Bài 1

a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)

                y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)

                          y= \(\frac{-3}{14}\)

b.5^x + 5^x+2=650

5^x . 1 + 5^x + 5²=650

5^x(1+25)=650

5^x.26=650

5^x=25

x=2

1:2:3:4:5:7:9:10

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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 \(x\in\left\{\frac{1}{12}\right\}\)

\(\hept{\begin{cases}x+y+z+t=1\\x+y+z=2\end{cases}}\)

\(\Rightarrow\left(x+y+z+t\right)-\left(x+y+z\right)=1-2\)

\(\Rightarrow t=-1\)

\(\hept{\begin{cases}x+y+z+t=1\\y+z+t=3\end{cases}}\)

\(\Rightarrow\left(x+y+z+t\right)-\left(y+z+t\right)=1-3\)

\(\Rightarrow x=-2\)

\(\hept{\begin{cases}x+y+z+t=1\\z+x+t=4\end{cases}}\)

\(\Rightarrow\left(x+y+z+t\right)-\left(z+x+t\right)=1-4\)

\(\Rightarrow y=-3\)

\(x+y+z+t=1\)

\(\Rightarrow\left(-2\right)+\left(-3\right)+\left(-1\right)+t=1\)

\(\Rightarrow\left(-6\right)+t=1\)

\(\Rightarrow t=7\)

a) Từ \(\frac{x}{4}=\frac{25}{x}=>x.x=25.4\)

=> \(x^2=100\)

=> x=10 hoặc -10

b) Từ \(\frac{y^2}{3}=\frac{12}{1}=>y^2.1=12.3\)

=> \(y^2=36\)

=> y=6 hoặc -6

a)x^2=25*4

x^2=100

suy ra x=10

b)y^2*1=12*3

y^2*1=36

y^2=36

suy ra y=6 nha 

\(\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)

\(\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)

\(\frac{3}{x}=\frac{-1}{2}\)

\(\Rightarrow3.2=\left(-1\right).x\)

\(\Rightarrow6=\left(-1\right).x\)

\(\Rightarrow x=6:\left(-1\right)\)

\(\Rightarrow x=-6\)

\(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)

\(\Rightarrow\frac{x}{2}-\frac{1}{2}=\frac{2}{y}\)

\(\Rightarrow\frac{x-1}{2}=\frac{2}{y}\)

\(\Rightarrow\hept{\begin{cases}x-1=2\\2=y\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}\)

\(b,\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)

\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)

\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{8}{6}\)

\(\Rightarrow\frac{3}{x}=\frac{-3}{6}\)

\(\Rightarrow x\cdot(-3)=18\Rightarrow x=-6\)