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Cho biết: \(\dfrac{m}{n}=\dfrac{p}{q}\) . Chứng minh rằng\(\dfrac{mp}{nq}=\dfrac{m^2+p^2}{n^2+q^2}\)
Đặt \(\dfrac{m}{n}=\dfrac{p}{q}=k\Rightarrow\left\{{}\begin{matrix}m=nk\\p=qk\end{matrix}\right.\)
\(\Rightarrow\dfrac{mp}{nq}=\dfrac{nk.qk}{nq}=k^2\left(1\right)\)
\(\Rightarrow\dfrac{m^2+p^2}{n^2+q^2}=\dfrac{\left(nk\right)^2+\left(qk\right)^2}{n^2+q^2}=\dfrac{n^2k^2+q^2k^2}{n^2+q^2}=\dfrac{k^2\left(n^2+q^2\right)}{n^2+q^2}=k^2\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{mp}{nq}=\dfrac{m^2+p^2}{n^2+q^2}\left(đpcm\right)\)
Đặt \(\dfrac{m}{n}=\dfrac{p}{q}=k\) ⇒ m=nk ; p=qk
Khi đó,
\(\dfrac{mp}{nq}=\dfrac{nk.qk}{nq}=\dfrac{k^2.nq}{nq}=k^2\) (1)
\(\dfrac{m^2+p^2}{n^2+q^2}=\dfrac{\left(nk\right)^2+\left(qk\right)^2}{n^2+q^2}=\dfrac{n^2.k^2+q^2+k^2}{n^2+q^2}=\dfrac{k^2.\left(n^2+q^2\right)}{n^2+q^2}=k^2\left(2\right)\)
Từ (1), (2) ⇒ \(\dfrac{mp}{nq}=\dfrac{m^2+p^2}{n^2+q^2}\)(đpcm)
CHÚC BẠN HỌC TỐT!!!!!!!!!
Đặt \(\dfrac{m}{n}=\dfrac{p}{q}=k=>m=kn,p=qk\)
Ta có \(\dfrac{mp}{nq}=\dfrac{kn.qk}{nq}=\dfrac{k^{2^{ }}\left(nq\right)}{nq}=k^2\left(1\right)\)
\(\dfrac{m^2+p^2}{n^2+q^2}=\dfrac{k^2.n^2+k^2.q^2}{n^2+q^2}=\dfrac{k^2\left(n^2+q^2\right)}{n^2+q^2}=k^2\left(2\right)\)
Từ (1), (2) => ..............
4/ \(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{24}\end{matrix}\right.\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}=k\) (đặt k)
Suy ra \(x=15k;y=20k;z=24k\)
Thay vào,ta có:
\(M=\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Giải:
Đặt \(\dfrac{m}{n}=\dfrac{p}{q}=k\Rightarrow m=nk;p=qk\left(k\ne0\right).\)
Ta có:
\(\dfrac{mp}{nq}=\dfrac{nkqk}{nq}=\dfrac{nqk^2}{nq}=k^2_{\left(1\right)}.\)
\(\dfrac{m^2+p^2}{n^2+q^2}=\dfrac{\left(nk\right)^2+\left(qk\right)^2}{n^2+q^2}=\dfrac{n^2k^2+q^2k^2}{n^2+q^2}=\dfrac{k^2\left(n^2+q^2\right)}{n^2+q^2}=k^2_{\left(2\right)}.\)
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\Rightarrow\dfrac{mp}{nq}=\dfrac{m^2+p^2}{n^2+q^2}\left(đpcm\right).\)
ừm Phạm Phú Hoàng Long