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5) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2\)
\(=\left(x-y-x-y\right)^2\)
\(=\left(-2y^2\right)\)
\(=4y^2\)
6) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x-5\right)\left(x+5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
7) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)
\(=\left(x-2\right)^2-2\left(x-2\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left[\left(x-2\right)-\left(x+1\right)\right]^2\)
\(=\left(-3\right)^2=9\)
8) \(-\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)
\(=\left(2x-3y\right)^2+2\left(2x+3y\right)\left(2x-3y\right)+\left(2x+3y\right)^2\)
\(=\left[\left(2x+3y\right)+\left(2x-3y\right)\right]^2\)
\(=\left(4x\right)^2=16x^2\)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
1,sai đề
2, \(\dfrac{x^2}{xy+x}+\dfrac{y}{y^2-1}-\dfrac{x}{x\left(y-1\right)}\)
\(=\dfrac{x^2}{x\left(y+1\right)}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}-\dfrac{x}{x\left(y-1\right)}\)
\(=\dfrac{x}{y+1}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}-\dfrac{1}{y-1}\)
\(=\dfrac{x\left(y-1\right)-y-1}{\left(y+1\right)\left(y-1\right)}+\dfrac{y}{\left(y-1\right)\left(y+1\right)}\)
\(=\dfrac{xy-x-y-1+y}{y^2-1}=\dfrac{xy-x-1}{y^2-1}\)
3, \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy x = 2 hoặc x = -5
Bài 1 :
Sửa đề :\(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+\left(x+7\right)\)
\(=2x^2+3x-10x-15-x^2+6x+x+7\)
\(=2x^2-2x^2-7x+7x-15+7\)
\(=-8\)
\(\Rightarrow\) Biểu thức trên bằng 8 nên giá trị của biểu thức ko phụ thuộc vào giá trị của biến x
Bài 2 , 3 : Tú làm ròi nghĩ làm
1: a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37\) (Vì \(x-y=7\))
\(=100\)
Vậy \(A=100\)
b) Ta có: \(B=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
Vậy \(B=25\)
c) Ta có : \(C=\left(x-y\right)^2\)
\(=x^2-2xy+y^2\)
\(=\left(x^2+y^2\right)-2xy\)
\(=26-2.5\) (Vì \(x^2+y^2=26\) ; \(xy=5\))
\(=16\)
Vậy \(C=16\)
2: a) \(\left(x+y\right)^2-y^2=x^2+2xy+y^2-y^2\)
\(=x^2+2xy\)
\(=x\left(x+2y\right)\) \(\left(dpcm\right)\)
b) \(\left(x^2+y^2\right)^2-2xy^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)^2\left(x+y\right)^2\) \(\left(dpcm\right)\)
c) \(\left(x+y\right)^2=x^2+2xy+y^2\)
\(=\left(x^2-2xy+y^2\right)+4xy\)
\(=\left(x-y\right)^2+4xy\) \(\left(dpcm\right)\)
Chúc bn học tốt ✔✔✔
