Gọi x là số học sinh lớp 7A (0 < x < 153, x € N*)

=> Số học sinh lớp 7B,C lần lượt là: 8x/9 và (17/16)(8x/9)

Theo đeef bài ta có:

x + 8x/9 + (17/16)(8x/9) = 153 (HS)

<=> ( 1 + 8/9 + (17/16)(8/9) )x = 153

<=> x = 153 : 17/6 = 54 (HS)

=> Số HS lơp 7B,C lần lượt là:

8x/9 = 48 (HS)

153 - 48 - 54 = 51 (HS)

Vậy (tự vậy nhé ^^ )

Hok tốt~

\(x^2=4\rightarrow x=\pm2\)
\(x^2=5\rightarrow x=\pm\sqrt{5}\)
\(x^2=0\rightarrow x=0\)
\(x^2=1\rightarrow x=\pm1\)
\(x^2-9=0\rightarrow x^2=9\rightarrow x=\pm3\)
\(x^2+1=0\rightarrow x^2=-1\rightarrow x\in\varnothing\)
\(x^2=2\rightarrow x=\pm\sqrt{2}\)
\(x^2-3=0\rightarrow x^2=3\rightarrow x=\pm\sqrt{3}\)
\(x^2+1=82\rightarrow x^2=81\rightarrow x=\pm9\)
\(7x^2=63\rightarrow x^2=9\rightarrow x=\pm3\)
\(x^2+\frac{7}{4}=\frac{23}{4}\rightarrow x^2=4\rightarrow x=\pm2\)
Chúc bạn hok tốt!!!

a, \(\frac{a}{b}=\frac{c}{d}=k\)

\(\frac{a}{b}=k\Rightarrow a=b\cdot k\)

\(\frac{c}{d}=k\Rightarrow c=d\cdot k\)

Theo bài ra ta có :

\(\frac{a+b}{b}=\frac{b\cdot k+b}{b}=\frac{b\cdot\left(k+1\right)}{v}=k+1\left(1\right)\)

\(\frac{c+d}{d}=\frac{d\cdot k+d}{d}=\frac{d\left(k+1\right)}{d}=k+1\left(2\right)\)

Từ (1) và (2) ta có

\(k+1=k+1\)

\(\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\left(đpcm\right)\)

b, Vì \(\frac{a}{b}=\frac{c}{d}\)nên :

\(\frac{a}{b}=\frac{c}{d}=\frac{3a}{3b}=\frac{2c}{2d}\)

\(\frac{a}{b}=\frac{c}{d}=-\frac{5a}{-5b}=\frac{3c}{3d}\)

\(\Rightarrow\frac{3a+2c}{3b+2d}=\frac{-5a+3c}{-5b+3d}\)

\(\frac{x}{8}=\frac{5,4}{3}=>x=\frac{5,4.8}{3}=14,4\)

x/8=5,4/3

=> x = (8 * 5,4 ) : 3

=> x =  14 , 4

\(\sqrt{81}=9\)

\(\sqrt{0,64}=0,8\)

\(\sqrt{\frac{49}{100}}=\frac{7}{10}\)

\(\sqrt{8100}=90\)

\(\sqrt{100=}10\)

\(\sqrt{0,01}=0,1\)

\(\sqrt{\frac{4}{25}}=\frac{2}{5}\)

\(\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}\)

\(\sqrt{81}=9\);\(\sqrt{0,64}=0,8\);\(\sqrt{\frac{49}{100}}=\frac{7}{10}\);\(\sqrt{8100}=90\); \(\sqrt{100}=10\); \(\sqrt{0,01}=0,1\); \(\sqrt{\frac{4}{25}}=\frac{2}{5}\); \(\sqrt{\frac{0,09}{121}}=\frac{0,3}{11}=\frac{3}{110}\)

Gọi A là\(\frac{\frac{2}{39}-\frac{1}{15}-\frac{2}{153}}{\frac{1}{34}+\frac{3}{20}-\frac{3}{20}}:\frac{\frac{2}{71}-\frac{5}{121}+1}{\frac{65}{121}-13-\frac{26}{71}}\)

\(=\frac{-\frac{4}{9}\cdot\left(\frac{1}{34}+\frac{3}{20}-\frac{3}{26}\right)}{-\left(\frac{1}{34}-\frac{3}{20}-\frac{3}{76}\right)}:\frac{\frac{2}{71}-\frac{5}{121}+1}{-13\left(\frac{2}{71}-1-\frac{5}{121}\right)}\)

\(=\frac{4}{-9}:-\frac{1}{3}\)

\(=\frac{52}{9}\)

\(\frac{\frac{2}{3}\left(\frac{1}{13}-\frac{1}{10}-\frac{1}{51}\right)}{\frac{3}{2}\left(\frac{1}{51}+\frac{1}{10}-\frac{1}{13}\right)}:\frac{\frac{2}{71}-\frac{5}{121}+1}{-13\left(\frac{5}{121}-1-\frac{2}{71}\right)}=\frac{-\frac{4}{9}\left(\frac{1}{13}-\frac{1}{10}-\frac{1}{51}\right)}{-\left(\frac{1}{51}+\frac{1}{10}-\frac{1}{13}\right)}.-13=\frac{52}{9}\)

mình ko viêt đề nha

Lần lượt từ trái qua phải, trên xuống dưới là: 13,12,15,17,20,1/2,1/3,1/4,1/4,1/5,1/6,2/4,3/5,5/3,7/6