Cho xy/x^2+y^=5/8. Rút gọn phân thức P= x^2-2xy+y^2/x^2+2xy+y^2
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x 2 +y 2 xy = 8 5 ⇒x 2 +y 2 = 5 8xy \Rightarrow P=\frac{\frac{8xy}{5}-2xy}{\frac{8xy}{5}+2xy}=\frac{8xy-10xy}{8xy+10xy}=\frac{-2}{18}=-\frac{1}{9}⇒P= 5 8xy +2xy 5 8xy −2xy = 8xy+10xy 8xy−10xy = 18 −2 =− 9 1
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1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)
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B) Ta có: 2x-2y-x2+2xy-y2
⇔ 2(x-y)-(x2-2xy+y2)
⇔ 2(x-y)-(x-y)2
⇔ (x-y)(2-x+y)
Đúng thì tick nhé
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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) \(\left(x^2-2xy+y^2\right)\left(x-y\right)-\left(x-y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2-2xy+y^2-x^2+2xy+y^2\right)\)
\(=\left(x-y\right)2x^2\)
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\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\dfrac{\left(-x+y-z\right)^2}{\left(x-y\right)^2-z^2}\)
\(=\dfrac{\left[-\left(x-y+z\right)\right]^2}{\left(x-y-z\right)\left(x-y+z\right)}\)
\(=\dfrac{x-y+z}{x-y-z}\)
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Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
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b) \(A+B=x^2+y^2+2x+3+2x^2+y^2+2x+1=3x^2+2y^2+4x+4\)
\(A-B=x^2+y^2+2x+3-2x^2-y^2-2x-1=-x^2+2\)
a) Ta có: \(A=x^2+y^2-2xy+2x+2xy+3\)
\(=x^2+y^2+2x-\left(2xy-2xy\right)+3\)
\(=x^2+y^2+2x+3\)
Ta có: \(B=2x^2+y^2-xy+2x+xy+1\)
\(=2x^2+y^2+2x+\left(xy-xy\right)+1\)
\(=2x^2+y^2+2x+1\)