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\(2\frac{1}{5}+\frac{3}{5}x=\frac{3}{4}\Rightarrow\frac{3}{5}x=\frac{3}{4}-2\frac{1}{5}=-\frac{29}{20}\Rightarrow x=-\frac{29}{20}:\frac{3}{5}=-\frac{29}{12}\)
Vậy \(x=-\frac{29}{12}\)
\(\left(2\frac{1}{5}+\frac{3}{5}x\right)=\frac{3}{4}\)
\(2\frac{1}{5}+\frac{3}{5}x=\frac{3}{4}\)
\(\frac{11}{5}+\frac{3}{5}x=\frac{3}{4}\)
\(\frac{3}{5}x=\frac{3}{4}-\frac{11}{5}=-\frac{29}{20}\)
\(x=\frac{-29}{20}:\frac{3}{5}=-\frac{29}{12}\)
a) \(\frac{3}{x-5}=\frac{-4}{x+2}\)(ĐKXĐ: \(x\ne5;x\ne-2\))
\(\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)(thỏa mãn ĐKXĐ)
b) \(2,4x-36=-\frac{7}{12}\)
\(\Leftrightarrow2,4x=\frac{425}{12}\)
\(\Leftrightarrow x=\frac{2125}{144}\)
c) \(\left(\frac{19}{5}-2x\right).\frac{4}{3}=\frac{40}{7}\)
\(\Leftrightarrow\frac{19}{5}-2x=\frac{30}{7}\)
\(\Leftrightarrow2x=-\frac{17}{35}\)
\(\Leftrightarrow x=-\frac{17}{70}\)
b) \(\left(2,4.x-36\right)\div1\frac{5}{7}=-1\)
\(\left(2,4.x-36\right)=-1.\frac{12}{7}\)
\(2,4.x-36=-\frac{12}{7}\)
\(2,4.x=-\frac{12}{7}+36\)
\(2,4.x=\frac{240}{7}\)
\(x=\frac{240}{7}\div2,4\)
\(x=\frac{100}{7}\)
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
\(\Rightarrow x-\frac{5}{21}=\left|1\frac{1}{6}\right|\)
\(\Rightarrow x-\frac{5}{21}=\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{6}+\frac{5}{21}=\frac{49}{42}+\frac{10}{42}=\frac{59}{42}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow x+\left|-1\frac{2}{3}\right|=\frac{3}{4}\)
\(\Rightarrow x-\frac{3}{4}=-\left|-1\frac{2}{3}\right|\)
\(\Rightarrow x-\frac{3}{4}=-1\frac{2}{3}\)
\(\Rightarrow x-\frac{3}{4}=-\frac{5}{3}\)
\(\Rightarrow x=-\frac{5}{3}+\frac{3}{4}=-\frac{11}{12}\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{2}\\x-\frac{1}{3}=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}+\frac{1}{3}=\frac{17}{6}\\x=-\frac{5}{2}+\frac{1}{3}=-\frac{13}{6}\end{matrix}\right.\)
4) \(\left|x+\frac{2}{3}\right|=0\)
\(\Rightarrow x+\frac{2}{3}=0\)
\(\Rightarrow x=0-\frac{2}{3}=-\frac{2}{3}\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
\(\Rightarrow\left|x+2\right|=\frac{2}{15}\)
\(\Rightarrow\left[{}\begin{matrix}x+2=\frac{2}{15}\\x+2=-\frac{2}{15}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{15}-2=-\frac{28}{15}\\x=-\frac{2}{15}-2=-\frac{32}{15}\end{matrix}\right.\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
\(\Rightarrow\left|x-4\right|=\frac{19}{20}\)
\(\Rightarrow\left[{}\begin{matrix}x-4=\frac{19}{20}\\x-4=-\frac{19}{20}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{19}{20}+4=\frac{99}{20}\\x=-\frac{19}{20}+4=\frac{61}{20}\end{matrix}\right.\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
Vì \(\left|x-\frac{5}{4}\right|\ge0\)
=> Không có giá trị x thỏa mãn với điều kiện trên
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
\(\left(2,4x-36\right)=-1.\frac{12}{7}\)
\(\left(2,4x-36\right)=\frac{-12}{7}\)
\(2,4x=\frac{-12}{7}+36\)
\(2,4x=\frac{240}{7}\)
\(x=\frac{240}{7}:2,4\)
\(x=\frac{100}{7}\)
\(\left(2,4x-36\right):1\frac{5}{7}=-1\)
2,4x - 36 = -1 . \(\frac{12}{7}=\) -12/7
2,4x = -12/7 + 36 = 240/7
x = 240/7 : 2,4 = 100/7