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\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
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Ta có : \(\left(5x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\2x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=3\\2x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{5}{2}\end{cases}}\)
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a) x-1/3=5/14.(-7/6)
x-1/3=-5/12
x=-5/12+1/3
x= -1/12
b) 3/4+1/4x=0,2
1/4x=0,2-3/4
1/4x=-11/20
x=-11/20:1-4
x=-11/5
c) 1/12.x^2=1.1/2
1/12.x^2=1/2
x^2= 1/2:1/12
x^2=6
=> x=căn bậc của 6
Chúc bn hok tốt
a)\(x-\frac{1}{3}=\frac{5}{14}\cdot\left(\frac{-7}{6}\right)\)
\(x-\frac{1}{3}=\frac{-5}{12}\)
\(x=\frac{-5}{12}+\frac{1}{3}\)
\(x=\frac{-5}{12}+\frac{4}{12}\)
\(x=\frac{-1}{12}\)
b)\(\frac{3}{4}+\frac{1}{4}\cdot x=0,2\)
\(\frac{3}{4}+\frac{1}{4}\cdot x=\frac{1}{5}\)
\(\frac{1}{4}\cdot x=\frac{1}{5}-\frac{3}{4}\)
\(\frac{1}{4}\cdot x=\frac{4}{20}-\frac{15}{20}\)
\(\frac{1}{4}\cdot x=\frac{-11}{20}\)
\(x=\frac{-11}{20}\cdot4\)
\(x=-\frac{11}{5}\)
c)\(\frac{1}{12}\cdot x^2=1\cdot\frac{1}{3}\)
\(\frac{1}{12}\cdot x^2=\frac{1}{3}\)
\(x^2=\frac{1}{3}\cdot12\)
\(x^2=4=\left(\pm2\right)^2\)
\(x=\pm2\)
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Bài 2:
b: x+25%x=-1,25
=>1,25x=-1,25
hay x=-1
c: x-75%x=1/4
=>1/4x=1/4
hay x=1
Bài 2:
a: =3/2-11/4=6/4-11/4=-5/4
b: =-49/6-17/2=-49/6-51/6=-100/6=-50/3
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1. 75% .x - 1/1/5 = 0,6x
=> 3/4x - 6/5 = 3/5x
=> 6/5 = 3/4x - 3/5x
=> 6/5 = 3/20x
=> x = 6/5 : 3/20 = 8
2.4/x=-y/6=0,5
=> 4/x=1/2
=> 4/x = 4/8
=> x = 8
-y/6 = 1/2
=> y = -3
ta có 4/8 = -(-3)/6 = 1/2
3. 2/5 + 2/7 - 2/9 = 146/315
6/5+6/7-6/9=146/105
mà 146/105 > 146/315
=> 2/5+2/7-2/9 > 6/5+6/7-6/9
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a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn