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\(\text{Đặt }\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\left(1\right)\)
\(\text{Thay (1) vào }\frac{ab}{cd}\text{ ta có :}\)
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\left(2\right)\)
\(\text{Thay (1) vào }\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\text{ ta có :}\)
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\frac{b^2}{d^2}\left(3\right)\)
\(\text{Từ (2) và (3) }\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(đat:\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(a,\frac{a^2-b^2}{ab}=\frac{b^2k^2-b^2}{bkb}=\frac{b^2\left(k^2-1\right)}{b^2k}=\frac{k^2-1}{k};\frac{c^2-d^2}{cd}=\frac{d^2\left(k^2-1\right)}{d^2k}=\frac{k^2-1}{k}\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\) \(b,\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left[b\left(k+1\right)\right]^2}{b^2k^2+b^2}=\frac{b^2\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{\left(k^2+1\right)};\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left[d\left(k+1\right)\right]^2}{d^2k^2+d^2}=\frac{d^2\left(k+1\right)^2}{d^2\left(k^2+1\right)}=\frac{\left(k+1\right)^2}{k^2+1}\Rightarrow\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\) \(c,\frac{a}{a+b}=\frac{bk}{bk+b}=\frac{bk}{b\left(k+1\right)}=\frac{k}{k+1};\frac{c}{c+d}=\frac{dk}{dk+d}=\frac{dk}{d\left(k+1\right)}=\frac{k}{k+1}\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a+b^2\right)}{\left(c+d\right)^2}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
Vậy \(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
CHÚC BẠN HỌC TỐT
a) áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-c^2}{b^2-d^2}\)
Do \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)=> đpcm
b) áp dụng tính chất của dãy tỉ số bằng nhau ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\left(\frac{a-c}{b-d}\right)^2\)=> đpcm
a) Ta có: \(\frac{a}{b}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
+)\(\frac{2a^2-3b^2}{2c^2-3d^2}=\frac{2.\left(bk\right)^2-3b^2}{2.\left(dk\right)^2-3d^2}=\frac{2.b^2.k^2-3.b^2}{2.d^2.k^2-3.d^2}\)
\(=\frac{2.b^2.\left(k^2-3\right)}{2.d^2.\left(k^2-3\right)}\)
\(=\frac{b^2}{d^2}\)(1)
+)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2), ta có: \(\frac{2a^2-3b^2}{2c^2-3d^2}=\frac{ab}{cd}\)
Học tốt nha!!!
Giải:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\)
a, Ta có: \(k^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\) (1)
\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
b, Ta có: \(k=\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow k^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (1)
\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)Thay vào \(\frac{a^2-b^2}{c^2-d^2}\) ta được:
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{b^2k^2-b^2}{d^2k^2-d^2}\Rightarrow\frac{b^2}{d^2}\Rightarrow\frac{b.b}{d.d}\left(1\right)\)
Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow a=b;c=d\left(2\right)\)
Từ (1) và (2) suy ra:\(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
Ta có : \(\frac{a}{b}=\frac{c}{d}=\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a}{c}.\frac{b}{d}=\frac{a+b}{c+d}.\frac{a+b}{c+d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
C1: Đặt a/b=c/d=k => a=bk, c=dk
Ta có (a-b)^2/(c-d)^2=(bk-b)^2/(dk-d)^2 =b^2 . (k-2)^2 / d^2 .(k-1)^2 =b^2/d^2
ab/cd=bk.b/dk.d =b^2/d^2
=> (a-b)^2/(c-d)^2=ab/cd