Question

a, \(\frac{2x+1}{-3}\)=\(\frac{-7}{11}\)

b, ( |x| + \(\frac{1}{3}\) )\(^2\) - 0,375 = (\(\frac{-1}{2}\))\(^3\)

Answer 1

\(a,\frac{2x+1}{-3}=\frac{-7}{11}\)

\(\Leftrightarrow11\left(2x+1\right)=21\)

\(\Leftrightarrow22x+11=21\)

\(\Leftrightarrow22x=10\)

\(\Leftrightarrow x=\frac{5}{11}\)

Vậy .................

\(b,\left(\left|x\right|+\frac{1}{3}\right)^2-0,375=\left(-\frac{1}{2}\right)^3\)

\(\Leftrightarrow\left(\left|x\right|+\frac{1}{3}\right)^2-\frac{3}{8}=-\frac{1}{8}\)

\(\Leftrightarrow\left(\left|x\right|+\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Leftrightarrow\left(\left|x\right|+\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow\left|x\right|+\frac{1}{3}=\frac{1}{2}\)

\(\Leftrightarrow\left|x\right|=\frac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{6}\\x=\frac{1}{6}\end{matrix}\right.\)

Answer 2

a) \(\left(2x+1\right).11=-7.\left(-3\right)\)

\(22x+11=21\)

\(22x=10\)

\(x=\frac{10}{22}=\frac{5}{11}\)

Answer 3

a) Ta có: \(\frac{2x+1}{-3}=\frac{-7}{11}\)

\(\Leftrightarrow2x+1=\frac{\left(-7\right)\cdot\left(-3\right)}{11}=\frac{21}{11}\)

\(\Leftrightarrow2x=\frac{21}{11}-1=\frac{21}{11}-\frac{11}{11}=\frac{10}{11}\)

\(\Leftrightarrow x=\frac{\frac{10}{11}}{2}=\frac{10}{11}\cdot\frac{1}{2}=\frac{10}{22}=\frac{5}{11}\)

Vậy: \(x=\frac{5}{11}\)

Answer 4

Answer 5

Đề b

Answer 6

a, Ta có : \(\frac{2x+1}{-3}=\frac{-7}{11}\)

=> \(11\left(2x+1\right)=\left(-7\right)\left(-3\right)=21\)

=> \(22x+11=21\)

=> \(22x=10\)

=> \(x=\frac{10}{22}=\frac{5}{11}\)

b, \(\left(\left|x\right|+\frac{1}{3}\right)^2-0,375=\left(\frac{-1}{2}\right)^3\)

=> \(\left(\left|x\right|+\frac{1}{3}\right)^2-0,25=0\)

=> \(\left(\left|x\right|+\frac{1}{3}+0,5\right)\left(\left|x\right|+\frac{1}{3}-0,5\right)=0\)

=> \(\left(\left|x\right|+\frac{5}{6}\right)\left(\left|x\right|-\frac{1}{6}\right)=0\)

=> \(\left[{}\begin{matrix}\left|x\right|+\frac{5}{6}=0\\\left|x\right|-\frac{1}{6}=0\end{matrix}\right.\)

TH₁ : \(x\ge0\)

=> \(\left|x\right|=x\)

Nên ta có phương trình : \(\left[{}\begin{matrix}x+\frac{5}{6}=0\\x-\frac{1}{6}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{5}{6}\left(KTM\right)\\x=\frac{1}{6}\left(TM\right)\end{matrix}\right.\)

=> \(x=\frac{1}{6}\)

TH₂ : \(x< 0\)

=> \(\left|x\right|=-x\)

Nên ta có phương trình : \(\left[{}\begin{matrix}-x+\frac{5}{6}=0\\-x-\frac{1}{6}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{5}{6}\left(KTM\right)\\x=-\frac{1}{6}\left(TM\right)\end{matrix}\right.\)

=> \(x=-\frac{1}{6}\)

Answer 7

Vậy phương trình có nghiệm là \(x=\pm\frac{1}{6}\) .

Answer 8

a \(\frac{2x+1}{-3}=\frac{-7}{11}=>\left(2x+1\right).11=21=>22x+11=21\)

=> 22x = 10 => x = \(\frac{5}{11}\)

b, \(\left(\left|x\right|+\frac{1}{3}\right)^2-0,375=\left(\frac{-1}{2}\right)^3\)

= \(\left(\left|x\right|+\frac{1}{3}\right)^2-0,375=\frac{-1}{8}\)

= \(\left(\left|x\right|+\frac{1}{3}\right)^2=\frac{1}{4}\)

=> \(\left|x\right|+\frac{1}{3}=\frac{1}{2}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> x = \(\frac{1}{6}\) hoặc x = \(\frac{-1}{6}\)