right)-\left(6x-1\right)\left(x+2\right)=25\)
\(\Rightarrow6x^2-12x-\left(6x^2+12x-x-2\right)=25\)
\(\Rightarrow6x^2-12x-6x^2-12x+x+2=25\)
\(\Rightarrow-23x+2=25\)
\(\Rightarrow-23x=25-2-23\)
\(\Rightarrow x=23:\left(-23\right)=-1\)
Vậy x = -1
b) \(\left(x^2-2xy+y^2\right)\left(x-y\right)-\left(x-y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2-2xy+y^2-x^2+2xy+y^2\right)\)
\(=\left(x-y\right)2x^2\)
sai đề rồi nhé , đề phải là :
\(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\frac{\left(x-y\right)^3+3xy.\left(x-y\right)+z^3+3xyz}{x^2+2xy+y^2+y^2+2yz+z^2+z^2-2xz+x^2}\)
\(=\frac{\left(x-y+z\right).\left[\left(x-y\right)^2-\left(x-y\right).z+z^2\right]+3xy.\left(x-y+z\right)}{2x^2+2y^2+2z^2+2xy+2yz-2xz}\)
\(=\frac{\left(x-y+z\right).\left(x^2-2xy+y^2-xz+yz+z^2+3xy\right)}{2.\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\frac{\left(x-y+z\right).\left(x^2+y^2+z^2+xy+yz-xz\right)}{2.\left(x^2+y^2+z^2+xy+yz-xz\right)}\)
\(=\frac{x-y+z}{2}\)
câu 1.
P= 2(x+y)(x-y)+(x-y)^2+(x+y)^2-4y^2
P= (x+y+x-y)^2-(2y)^2
P=(2x-2y)(2x+2y)
P=4(x^2-y^2)
câu 2.
a, x^3-2x^2-4xy^2+x= x(x^2-2x+1)-4xy^2
=x(x-1)^2-4xy^2
=x(x-1-2y)(x-1+2y)
b, (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4)(x^2+5x+6)-24
Đặt x^2+5x+4= a
Lúc đó: (x+1)(x+2)(x+3)(x+4)-24= a(a+2)-24
= a^2+2a-24
=a^2+2a+1-25
= (a+1)^2-5^2
= (a+1-5)(a+1+5)
= (a-4)(a+6)
mà ta đặt x^2+5x+4=a => (x+1)(x+2)(x+3)(x+4)-24= (x^2+5x+4-4)(x^2+5x+4+6)
= (x^2+5x)(x^2+5x+10)
câu3. (x+2)^2= 4-x^2
=> (x+2)^2-4+x^2=0
=>. (x+2)^2-(2-x)(2+x)=0
=> (x+2)(x+2-2+x)=0
=> (x+2)2x=0
=> x+2=0 hoặc 2x=0
=> x=-2 hoặc x=0
1)P=2(x^2-y^2)+x^2-2xy+y^2+x^2+2xy+y^2-4y^2=2x^2-2y^2+2x^2+2y^2-4y^2=4x^2-4y^2 . 3) <=> x^2+4x+4-4+x^2=0
<=> 2x^2+4x=0 <=>2x(x+2)=0 <=>2x=0 hay x+2=0 <=>x=0 hay x=-2
\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(=\frac{\left(x+y\right)^2-1}{\left(x-1\right)^2-y^2}\)
\(=\frac{\left(x+y-1\right)\left(x+y+1\right)}{\left(x-1-y\right)\left(x-1+y\right)}\)
\(=\frac{x+y+1}{x-y-1}\)
1. a) ... \(=\left(x+y\right)^2-4^2\)
b) ... \(=x^2-\left(y-6\right)^2\)
2. a) ...\(=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=\left(1-4x\right)\left(1+4x\right)\)
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
\(x.\left(x-y\right)-\left(x+y\right).\left(x-y\right)=\left(x-y\right).\left(x-x-y\right)=-y.\left(x-y\right)\)
\(2a\left(a-1\right)-2.\left(a+1\right)^2=2.\left[a.\left(a-1\right)-\left(a+1\right)^2\right]=2.\left(a^2-a-a^2-2a-1\right)=-2.\left(3a+1\right)\)\(\left(x+2\right)^2-\left(x-1\right)^2=\left(x+2-x+1\right).\left(x+2+x-1\right)=3.\left(2x+1\right)\)
\(x.\left(x-3\right)^2-x.\left(x+5\right).\left(x-2\right)=x.\left[\left(x-3\right)^2-\left(x+5\right).\left(x-2\right)\right]=x.\left(x^2-6x+9-x^2-3x+10\right)=x.\left(19-9x\right)\)
\(a,P=x^2-16-x^2+8x-16=8x-32\\ b,=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\\ =2y^2-10xy=2\cdot9-10\left(-3\right)\cdot2=78\)
bn nào jup mk nhanh vs mk đag cần gấp
(x+y)^2-(x-y)^2-4(x-1)y
=(x+y-x+y)(x+y+x-y)-4xy+4y
=2x*2y-4xy+4y
4xy-4xy+4y
=4y