kết quả thì mình ko chắc

a) \(\frac{53}{101}\cdot-\frac{13}{97}+\frac{53}{101}\cdot-\frac{84}{97}\)

\(=\frac{53}{101}\cdot\left(-\frac{13}{97}-\frac{84}{97}\right)\)

\(=\frac{53}{101}\cdot\left(-1\right)\)

\(=-\frac{53}{101}\)

giống huy nhé

Ta có : \(\frac{3}{x}=\frac{-36}{84}=\frac{3}{7}\) => \(x=7\)

  \(\frac{y}{35}=\frac{-36}{84}=\frac{3}{7}\)=> \(7y=3\cdot35\)=> \(y=\frac{3\cdot35}{7}=15\)

\(y=\frac{-36\times35}{84}=-15\)

\(x=\frac{35\times3}{-15}=-7\)

Vậy................

a) \(\left(6\frac{2}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)

=> \(\left(\frac{44}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}=-\frac{11}{7}\)

=> \(\frac{44}{7}x+\frac{3}{7}=-\frac{5}{7}\)

=> \(\frac{44}{7}x=-\frac{8}{7}\)

=> \(\frac{44x}{7}=-\frac{8}{7}\)

=> 44x = -8 => 11x = -2 => \(x=-\frac{2}{11}\)

b) \(3\frac{1}{4}x+\left(-\frac{7}{6}\right)-1\frac{2}{3}=\frac{5}{12}\)

=> \(\frac{13}{4}x-\frac{7}{6}-1\frac{2}{3}=\frac{5}{12}\)

=> \(\frac{13}{4}x-\frac{7}{6}=\frac{25}{12}\)

=> \(\frac{13}{4}x=\frac{13}{4}\)

=> x = 1

c) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

=> \(\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)

d) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

=> \(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

e) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{-24}{27}\)

=> \(\left(3x-\frac{7}{9}\right)^3=-1\frac{5}{27}-\left(-\frac{24}{27}\right)=-\frac{32}{27}+\frac{24}{27}=-\frac{8}{27}\)

=> \(\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

=> \(3x-\frac{7}{9}=-\frac{2}{3}\)

=> \(x=\frac{-\frac{2}{3}+\frac{7}{9}}{3}=\frac{1}{27}\)

g) \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{99\cdot100}=\frac{99}{100}\)

=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{99}-\frac{x}{100}=\frac{99}{100}\)

=> \(\frac{x}{1}-\frac{x}{100}=\frac{99}{100}\)

=> \(\frac{100x-x}{100}=\frac{99}{100}\)

=> \(\frac{99x}{100}=\frac{99}{100}\)

=> x = 1

h) \(\frac{x}{3}+\frac{x}{6}+\frac{x}{10}+\frac{x}{15}=3x-1\)

=> \(\frac{2x}{6}+\frac{2x}{12}+\frac{2x}{20}+\frac{2x}{30}=3x-1\)

=> \(\frac{2x}{2\cdot3}+\frac{2x}{3\cdot4}+\frac{2x}{4\cdot5}+\frac{2x}{5\cdot6}=3x-1\)

=> \(2\left(\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}\right)=3x-1\)

=> \(2\left(\frac{x}{2}-\frac{x}{6}\right)=3x-1\)

=> \(2\left(\frac{3x}{6}-\frac{x}{6}\right)=3x-1\)

=> \(2\cdot\frac{2x}{6}=3x-1\)

=> \(\frac{x}{3}=\frac{3x-1}{2}\)

=> 2x = 3(3x - 1)

=> 2x - 9x + 3  = 0

=> -7x = -3

=> x = 3/7

\(\frac{x}{5}=\frac23\)

\(x\) = \(\frac23\times5\)

\(x=\frac{10}{3}\)

Vậy \(x=\frac{10}{3}\)

\(\frac{x}{3}-\frac12=\frac15\)

\(\frac{x}{3}\) = \(\frac15\) + \(\frac12\)

\(\frac{x}{3}\) = \(\frac{2}{10}+\frac{5}{10}\)

\(\frac{x}{3}=\frac{7}{10}\)

\(x=\frac{7}{10}\times3\)

\(x=\frac{21}{10}\)

Vậy \(x=\frac{21}{10}\)

\(\frac{x}{5}+\frac12=\frac{6}{10}\)

\(\frac{x}{5}=\frac{6}{10}-\frac12\)

