Question

[truc linh nguyen]

cho tỉ lệ thức : \(\frac{a}{b}\) =\(\frac{c}{d}\) CM:

1)\(\frac{a+c}{a}\)=\(\frac{b+d}{b}\)

2)\(\frac{2A+3B}{2A-3B}\)= \(\frac{2C+3D}{2C-3D}\)

3)\(\frac{A.B}{C.D}\)= \(\frac{a^2-b^2}{c^2-d^2}\)

4)(\(\frac{a+b}{c+d}\))²=\(\frac{a^2+b^2}{c^2+d^2}\)

ai giúp mk vs ạ ! mình đang cần gấp ạ!

Answer 1

[Nguyễn Lê Phước Thịnh]

1) Ta có: \(\frac{a}{b}=\frac{c}{d}\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Leftrightarrow\frac{a}{c}+1=\frac{b}{d}+1\)

\(\Leftrightarrow\frac{a+c}{c}=\frac{b+d}{d}\)(đpcm)

2) Để \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\) thì \(\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)

\(\Leftrightarrow\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a}{2c}=\frac{3b}{3d}\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{a}{c}=\frac{b}{d}\)

\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

hay \(\frac{a}{b}=\frac{c}{d}\)(đpcm)

3) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)(1)

Ta có: \(\frac{a^2-b^2}{c^2-d^2}\)

\(=\frac{k^2\cdot b^2-b^2}{k^2\cdot d^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(2)

Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

4) Ta có: \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

nên \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2\cdot k^2+b^2}{d^2\cdot k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(3)

Ta có: \(\left(\frac{a+b}{c+d}\right)^2\)

\(=\left(\frac{bk+b}{dk+d}\right)^2\)

\(=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2\)

\(=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\)(4)

Từ (3) và (4) suy ra \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)