Trả lời:

c, \(\left(x-\frac{2}{7}\right)\left(0,5x^2+\frac{3}{4}\right)=0\) 

\(\Leftrightarrow x-\frac{2}{7}=0\) ( vì \(0,5x^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\) )

\(\Leftrightarrow x=\frac{2}{7}\)

Vậy x = 2/7 

d, \(\left(-\frac{5}{4}x+3,25\right)\left[\frac{3}{5}-\left(-\frac{5}{2}x\right)\right]=0\)

\(\Leftrightarrow\left(-\frac{5}{4}x+3,25\right)\left(\frac{3}{5}+\frac{5}{2}x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{4}x+3,25=0\\\frac{3}{5}+\frac{5}{2}x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{5}\\x=-\frac{6}{25}\end{cases}}}\)

Vậy x = 13/5; x = - 6/25

c)(x−27 )(0,5x²+34 )=0

d)(−54 x+3,25)[35 −(−52 x)]=0

Trả lời:

a, \(3\left|x\right|+2=5\)

\(\Leftrightarrow3\left|x\right|=3\)

\(\Leftrightarrow\left|x\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy x = 1; x = - 1

b, \(4-\left|x\right|=\frac{1}{4}\)

\(\Leftrightarrow\left|x\right|=\frac{15}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{4}\\x=-\frac{15}{4}\end{cases}}\)

Vậy x = 15/4; x = - 15/4

c, \(4+\left|2x\right|=\frac{2}{3}\)

\(\Leftrightarrow\left|2x\right|=\frac{10}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{10}{3}\\2x=-\frac{10}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-\frac{5}{3}\end{cases}}}\)

Vậy x = 5/3; x = - 5/3

d, \(3\left|x\right|=\frac{1}{7}-\frac{1}{8}\)

\(\Leftrightarrow3\left|x\right|=\frac{1}{56}\)

\(\Leftrightarrow\left|x\right|=\frac{1}{168}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{168}\\x=-\frac{1}{168}\end{cases}}\)

Vậy x = 1/168; x = - 1/168

a, \(|-2,1|\)\(=2,1\)

b,\(|-\)\(\frac{2}{9}|\)\(=\frac{2}{9}\)

c,\(|12,2|\)\(=12,2\)

d,\(|0|\)\(=0\)

e,\(|\frac{3}{2}|\)=\(\frac{3}{2}\)

Hok tốt~

\(\frac{3}{11}:-\frac{9}{5}+\frac{3}{11}:-\frac{18}{13}\)

\(=\frac{3}{11}\cdot-\frac{5}{9}+\frac{3}{11}\cdot-\frac{13}{18}\)

\(=\frac{3}{11}\cdot\left(-\frac{5}{9}+-\frac{13}{18}\right)\)

\(=\frac{3}{11}\cdot-\frac{23}{18}\)

\(=-\frac{23}{66}\)

- \(\frac{13}{66}\)

\(\frac{2x-5}{y+1}=\frac{x-1}{3y}=\frac{1}{3}\Rightarrow\hept{\begin{cases}3\left(2x-5\right)=y+1\\x-1=y\end{cases}}\)

\(\Rightarrow6x-15=x-1+1\)

\(\Leftrightarrow x=3\)

suy ra \(y=2\).