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Bài 7:
\(a,A=\dfrac{2a+a-3}{a-3}\cdot\dfrac{\left(a-3\right)\left(a+3\right)}{3}=\dfrac{3\left(a-1\right)\left(a+3\right)}{3}=\left(a-1\right)\left(a+3\right)\\ b,B=\dfrac{b+3-6}{b+3}:\dfrac{b^2-9-b^2+10}{\left(b-3\right)\left(b+3\right)}\\ B=\dfrac{b-3}{b+3}\cdot\left(b-3\right)\left(b+3\right)=\left(b-3\right)^2\)
Bài 8:
\(a,M=\dfrac{4m^2-4mn+n^2}{m^2}:\dfrac{n-2m}{mn}=\dfrac{\left(n-2m\right)^2}{m^2}\cdot\dfrac{mn}{n-2m}=\dfrac{n\left(n-2m\right)}{m}\\ b,N=\dfrac{1}{3}+x:\dfrac{x+3-x}{x+3}=\dfrac{1}{3}+x\cdot\dfrac{x+3}{3}=\dfrac{1+x^2+3x}{3}\)
Bài 8:
b: \(N=\dfrac{1}{3}+\dfrac{x}{\dfrac{x+3-x}{x+3}}=\dfrac{1}{3}+\dfrac{x}{\dfrac{3}{x+3}}=\dfrac{1}{3}+\dfrac{x+3}{3x}=\dfrac{x+x+3}{3x}=\dfrac{2x+3}{3x}\)
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Bài 5:
a) \(x^2-xy+x-y\)
\(=\left(x^2-xy\right)+\left(x-y\right)\)
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x+1\right)\left(x-y\right)\)
b) \(xz+yz+4x+4y\)
\(=\left(xz+yz\right)+\left(4x+4y\right)\)
\(=z\left(x+y\right)+4\left(x+y\right)\)
\(=\left(z+4\right)\left(x+y\right)\)
c) \(x^2-x-y^2+y\)
\(=\left(x^2-y^2\right)-\left(x-y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
d) \(x^2+2x+2z-z^2\)
\(=\left(x^2-z^2\right)+\left(2x+2z\right)\)
\(=\left(x+z\right)\left(x-z\right)+2\left(x+z\right)\)
\(=\left(x+z\right)\left(x-z+2\right)\)
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4.2:
a: x^2-x+1=x^2-x+1/4+3/4
=(x-1/2)^2+3/4>=3/4>0 với mọi x
=>x^2-x+1 ko có nghiệm
b: 3x-x^2-4
=-(x^2-3x+4)
=-(x^2-3x+9/4+7/4)
=-(x-3/2)^2-7/4<=-7/4<0 với mọi x
=>3x-x^2-4 ko có nghiệm
5:
a: x^2+y^2=25
x^2-y^2=7
=>x^2=(25+7)/2=16 và y^2=16-7=9
x^4+y^4=(x^2)^2+(y^2)^2
=16^2+9^2
=256+81
=337
b: x^2+y^2=(x+y)^2-2xy
=1^2-2*(-6)
=1+12=13
x^3+y^3=(x+y)^3-3xy(x+y)
=1^3-3*1*(-6)
=1+18=19
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a: Xét ΔABC có BM/BC=BD/BA
nên MD//AC
=>MM' vuông góc AB
=>M đối xứngM' qua AB
b: Xét tứ giác AMBM' có
D là trung điểm chung của AB và MM'
MA=MB
Do đó: AMBM' là hình thoi