1.Cho \(\dfrac{m-n}{p-q}\)=\(\dfrac{n}{q}\). Chứng minh\(\dfrac{m^2+n^2}{p^2+q^2}=\dfrac{\left(m+n\right)^2}{\left(p+q\right)^2}\)(Giả thiết các tỉ số đều có nghĩa)
2.Cho \(\dfrac{2}{a}=\dfrac{1}{b}+\dfrac{1}{c}\)(a,b,c\(\ne\)0,a\(\ne\)c). Chứng minh rằng:\(\dfrac{b}{c}=\dfrac{b-a}{a-c}\)
3.Cho b2=ac.Chứng minh:\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)
4.Cho \(\dfrac{x}{3}\)=\(\dfrac{y}{4}\) và \(\dfrac{y}{5}=\dfrac{z}{6}\). Tính \(M=\dfrac{2x+3y+4z}{3x+4y+5z}\)
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:
\(M=\dfrac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
3. \(b^2=ac\Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}^{\left(đpcm\right)}\)
Bài 2 hơi vất vả đấy! =)"
2/ Ta có: \(\dfrac{2}{a}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{b+c}{bc}\)
\(\Leftrightarrow\dfrac{2}{b+c}=\dfrac{a}{bc}\Leftrightarrow a=\dfrac{2bc}{b+c}\)
Thay vào,ta có: \(\dfrac{b-a}{a-c}=\dfrac{b-\dfrac{2bc}{b+c}}{\dfrac{2bc}{b+c}-c}\)
\(=\dfrac{\dfrac{b^2+bc-2bc}{b+c}}{\dfrac{2bc-bc-c^2}{b+c}}=\dfrac{\left(\dfrac{b^2-bc}{b+c}\right).\left(b+c\right)}{bc-c^2}\)
\(=\dfrac{b^2-bc}{bc-c^2}=\dfrac{b\left(b-c\right)}{c\left(b-c\right)}=\dfrac{b}{c}^{\left(đpcm\right)}\)