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a: Xét ΔADH vuông tại H và ΔBCI vuông tại I có
AD=BC
\(\widehat{D}=\widehat{C}\)
Do đó: ΔADH=ΔBCI
Suy ra: DH=CI
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\dfrac{1-3x}{2x}-\dfrac{2-3x}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)-2x\left(2-3x\right)-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1+6x^2+3x-4x+6x^2-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{12x^2-2x+1}{4x^2-2x}\)
b: Ta có: \(\dfrac{x+2}{x^3-1}-\dfrac{-2}{x^2+x+1}-\dfrac{1}{x+1}\)
\(=\dfrac{x+2+2x-2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^3-1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x^2+3x-x^3+1}{\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
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\(\dfrac{x-1}{x^2+5x}=\dfrac{\left(x-1\right)\left(x-5\right)}{x\left(x-5\right)\left(x+5\right)};\dfrac{x+1}{x^2-25}=\dfrac{x\left(x+1\right)}{x\left(x-5\right)\left(x+5\right)}\)
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Bài 3:
c: \(5x\left(x-1\right)+3y\left(x-1\right)=\left(x-1\right)\left(5x+3y\right)\)
d: \(x\left(x+y\right)-12x-12y=\left(x+y\right)\left(x-12\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2b,=\left(2x^3-4x^2-4x^2+8x-2x+4-9\right):\left(2x-4\right)\\ =\left[\left(2x-4\right)\left(x^2-2x-2\right)-9\right]:\left(2x-4\right)\\ =x^2-2x-2\left(\text{ dư -9}\right)\)
Bài 1: Ta có:
\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{a^3+b^3+c^3}{abc}\) (2)
Mà: \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^3=0\)
\(\Rightarrow a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2+6abc=0\)
\(\Rightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)-3abc=0\)
\(\Rightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3ac\left(a+b+c\right)+3bc\left(a+b+c\right)-3abc=0\)
\(\Rightarrow a^3+b^3+c^3+\left(a+b+c\right)\left(3ab+3ac+3bc\right)-3abc=0\) (1)
Thay \(a+b+c=0\) (1) ta có:
\(a^3+b^3+c^3+0\cdot\left(3ab+3ac+3bc\right)-3abc=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Thay vào (2) ta có:
\(\dfrac{3abc}{abc}=3\)
ậy
1:
a+b=c=0
=>a+b=-c; a+c=-b; b+c=-a
\(A=\dfrac{a^3+b^3+c^3}{abc}\)
\(=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{abc}=\dfrac{\left(-c\right)^3+3bac+c^3}{abc}\)
=3abc/abc=3