\(\frac{x}{5}=\frac{6}{10}-\frac{5}{10}\)

\(\frac{x}{5}=\frac{1}{10}\)

\(x=\frac{1}{10}\times5\)

\(x=\frac12\)

Vậy \(x=\frac12\)

\(\frac{x+3}{15}\) = \(\frac13\)

\(x+3=\frac13\times15\)

\(x+3=5\)

\(x=5-3\)

\(x=2\)

Vậy \(x=2\)

3^x+2=3⁶⁹

=>x+2=69

x=69-2

x=67

2^x-5=8¹⁰

2^x-5=2³⁰

=>x-5=30

x=30+5

x=35

3^x+2+3^x=810

3^x.3²+3^x=810

3^x.(3²+1)=810

3^x.10=810

3^x=810:10

3^x=81

3^x=3⁴

=>x=4

5^x+1-5^x=500

5^x.5-5^x=500

5^x.(5-1)=500

5^x.4=500

5^x=500:4

5^x=125

5^x=5³

=>x=3

a) 3^x+2 = 3⁶⁹

x + 2 = 69

x = 69 - 2

x = 67

b) 2^x-5 = 8¹⁰

2^x-5 = 2³⁰

x  - 5 = 30

x = 30 + 5

x = 35

c) 3^x+2 + 3^x = 810

3^x . 9 + 3^x  . 1 = 810

3^x . ( 9 + 1 ) = 810

3^x . 10 = 810

3^x = 810 : 10

3^x = 81

3^x = 3⁴

=> x = 4

d) 5^x+1 - 5^x = 500

5^x . 5 - 5^x . 1 = 500

5^x . ( 5 - 1 ) = 500

5^x . 4 = 500

5^x = 500 : 4

5^x = 125

5^x = 5³

=> x = 3

dài :vv

a) \(\left|2x-5\right|=4\Leftrightarrow\hept{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Leftrightarrow\hept{\begin{cases}2x=9\\2x=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)

b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)

\(\Leftrightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\Leftrightarrow\hept{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}\Leftrightarrow\hept{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)

Bài 1 :

a) \(|2x-5|=4\)

\(\Rightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}\Rightarrow}\orbr{\begin{cases}2x=9\\2x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}}\)

b) \(\frac{1}{3}-\left|\frac{5}{4}-2x\right|=\frac{1}{4}\)

\(\Rightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{12}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{12}\\\frac{5}{4}-2x=-\frac{1}{12}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{7}{6}\\2x=\frac{4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=\frac{2}{3}\end{cases}}}\)

c) \(\left|\frac{-2}{3}\right|+\left|x-\frac{1}{3}\right|=\left|-1\right|-\left|\frac{-1}{3}\right|\)

\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=1-\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}+\left|x-\frac{1}{3}\right|=\frac{2}{3}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=0\Rightarrow x-\frac{1}{3}=0\Rightarrow x=\frac{1}{3}\)

d) \(\left|-\frac{1}{2}\right|-\left|x+\frac{1}{4}\right|=\left|-\frac{3}{4}\right|\)

\(\Rightarrow\frac{1}{2}-\left|x+\frac{1}{4}\right|=\frac{3}{4}\)

\(\Rightarrow\left|x+\frac{1}{4}\right|=-\frac{1}{4}\)

Vì \(\left|x\right|\ge0\Rightarrow\)ko có gtri nào của x thỏa mãn đề bài

Bài 2 :

a) \(\left|x-1\right|=3x+2\)

\(\Rightarrow\orbr{\begin{cases}x-1=3x+2\\x-1=-3x-2\end{cases}\Rightarrow\orbr{\begin{cases}x-3x=2+1\\x+3x=-2+1\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}-2x=3\\4x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{-1}{4}\end{cases}}\)

b|) \(\left|9+x\right|=2x\Rightarrow\orbr{\begin{cases}9+x=2x\\9+x=-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-2x=-9\\x+2x=-9\end{cases}\Rightarrow\orbr{\begin{cases}-x=-9\\3x=-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=9\\x=-3\end{cases}}}\)

c) \(\left|x+6\right|-9=2x\Rightarrow\left|x+6\right|=2x+9\)

\(\Rightarrow\orbr{\begin{cases}x+6=2x+9\\x+6=-2x-9\end{cases}\Rightarrow}\orbr{\begin{cases}x-2x=9-6\\x+2x=-9-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=3\\3x=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}}\)

Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^

các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho