Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(VT=\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\\ =\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5c+3d}{5c-3d}=VP\)

hết cứu

\(\dfrac{2}{15}-\dfrac{7}{10}\\ =\dfrac{4}{30}-\dfrac{21}{30}\\ =\dfrac{4-21}{30}\\ =\dfrac{-17}{30}\)

    Em ơi chia hết cho 4 thì làm sao lại dư 1 được nữa em.

Gọi a là số cần tìm

Vì a chia 4 dư 1 nên a là số lẻ

Nhưng theo đề bài, a là số chẵn

nên không có số nào thỏa đề bài

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\b=dk\end{matrix}\right.\)

Ta có: 

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

\(VP=\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=\dfrac{bd\cdot k^2}{bd}=k^2\left(2\right)\)

Từ (1) và (2) => \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)

ét o ét

tự mình trả lời D.Không đổi

\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\\ x+1+x+4+x+7+...+x+28=155\\ \left(x+x+...+x\right)+\left(1+4+7+...+28\right)=155\\ 10\times x+\left[\left(28-1\right):3+1\right]\times\left(28+1\right):2=155\\ 10\times x+10\times29:2=155\\ 10\times x+145=155\\ 10\times x=155-145=10\\ x=10:10=1\)

cứu với

\(500-\left\{5\cdot\left[409-\left(2^3\cdot3-21\right)^2\right]-1724\right\}\\ =500-\left\{5\cdot\left[409-\left(8\cdot3-21\right)^2\right]-1724\right\}\\ =500-\left\{5\cdot\left[409-\left(24-21\right)^2\right]-1724\right\}\\ =500-\left[5\cdot\left(409-3^2\right)-1724\right]\\ =500-\left[5\cdot\left(409-9\right)-1724\right]\\ =500-\left(5\cdot400-1724\right)\\ =500-\left(2000-1724\right)\\ =500-276\\ =224\)

\(500-\left\{5\left[409-\left(2^3\times3-21\right)^2\right]-1724\right\}\)

\(=500-\left\{5\left[409-\left(24-21\right)^2\right]-1724\right\}\)

\(=500-\left\{5\left[409-9\right]-1724\right\}\)

\(=500-\left\{5.400-1724\right\}\)

\(=500-\left\{2000-1724\right\}\)

\(=500-2000+1724\)

\(=224\)

Khi để cả sợi dây thì lúc đốt sợi dây sẽ cháy từ đầu đến cuối sợi dây nên cần nhiều thời gian. muốn đốt cháy nhanh thì cần chia nhỏ sợi dây và đốt cùng một lúc tại cùng một thời điểm. 

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{100}\\ =\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{25}{100}-\dfrac{10}{100}+\dfrac{1}{100}\\ =\dfrac{16}{100} =\dfrac{4}{25}\)

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{100}\\ =\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{100}\\ =\dfrac{4}{25}\)

\(a.\dfrac{2}{3}-\left(-\dfrac{1}{2}-x\right)=-\dfrac{4}{5}\\ \dfrac{2}{3}+\dfrac{1}{2}+x=-\dfrac{4}{5}\\ x=-\dfrac{4}{5}-\dfrac{2}{3}-\dfrac{1}{2}\\ x=-\dfrac{59}{30}\\ b.\left(-x-3\dfrac{1}{4}\right)-\left(1\dfrac{2}{3}-2\dfrac{3}{4}\right)=\dfrac{-5}{6}\\ \left(-x-\dfrac{13}{4}\right)-\left(\dfrac{5}{3}-\dfrac{11}{4}\right)=\dfrac{-5}{6}\\ -x-\dfrac{13}{4}-\dfrac{5}{3}+\dfrac{11}{4}=-\dfrac{5}{6}\\ -x-\dfrac{5}{3}-\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{4}{3}\\ c.\dfrac{8}{23}\cdot\dfrac{46}{24}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{2}{3}-\dfrac{1}{2}x=\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ d.\dfrac{x-1}{16}=\dfrac{3}{x+1}\\ \left(x-1\right)\left(x+1\right)=3\cdot16=48\\ x^2-1=48\\ x^2=49\\ x^2=7^2\\ x=\pm7\)

\(e.\left(1,2\right)^3x^2=\left(1,2\right)^5\\ x^2=\dfrac{\left(1,2\right)^5}{\left(1,2\right)^3}\\ x^2=\left(1,2\right)^2\\ x=\pm1,2\\ f.\left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=4\\ \left(\dfrac{2}{3}x-\dfrac{1}{4}\right)^2=2^2\\TH1:\dfrac{2}{3}x-\dfrac{1}{4}=2\\ \dfrac{2}{3}x=2+\dfrac{1}{4}=\dfrac{9}{4}\\ x=\dfrac{9}{4}:\dfrac{2}{3}=\dfrac{27}{8}\\ TH2:\dfrac{2}{3}x-\dfrac{1}{4}=-2\\ \dfrac{2}{3}x=-2+\dfrac{1}{4}=-\dfrac{7}{4}\\ x=\dfrac{-7}{4}:\dfrac{2}{3}=-\dfrac{21}{8}\\ g.\left(\dfrac{1}{6}x-3\right)^2=\dfrac{4}{9}\\ \left(\dfrac{1}{6}x-3\right)^2=\left(\dfrac{2}{3}\right)^2\\ TH1:\dfrac{1}{6}x-3=\dfrac{2}{3}\\ \dfrac{1}{6}x=\dfrac{2}{3}+3=\dfrac{11}{3}\\ x=\dfrac{11}{3}:\dfrac{1}{6}=22\\ TH2:\dfrac{1}{6}x-3=-\dfrac{2}{3}\\ \dfrac{1}{6}x=-\dfrac{2}{3}+3=\dfrac{7}{3}\\ x=\dfrac{7}{3}:\dfrac{1}{6}=14\)