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

a: \(sin75^0=sin\left(30^0+45^0\right)\)

\(=sin30^0\cdot cos45^0+cos30^0\cdot sin45^0\)

\(=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}\)

\(=\dfrac{\sqrt{2}+\sqrt{6}}{4}\)

b: \(tan75=tan\left(30+45\right)\)

\(=\dfrac{tan30+tan45}{1-tan30\cdot tan45}\)

\(=\dfrac{\dfrac{\sqrt{3}}{3}+1}{1-\dfrac{\sqrt{3}}{3}\cdot1}=\dfrac{3+\sqrt{3}}{3}:\dfrac{3-\sqrt{3}}{3}=\dfrac{3+\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{\left(3+\sqrt{3}\right)^2}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}=\dfrac{12+6\sqrt{3}}{9-3}=2+\sqrt{3}\)

c: \(\dfrac{1}{\sqrt{3}\cdot sin250^0}-\dfrac{1}{cos110^0}\)

\(=\dfrac{1}{\sqrt{3}\cdot sin\left(360^0-110^0\right)}-\dfrac{1}{cos110^0}\)

\(=\dfrac{1}{\sqrt{3}\cdot sin\left(-110^0\right)}-\dfrac{1}{cos110^0}=\dfrac{-1}{\sqrt{3}\cdot sin110^0}-\dfrac{1}{cos110^0}\)

\(=\dfrac{-cos110^0-\sqrt{3}\cdot sin110^0}{\sqrt{3}\cdot sin110^0\cdot cos110^0}=\dfrac{-\dfrac{1}{2}\cdot cos110^0-\dfrac{\sqrt{3}}{2}\cdot sin110^0}{\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}\cdot2\cdot sin110^0\cdot cos110^0}\)

\(=-\dfrac{cos110^0\cdot sin30^0+sin110^0\cdot cos30^0}{\dfrac{\sqrt{3}}{2}\cdot sin220^0}\)

\(=-\dfrac{sin140^0}{\dfrac{\sqrt{3}}{2}\cdot sin220^0}\)

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

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

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

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

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

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

\(=500-2000+1724\)

\(=224\)

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. 

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

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