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\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{c}{d}.\dfrac{c}{d}=\dfrac{a}{b}.\dfrac{c}{d}\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2+c^2}{b^2+d^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\left(1\right)\)
\(VT=a^2c^2+2abcd+b^2d^2+a^2d^2-2adbc+b^2c^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(VP=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
Ta thấy: \(VT=VP\)
\(\Rightarrow\left(1\right)\) luôn đúng.
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a) 3,5(15) = 3,5 + 0,0(15) = 3,5 + 1,5. 0,(01) = 3,5 + 1,5.1/99 = 3,5 + 1/66 = 116/33
b) Ta có: \(\frac{2x-y}{x+y}=\frac{2}{3}\)
=> (2x - y).3 = 2(x + y)
=> 6x - 3y = 2x + 2y
=> 6x - 2x = 2y + 3y
=> 4x = 5y
=> \(\frac{x}{y}=\frac{5}{4}\)
c) Đặt : \(\frac{a}{b}=\frac{c}{d}=k\) => \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó, ta có:
\(\frac{\left(bk\right)^2+bk.dk}{\left(dk\right)^2+dk.bk}=\frac{b^2k^2+bdk^2}{d^2k^2+bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2+bd\right)}=\frac{b^2+bd}{d^2+bd}\)
=> Đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn đánh lại đề đi, Để ghi dấu mũ bạn ấn nút "x2" trên thanh công cụ, sau khi bạn gõ xong dấu mũ rồi bạn ấn lại nó để đưa về trạng thái thường
\(\frac{\left(a+b\right)2}{\left(c+d\right)2}=\frac{2a+2b}{2c+2d}\)
Vậy \(\frac{\left(a+b\right)2}{\left(c+d\right)2}=\frac{2a+2b}{2c+2d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{b^2}{c}\\d=\dfrac{c^2}{b}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{b^2}{c}:\dfrac{c^2}{b}\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{b^2}{c}.\dfrac{b}{c^2}\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{b^3}{c^3}=\dfrac{8b^3}{8c^3}=\dfrac{a^3}{b^3}=\dfrac{125c^3}{125d^3}\)
\(\Rightarrow\dfrac{a}{d}=\dfrac{a^3+8b^3+125c^3}{b^3+8c^3+125d^3}\left(dpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt a/b=c/d=k=>a=bk;c=dk
=>\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left(b\left(k+1\right)\right)^2}{\left(d\left(k+1\right)\right)^2}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\) (2)
từ (1) và (2)=>đpcm
tick nhé
oh minecratf kìa
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)=> \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó, ta có: \(\frac{\left(bk\right)^2+\left(bk\right)\left(dk\right)}{\left(dk^2\right)-\left(bk\right)\left(dk\right)}=\frac{b^2.k^2+b.d.k^2}{d^2.k^2-b.d.k^2}=\frac{\left(b^2+bd\right).k^2}{\left(d^2-bd\right).k^2}=\frac{b^2+bd}{d^2-bd}\)
Vậy ...