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Ta có:
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a) \(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b) \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c) \(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2.\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2\right)+1}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
c) có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a^2}{^{c^2}}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
Lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(2\right)\)
Từ (1) và (2) có \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
các câu còn lại bạn tự làm đi! HI.......
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(2)
Từ (1) và (2) \(\Rightarrow\)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c)\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(ck+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
k cho mình nhé
\(\frac{a}{b}=\frac{c}{d}=\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{2a^2}{2c^2}=\frac{3b^2}{3d^2}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{2a^2}{2c^2}=\frac{3b^2}{3d^2}=\frac{2a^2-3b^2}{2c^2-3d^2}\)(1)
\(\frac{a^2}{c^2}=\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)(2)
từ (1) và (2) => đpcm
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}.\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\Rightarrow\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)(T/c dãy tỷ số bằng nhau)
\(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
\(\frac{a}{b}=\frac{c}{d}\)
Suy ra \(\frac{2a}{2b}=\frac{2c}{2d}=\frac{2a+2b}{2c+2d}=\frac{2a-2b}{2c-2d}\)
Suy ra\(\frac{2a+2b}{2a-2b}=\frac{2c+2d}{2c-2d}\)
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