a) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)

b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)

c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)

e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

dễ mà

x+7=-16-2

x=7=-18

x=-18:7

x=-18/7

x-9=7 

x= 7+9

x=16

x-5 = 0 hoặc x+7= 0

x=5 hoặc x=-7

=> x thuộc { 5,-7}

tức vờ lờ

a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-13}{20}\)

Vậy \(x=\frac{-13}{20}\)

b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)

  \(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)

\(\frac{-1}{2}x=\frac{15}{8}\)

\(x=\frac{-15}{4}\)

Vậy \(x=\frac{-15}{4}\)

\(a,2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{11}{20}-\frac{5}{4}\)

\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)

\(x+\frac{2}{5}=\frac{-7}{10}:2\)

\(x+\frac{2}{5}=\frac{-7}{20}\)

\(x=\frac{-7}{20}-\frac{2}{5}\)

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

app hay 

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

1) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

                  \(\frac{x}{y}=\frac{17}{3}\) => \(\frac{x}{17}=\frac{y}{3}=\frac{x+y}{17+3}=\frac{-60}{20}=-3\)

=> \(\hept{\begin{cases}\frac{x}{17}=-3\\\frac{y}{3}=-3\end{cases}}\) => \(\hept{\begin{cases}x=-51\\y=-9\end{cases}}\)

Vậy ....

2) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

           \(\frac{x}{19}=\frac{y}{21}\)=> \(\frac{2x}{38}=\frac{y}{21}=\frac{2x-y}{38-21}=\frac{34}{17}=2\)

=> \(\hept{\begin{cases}\frac{x}{19}=2\\\frac{y}{21}=2\end{cases}}\) => \(\hept{\begin{cases}x=38\\y=42\end{cases}}\)

vậy ...

3) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

       \(\frac{x^2}{9}=\frac{y^2}{16}=\frac{x^2+y^2}{9+16}=\frac{100}{25}=4\)

=> \(\hept{\begin{cases}\frac{x^2}{9}=4\\\frac{y^2}{16}=4\end{cases}}\) => \(\hept{\begin{cases}x^2=36\\y^2=64\end{cases}}\) => \(\hept{\begin{cases}x=\pm6\\y=\pm8\end{cases}}\)

Vậy ...

4) Ta có: \(\frac{x}{y}=\frac{10}{9}\) => \(\frac{x}{10}=\frac{y}{9}\)

         \(\frac{y}{z}=\frac{3}{4}\) => \(\frac{y}{3}=\frac{z}{4}\) => \(\frac{y}{9}=\frac{z}{12}\)

=> \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

     \(\frac{x}{10}=\frac{y}{9}=\frac{z}{12}=\frac{x-y+z}{10-9+12}=\frac{78}{13}=6\)

=> \(\hept{\begin{cases}\frac{x}{10}=6\\\frac{y}{9}=6\\\frac{z}{12}=6\end{cases}}\) => \(\hept{\begin{cases}x=60\\y=54\\z=72\end{cases}}\)

Vậy ...