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ta có :
(a+b)^2 = (a+b) * (a+b)
=a^2+ab+ba+b^2
=13+6*2
=13+12
=25
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:a2+b2=13
(a+b)2-2ab=13(Mà ab=6)
(a+b)2-12=13
(a+b)2=25=52=(-5)2)
Vậy |a+b|=5
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Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
=> \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\left(\frac{a+b}{c+d}\right)^2=\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\)
=> \(\frac{ab}{cd}+\left[\left(\frac{a+b}{c+d}\right)^2:\left(\frac{a^2+b^2}{c^2+d^2}\right)\right]-\frac{a^2-b^2}{c^2-d^2}\)
= \(\frac{ab}{cd}+1-\frac{a^2-b^2}{c^2-d^2}\)
= \(1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)
Thay:
\(\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)
=> đpcm
Ta có : (a+b)2= a2+2ab+b2 =(a2+b2) +2ab = 13+12=25
Lưu ý : 2ab=2.6=12 nhé !!