Câu hỏi của VAN PHAN

Question

$\left(x-2\right)^3+\frac{8}{27}=0$

$4\frac{1}{3}:\frac{x}{4}=6:0,3$

(0,25-30%x).$\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}$

$\frac{x-2}{\frac{-2}{9}}=\frac{-2}{x-2}$

Nguyễn Lê Phước Thịnh · Accepted Answer

a) Ta có: $\left(x-2\right)^3+\frac{8}{27}=0$

$\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}$

$\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3$

$\Leftrightarrow x-2=\frac{-2}{3}$

hay $x=\frac{-2}{3}+2=\frac{4}{3}$

Vậy: $x=\frac{4}{3}$

b) Ta có: $4\frac{1}{3}:\frac{x}{4}=6:0,3$

$\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20$

$\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}$

hay $x=\frac{13\cdot4}{60}=\frac{13}{15}$

Vậy: $x=\frac{13}{15}$

c) Ta có: $\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}$

$\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}$

$\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}$

$\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16$

$\Leftrightarrow3x=-160$

hay $x=\frac{-160}{3}$

Vậy: $x=\frac{-160}{3}$

d) Ta có: $\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}$

$\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}$

$\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\x=\frac{4}{3}\end{matrix}\right.$

Vậy: $x\in\left\{\frac{8}{3};\frac{4}{3}\right\}$Câu trả lời: a) Ta có: $\left(x-2\right)^3+\frac{8}{27}=0$

$\Leftrightarrow\left(x-2\right)^3=\frac{-8}{27}$

$\Leftrightarrow\left(x-2\right)^3=\left(-\frac{2}{3}\right)^3$

$\Leftrightarrow x-2=\frac{-2}{3}$

hay $x=\frac{-2}{3}+2=\frac{4}{3}$

Vậy: $x=\frac{4}{3}$

b) Ta có: $4\frac{1}{3}:\frac{x}{4}=6:0,3$

$\Leftrightarrow\frac{13}{3}\cdot\frac{4}{x}=20$

$\Leftrightarrow\frac{4}{x}=20:\frac{13}{3}=20\cdot\frac{3}{13}=\frac{60}{13}$

hay $x=\frac{13\cdot4}{60}=\frac{13}{15}$

Vậy: $x=\frac{13}{15}$

c) Ta có: $\left(0,25-30\%x\right)\cdot\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}$

$\Leftrightarrow\left(\frac{1}{4}-\frac{3x}{10}\right)\cdot\frac{1}{3}=\frac{31}{6}+\frac{1}{4}=\frac{65}{12}$

$\Leftrightarrow\frac{1}{4}-\frac{3x}{10}=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}\cdot3=\frac{65}{4}$

$\Leftrightarrow\frac{3x}{10}=\frac{1}{4}-\frac{65}{4}=-16$

$\Leftrightarrow3x=-160$

hay $x=\frac{-160}{3}$

Vậy: $x=\frac{-160}{3}$

d) Ta có: $\frac{x-2}{-\frac{2}{9}}=\frac{-2}{x-2}$

$\Leftrightarrow\left(x-2\right)^2=-2\cdot\left(-\frac{2}{9}\right)=\frac{4}{9}$

$\Leftrightarrow\left[{}\begin{matrix}x-2=\frac{2}{3}\x-2=-\frac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}+2\x=\frac{-2}{3}+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\x=\frac{4}{3}\end{matrix}\right.$

Vậy: $x\in\left\{\frac{8}{3};\frac{4}{3}\right\}$

Trúc Giang · Answer

a/ (x - 2)3 + $\frac{8}{27}$ = 0

=> (x - 2)3 = 0 - $\frac{8}{27}$ = $\frac{-8}{27}$

=> x - 2 = $-\frac{2}{3}$

=> x = $-\frac{2}{3}+2=\frac{4}{3}$

b/ $4\frac{1}{3}:\frac{x}{4}=6:0,3$

=> $4\frac{1}{3}:\frac{x}{4}=6:\frac{3}{10}=6.\frac{10}{3}=20$

=> $\frac{x}{4}=4\frac{1}{3}:20=\frac{13}{3}.\frac{1}{20}=\frac{13}{60}$

=> $x=\frac{13}{60}.4=\frac{13}{15}$

c/ $\left(0,25-30\%x\right).\frac{1}{3}-\frac{1}{4}=5\frac{1}{6}$

=> $\left(0,25-30\%x\right).\frac{1}{3}=5\frac{1}{6}+\frac{1}{4}=\frac{65}{12}$

=> $0,25-\frac{30}{100}x=\frac{65}{12}:\frac{1}{3}=\frac{65}{12}.3=\frac{65}{4}$

=> $\frac{3}{10}x=0,25-\frac{65}{4}=\frac{1}{4}-\frac{65}{4}=-\frac{64}{4}=-16$

=> $x=-16:\frac{3}{10}=-16.\frac{10}{3}=-\frac{160}{3}$Câu trả lời: a/ (x - 2)3 + $\frac{8}{27}$ = 0