\(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+x^4\right)\left(x^8+y^8\right)\)
rút gọn biểu thức
Rút gọn biểu thức với x - y = 1
\(\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x^{16}+y^{16}\right)\)
Ta có \(x-y=1\)
\(=>x+y=\left(x+y\right).\left(x-y\right)\)
\(A=\left(x+y\right).\left(x-y\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^2-y^2\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^4-y^4\right).\left(x^4+y^4\right)\)
\(A=x^8-y^8\)
= \(-\left[\left(x-y\right)\left(x^2-y^2\right)\left(x^4-y^4\right)\left(x^8-y^8\right)\left(x^{16}-y^{16}\right)\right]\)
= \(-\left[\left(x-y\right)\left(x-y\right)^2\left(x-y\right)^4\left(x-y\right)^8\left(x-y\right)^{16}\right]\)
= \(-\left(1\cdot1^2\cdot1^4\cdot1^8\cdot1^{16}\right)\)
= -1
Cho biểu thức:
\(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\left(x\ne\pm y\right)\)
1. Rút gọn biểu thức \(C\) ;
2. Khi cho \(\left(x^2-y^2\right)\cdot C=-8\), hãy tính giá trị của biểu thức:
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\).
1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
Rút gọn biểu thức B= \(2\left(X^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
Rút gọn biểu thức sau:
\(\left(x+y\right)^2+\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)-3x^2\)
\(\left(x+y\right)^2+\left(x-y\right)^2+\left(x-y\right)\left(x+y\right)-3x^2\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-2xy+y^2\right)+\left(x^2-y^2\right)-3x^2\)
\(=x^2+2xy+y^2+x^2-2xy+y^2+x^2-y^2-3x^2\)
\(=3x^2+y^2-3x^2\)
\(=y^2\)
1. Viết biểu thức dưới dạng bình phương của một tổng
\(2xy^2+x^2y^4+1\)
2, Rút gọn biểu thức :
a, \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
b, \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
1) 2xy2+x2y4+1=(xy2)2+2xy2.1+12=(xy2+1)2
2)
a)2(x-y)(x+y)+(x+y)2+(x-y)2=(x+y+x-y)2=(2x)2=4x2
b)(x-y+z)2+(z-y)2+2(x-y+z)(y-z)
=(x-y+z)2+(y-z)2+2(x-y+z)(y-z)
=(x-y+z+y-z)2
=x2
Rút gọn các biểu thức :
a) \(P=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
b) \(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
Rút gọn biểu thức :
a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
b) P=\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
c) Q=\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
d) P = \(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)
c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)
a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(\Leftrightarrow2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)
\(\Leftrightarrow2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2\)
\(\Leftrightarrow4x^2\)
Cho x, y, z là các số thực dương thỏa mãn: \(x+y+z+\sqrt{xyz}=4\). Rút gọn biểu thức: \(A=\sqrt{x.\left(4-y\right).\left(4-z\right)}+\sqrt{y.\left(4-z\right).\left(4-x\right)}+\sqrt{z.\left(4-x\right).\left(4-y\right)}-\sqrt{xyz}\)
Rút gọn các biểu thức sau:
a/ \(\left(x-2y^{ }\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
b/ \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
\(=2x^2-4xy+\dfrac{15}{4}y^2\)
b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
=2x+15
a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)
b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)
\(=2x+15\)
a; \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
= \(x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
= \(2x^2-4xy+\dfrac{15}{4}y^2\)
b; \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
= \(x^2-4x+4+x^2+6x+9-2x^2+2\)
= \(2x+15\)