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GỢI Ý:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)
\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2\left(bk^2\right)-3bkb+5b^2}{2b^2+3bkb}=\frac{2b^2.k^2-3kb^2+5b^2}{2b^2+3b^2.k}\)\(=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+5}{2+3k}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)\(=\frac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3dkd}\)
Tương tự nhóm tiếp là ra
=>bằng nhau
a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)
\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)
\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow a=bk;c=dk\)
\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+5}{3k+2}\)
\(\frac{2c^2-3cd+5d^2}{2d^2+3cd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\frac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\frac{2k^2-3k+5}{3k+2}\)
nên 2 phân số trên bằng nhau (đpcm)
Đặt: \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có : \(\frac{2a^2-3ab+5b^2}{2b^2+3ab}\)
<=> \(\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}\)
<=> \(\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}\)
<=> \(\frac{2k^2-3k+5}{2+3k}\left(1\right)\)
Ta có: \(\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)
<=> \(\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}\)
<=> \(\frac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}\)
<=> \(\frac{2k^2-3k+5}{2+3k}\left(2\right)\)
Từ 1 và 2 => đpcm
Mình hướng dẫn thôi. Chứ giờ đang bận.
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\).Rồi thay a = kb; c=kd vào từng vế. Thấy hai vế bằng nhau => đpcm
\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}=\frac{2a^2-3ab+5b^2}{2b^2-3cd+5d^2}=\frac{2b^2+3ab}{2d^2+3cd}\)
\(=>\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b, \(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
c, \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
1. Ta có: \(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được:
\(\frac{a}{c}=\frac{b}{d}=\frac{2a^2}{2c^2}=\frac{3ab}{3cd}=\frac{4b^2}{4d^2}=\frac{2a^2-3ab+4b^2}{2c^2-3cd+4d^2}=\frac{5b^2}{5d^2}=\frac{6ab}{6cd}=\frac{5b^2+6ab}{5d^2+6cd}\)
Suy ra : \(\frac{2a^2-3ab+4b^2}{2c^2-3cd+4d^2}=\frac{5b^2+6ab}{5d^2+6cd}\)
\(\Rightarrow\frac{2a^2-3ab+4b^2}{5b^2+6ab}=\frac{2c^2-3cd+4d^2}{5d^2+6cd}\) \(\left(dpcm\right)\)
ths bn nhiều