Question

[Thảo Linh]

Ai giúp mình với mình like cho

Answer 1

[Đức Hiếu]

a, \(\left|3x+\dfrac{7}{4}\right|=1\dfrac{1}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}3x+\dfrac{7}{4}=1\dfrac{1}{3}\\3x+\dfrac{7}{4}=-1\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x=-\dfrac{5}{12}\\3x=-\dfrac{37}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{36}\\x=-\dfrac{37}{6}\end{matrix}\right.\)

Vậy...........

b, \(\left|5\dfrac{1}{2}-4x\right|=\left|2\dfrac{1}{2}-1\dfrac{1}{3}\right|\)

\(\Rightarrow\left\{{}\begin{matrix}5\dfrac{1}{2}-4x=2\dfrac{1}{2}-1\dfrac{1}{3}\\5\dfrac{1}{2}-4x=1\dfrac{1}{3}-2\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4x=2\dfrac{1}{2}-1\dfrac{1}{3}-5\dfrac{1}{2}\\4x=1\dfrac{1}{3}-2\dfrac{1}{2}-5\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{\dfrac{13}{3}}{4}\\x=-\dfrac{\dfrac{20}{3}}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{13}{12}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

Vậy..............

c, \(\left|x+\dfrac{1}{3}\right|+4=5\dfrac{2}{3}\)

\(\Rightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{5}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{3}=-\dfrac{5}{3}\\x+\dfrac{1}{3}=\dfrac{5}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy...........

d, \(\left|3x+\dfrac{3}{4}\right|-\dfrac{1}{3}=\left|1\dfrac{1}{4}-\dfrac{1}{5}\right|\)

\(\Rightarrow\left|3x+\dfrac{3}{4}\right|=\dfrac{21}{20}+\dfrac{1}{3}\)

\(\Rightarrow\left|3x+\dfrac{3}{4}\right|=\dfrac{83}{60}\)

\(\Rightarrow\left\{{}\begin{matrix}3x+\dfrac{3}{4}=-\dfrac{83}{60}\\3x+\dfrac{3}{4}=\dfrac{83}{60}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=-\dfrac{32}{15}\\3x=\dfrac{19}{30}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{32}{45}\\x=\dfrac{19}{90}\end{matrix}\right.\)

Vậy............

e, \(3-\left|2x-\dfrac{11}{5}\right|=-\dfrac{1}{3}\)

\(\Rightarrow\left|2x-\dfrac{11}{5}\right|=\dfrac{10}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}2x-\dfrac{11}{5}=\dfrac{10}{3}\\2x-\dfrac{11}{5}=-\dfrac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{83}{30}\\x=-\dfrac{17}{30}\end{matrix}\right.\)

Vậy..............

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