giúp mik với đg cần gấp lắm r.
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Xét ΔABD và ΔAED có
AB=AE
\(\widehat{BAD}=\widehat{EAD}\)
AD chung
Do đó: ΔABD=ΔAED
Suy ra: BD=ED
hay D nằm trên đường trung trực của BE(1)
Ta có: AB=AE
nên A nằm trên đường trung trực của BE(2)
Từ (1) và (2) suy ra AD⊥BE
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
11)\(\dfrac{3x+1}{x-5}+\dfrac{2x}{x-5}=\dfrac{3x+2x+1}{x-5}=\dfrac{5x+1}{x-5}\)
12)\(\dfrac{4-x^2}{x-3}+\dfrac{2}{x^2-9}=\dfrac{4-x^2}{x-3}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(4-x^2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2+\left(2-x\right)\left(2+x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
13)
\(\dfrac{3}{4x-2}+\dfrac{2x}{4x^2-1}=\dfrac{3}{2\left(2x-1\right)}+\dfrac{2x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{3\left(2x+1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{2.2x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{6x+3+4x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{10x+3}{2\left(2x-1\right)\left(2x+1\right)}\)
14)
\(\dfrac{2x+1}{2x-4}+\dfrac{5}{x^2-4}=\dfrac{2x+1}{2\left(x-2\right)}+\dfrac{5}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{5.2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+5x+12}{2\left(x-2\right)\left(x+2\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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