-7 1/3 = -22/3

55/(-22/3) =  -15/2 = -7,5

====> -12/1,6 = 55/-7,5

======> 1,6/-12 = -7,5/55 ; 1,6/-7,5 = -12/55 ; -7,5/1,6 = 55/-12

( Mk gj hơi khó hỉu, moq pn pỏ wa nhé!! )

câu hỏi tương tự nha bn 

tick mk nha

KHÓ NHỈ

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

a)Xét \(VT=\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)

Xét \(VP=\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(2\right)\)

Từ (1) và (2) =>Đpcm

b)Xét \(VT=\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\left(1\right)\)

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

Từ (1) và (2) =>Đpcm

c)Xét \(VT=\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bk+b}{dk+d}\right)^2=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^2=\left[\frac{b}{d}\right]^2=\frac{b^2}{d^2}\left(1\right)\)

Xét \(VP=\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k+1\right)}{d^2\left(k+1\right)}=\frac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) =>Đpcm

a/ theo bài ra, ta có:

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

áp dụng tính caahts dã y tỉ số bằng nhau ta có :

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

=> \(\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\\ \Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\left(đpcm\right)\)

b/ theo bài ra, ta có:

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\\ \Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{ab}{cd}\left(1\right)\)

ta có:

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

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\) (2)

từ 1 và 2 => đpcm

c/ theo bài ra, ta có:

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

đặt \(\frac{a}{c}=\frac{b}{d}=k\)

ta có: a = kc

b = kd

=> \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{kc+kd}{c+d}\right)^2=\left(\frac{k\left(c+d\right)}{c+d}\right)^2=k^2\) (1)

=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kc\right)^2+\left(kd\right)^2}{c^2+d^2}=\frac{k^2c^2+k^2d^2}{c^2+d^2}=\frac{k^2\left(c^2+d^2\right)}{c^2+d^2}=k^2\left(2\right)\)

từ 1 và 2 => đpcm

\(6:\left(-27\right)=\left(-6\frac{1}{2}\right):29\frac{1}{4}\)

các tỉ lệ thức là:

\(\frac{6}{-27}=\frac{-6\frac{1}{2}}{29\frac{1}{4}};\frac{-27}{6}=\frac{29\frac{1}{4}}{-6\frac{1}{2}};\frac{6}{-6\frac{1}{2}}=\frac{-27}{29\frac{1}{4}};\frac{-6\frac{1}{2}}{6}=\frac{29\frac{1}{4}}{-27}\)