Question

1/Tìm x,y,z biết: 2x+3y=5z và x+y-z=190

2/Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) .Chứng minh rằng:

a/\(\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)

b/\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Giups mk vs các bn ơi

Answer 1

Bài 2:

a) Ta có : Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)

Theo tính chất dãy tỉ số bằng nhau, ta có :

\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}\left(1\right)\)

Và \(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a-7b}{5c-7d}\left(2\right)\)

Từ (1) và (2)=> \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)Vậy...

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Thay các đẳng thức vừa tìm được , ta có :

\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(1\right)\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\)

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

từ (1) và (2)=> đpcm

tik mik nha !!!

Answer 2

1. Bạn xem lại đề bài nhé! Mình nghĩ là \(2x=3y=5z\) thì đúng hơn!

2.

a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\)

Từ \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)(đpcm)

Vậy \(\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

Ta có:

\(VT=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\left(1\right)\)

\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)

Vậy \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

Answer 3

Câu 1 :

Thấy lạ lạ!

Câu 2 :

a) Đặt :

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\Leftrightarrow\dfrac{5.b.k+7b}{5.b.k-7b}=\dfrac{5.d.k+7d}{5.d.k-7d}\)

Xét :

\(\dfrac{5.b.k+7b}{5.b.k-7b}=\dfrac{b\left(5k+7\right)}{b\left(5k-7\right)}=\dfrac{5k+7}{5k-7}\left(1\right)\)

\(\dfrac{5.d.k+7d}{5.d.k-7d}=\dfrac{d\left(5k+7\right)}{d\left(5k-7\right)}=\dfrac{5k+7}{5k-7}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)