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a, 15-3(x+2)=-21
\(\Rightarrow3\left(x+2\right)=15-\left(-21\right)\)
\(\Rightarrow3\left(x+2\right)=36\)
\(\Rightarrow x+2=12\)
\(\Rightarrow x=10\)
vậy x=10
b,\(\frac{x-2}{3}=\frac{x+3}{2}\)
\(\Rightarrow\left(x-2\right).2=\left(x+3\right).3\)
\(\Rightarrow2x-4=3x+9\)
\(\Rightarrow2x-3x=9+4\)
\(\Rightarrow x=-13\)
k mình nha bn
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Ta có :
\(\frac{x}{3}:\frac{2}{3}=1\frac{3}{4}:\frac{2}{5}\)
\(\Rightarrow\frac{x}{3}:\frac{2}{3}=\frac{7}{4}:\frac{2}{5}\)
\(\Rightarrow\frac{x}{3}:\frac{2}{3}=\frac{35}{8}\)
\(\Rightarrow\frac{x}{3}=\frac{35}{8}.\frac{2}{3}=\frac{35.2}{8.3}=\frac{35}{12}\)
\(\Rightarrow\frac{x}{3}=\frac{35}{12}\)
\(\Rightarrow x=35:12.3=\frac{35}{4}\)
\(1;A=\frac{x+7}{x+1}=\frac{x+1+6}{x+1}=1+\frac{6}{x+1}\)
Vậy x + 1 là ước của 6 \(\Rightarrow x+1\in\left(1;-1;2;-2;3;-3;6;-6\right)\)
\(\Rightarrow x\in\left(0;-2;1;-3;2;-4;5;-7\right)\)
\(2;A=\frac{6x-2}{2x-3}=\frac{6x-9+7}{2x-3}=3+\frac{7}{2x-3}\)
Vậy 2x - 3 là ước của 7 \(\Rightarrow2x-3\in\left(1;-1;7;-7\right)\)
\(\Rightarrow x\in\left(2;1;5;-2\right)\)
\(3;A=\frac{4x-8}{2x+1}=\frac{4x+2-10}{2x+1}=2-\frac{10}{2x+1}\)
Vậy 2x + 1 là ước của 10 => .........
\(|3x-5|=|\frac{1}{2}+3|\)
\(\Rightarrow|3x-5|=|\frac{7}{2}|=\frac{7}{2}\)
\(\orbr{\begin{cases}3x-5=\frac{7}{2}\\3x-5=-\frac{7}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}3x=\frac{17}{2}\\3x=-\frac{3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{17}{6}\\x=-\frac{1}{2}\end{cases}}}\)
a)15-3 x (x+2)=-21
3 x (x +2)=15-(-21)=36
x+2=36/3=12
x=12-2
x=10
b)\(\left(x-2\right).2=\left(x+3\right).3\)
\(\left(x+1\right).2=3\)
\(x+1=6\)
\(x=6-1=5\)
a) 15-3(x+2)=-21
3(x+2)=15-(-21)=36
x+2=12
x=10
b) \(\frac{x-2}{3}=\frac{x+3}{2}\)
=>2(x-2)=3(x+3)
=>2x-4=3x+9
=>3x-2x=-4-9
=>x=-13