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Ta có: a/b = c/d => a^2/b^2 = c^2/d^2
Áp dụng tính chất của dãy tỉ số = nhau ta có:
a/b = c/d = a+c/b+d => a^2/b^2 =c^2/d^2 = (a+c/b+d)^2 (1)
a^2/b^2 = c^2/d^2 = a^2+c^2/b^2+d^2 (2)
Từ (1) và (2) => a^2+c^2/b^2+d^2 = (a+c/b+d)^2 (đpcm)
Vì \(\frac{a}{b}=\frac{c}{d}\)=>\(\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2=\frac{ac}{bd}\)hay \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{2ac}{2bd}\)
Aps dụng tính chất dãy tỉ số = nhau ta có:
\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+2ac+c^2}{b^2+2bd+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\left(\frac{a+c}{b+d}\right)^2\)
=>đpcm
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\left(đpcm\right)\)
\(\frac{a}{b}\) =\(\frac{c}{d}\) =>\(\frac{a}{c}\) =\(\frac{b}{d}\) =\(\frac{a-b}{c-d}\) =>\(\frac{ab}{cd}\) = \(\frac{a}{c}\) x\(\frac{b}{d}\) = \(\frac{a-b}{c-d}\) x \(\frac{a-b}{c-d}\) = \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Còn với\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) thì bạn chỉ cần thay dấu trừ thành dấu công là được
Chúc bạn học tốt
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: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Đpcm)
Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
=> \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)(Đpcm)
ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\left(1\right)\)
mà \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
Từ (1) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\)
ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
Lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\Rightarrow\frac{\left(a+b^2\right)}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Lại có: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{kb^2}{kd^2}=\frac{b^2}{d^2}\)
Tương tự: \(\frac{a^2+b^2}{c^2+d^2}=\frac{k^2b^2+b^2}{k^2d^2+d^2}=\frac{b^2\left(k+1\right)}{d^2\left(k+1\right)}=\frac{b^2}{d^2}\)
=> đpcm
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k,c=d.k\)
a) Ta có:
\(\frac{a}{3a+b}=\frac{b.k}{3.b.k+b}=\frac{b.k}{b\left(3k+1\right)}=\frac{k}{3k+1}\) (1)
\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
b) Ta có:
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
Cho \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k;c=d.k\)
Vế trái:
\(\frac{a}{b}=\frac{c}{d}=\frac{b.k.b}{d.k.d}=\frac{b^2}{d^2}\)(1)
Vế phải:
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right).2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2) ta có:
\(\frac{ab}{c\text{d}}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(đpcm)
ta có \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{a}=\frac{a+b}{c+d}\\\Rightarrow\frac{a}{c}.\frac{a}{c}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\\ \Rightarrow\frac{a}{c}.\frac{b}{d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}hay\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2} \)
a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2), ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2), ta có: \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
a) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)