a) ( x + y ) + ( x - y ) = ( x + x ) + ( y - y ) = 2x
b) ( x + y ) - ( x - y ) = ( x - x ) + ( y + y ) = 2y
a) ( x + y ) + ( x - y ) = ( x + x ) + ( y - y ) = 2x
b) ( x + y ) - ( x - y ) = ( x - x ) + ( y + y ) = 2y
tính:A=\(x^2\left(x+y\right)-y^2\left(x+y\right)+x^2\) \(-y^2+2\left(x+y\right)+3\) biết rằng x+y+1=0
a) CMR : \(\frac{\left|x\right|}{\left|y\right|+2}+\frac{\left|y\right|}{\left|x\right|+2}\ge\frac{\left|x\right|+\left|y\right|}{\left|x\right|+\left|y\right|+2}\)
b) CMR \(\frac{\left|x\right|}{\left|y\right|+2}+\frac{\left|y\right|}{\left|x\right|+2}\ge\frac{\left|x+y\right|}{\left|x+y\right|+2}\)
C=\(x\)\(\left[x^2-y\right]\)x\(\left[x^3-2y^2\right]\)x\(\left[x^4-3y^3\right]\)x\(\left[x^5-4y^4\right]\)tại \(x=2,y=-2\)
D=\(x^2\left[x+y\right]\)-\(y^2\)\(\left[x+y\right]\)+\(\left[x^2-y^2\right]\)+2\(\left[x+y\right]\)+3 biết rằng x+y+1=0
M=\(\left[x+y\right]\)x\(\left[y+z\right]\)x\(\left[x+z\right]\)biết ranhwfx+y+z=xyz=2
viết các biểu thứ sau dưới dạng tổng
e. \(\left(x+1\right)\)\(\left(x-1\right)\)
f .\(\left(x-2y\right)\left(x-2y\right);56.64\)
g. \(\left(x+y+z\right)\left(x-y-z\right)\)
h. \(\left(x-y+z\right)\left(x+y+z\right)\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
Chứng minh rằng:\(x^{\left(2^{y+1}\right)}+x^{\left(2^y\right)}+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)...\left(x^{\left(2^{y-1}\right)}+x^{\left(2^{y-2}\right)}+1\right)\left(x^{\left(2^y\right)}+x^{\left(2^{y-1}\right)}+1\right)\)với mọi \(x\in N;x>0\)và \(y\in N;y>1\)
a) \(\left|x\right|+\left|y\right|=4\)
b) \(\left|x\right|+\left|y\right|< 4\)
Rút gọn các biểu thức :
a, \(\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)\)
b, \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(c,\left(x+y-z\right)^2+2\left(z-x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
cmr nếu\(a\left(z+y\right)=b\left(z+x\right)=c\left(x+y\right);a\ne b\ne c\ne0\Rightarrow\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
cho 3 số x,y,z đôi 1 khác nhau và chứng minh rằng :
\(\dfrac{y-z}{\left(x-y\right)\cdot\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{y-x}{\left(z-x\right)\cdot\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)