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a: Xét hình thang ABCD có MN//AB//CD
nên AM/AD=BN/BC
b: MA/AD+NC/BC
=BN/BC+NC/BC
=1
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{2x^2+6x-x^2+2x-3-x^2-1}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+3-x+1}{x+3}\)
\(=\dfrac{8x-4}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{4}\)
\(=\dfrac{2x-1}{x-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-2x\right)^2=\left(3x-2\right)^2\)
\(=\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
\(\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
\(\left(3-5x\right)\left(x-1\right)=0\)
\(\Rightarrow3-5x=0\) \(x-1=0\)
\(\Rightarrow x=\frac{3}{5}\) or \(x=1\)
b)\(\left(x-2\right)^3+\left(5-2x\right)^3\)
=\(\left(x-2+5-2x\right)\left(\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right)\)
\(\left(3-x\right)\left(x^2-4x+4-5x+2x^2+10-4x+25-20x+4x^2\right)\)
(\(\left(3-x\right)\left(7x^2-33x+39\right)\)
..............
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![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(27x^3-54x^2+36x=8\)
\(\Rightarrow27x^3-54x^2+36x-8=0\)
\(\Rightarrow\left(3x\right)^3-3.\left(3x\right)^2.2+3.3x.2^2-2^3=0\)
\(\Rightarrow\left(3x-2\right)^3=0\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow3x=2\)
\(\Rightarrow x=\dfrac{2}{3}\)
(2x-5)^2-(5+2x)^2=0
<=>(2x-5-5-2x)(2x-5+5+2x)=0
<=>(-10).(4x)=0
<=>(-40x)=0
<=>x =0
27x^3-54x^2+36x=8
<=>27x^3-54x^2+36x-8=0
<=>(3x-2)^3=0
<=>3x-2=0
<=>3x=2
<=>x=2/3
a: =>\(\dfrac{3-2x}{4}+\dfrac{x-5}{3}>-2\)
=>\(\dfrac{9-6x+4x-20}{12}>-2\)
=>-2x-11>-24
=>2x+11<24
=>x<13/2
b: =>2(2x-1)-(2x-1)(2x+1)<0
=>(2x-1)(2-2x-1)<0
=>(2x-1)(1-2x)<0
=>(2x-1)^2>0
=>x<>1/2
c: =>3x^2-6x-20-4x^2+9<=0
=>-x^2-6x-11<=0
=>x^2+6x+11>=0(luôn đúng)
câu a cái mẫu là có -3 mà bạn