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a) \(\frac{3x-6}{x+4}=\frac{2\left(x+5\right)+\left(x-3\right)}{x-2}\)
\(\frac{3\left(x-2\right)}{x+4}=\frac{2\left(x+5\right)+x-3}{x-2}\)
\(\frac{3\left(x-4\right)}{x+4}=\frac{3x+7}{x-2}\)
\(3\left(x-2\right)\left(x-2\right)=\left(3x+7\right)\left(x+4\right)\)
\(3\left(x-2\right)^2=\left(3x+7\right)\left(x+4\right)\)
\(3x^2-12x+12=3x^2+12x+7x+28\)
\(3x^2-12x+12=3x^2+19x+28\)
\(-12x+12=19x+28\)
\(12=19x+28+12x\)
\(19x+28+12x=12\) (chuyển vế)
\(31x+28=12\)
\(31x=12-28\)
\(31x=-16\)
\(x=-\frac{16}{31}\)
\(\Rightarrow x=-\frac{16}{31}\)
\(a,\left(4\frac{1}{2}-\frac{2}{5}x\right):1\frac{3}{4}=\frac{11}{14}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{4}\cdot\frac{7}{4}\)
\(\Rightarrow\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{77}{16}\)
\(\Rightarrow\frac{9}{2}-\frac{2}{5}x=\frac{77}{16}\)
\(\Rightarrow-\frac{2}{5}x=\frac{77}{16}-\frac{9}{2}\)
\(\Rightarrow-\frac{2}{5}x=\frac{5}{16}\)
\(\Rightarrow x=\frac{5}{16}:\left(-\frac{2}{5}\right)\)
\(\Rightarrow x=-\frac{25}{32}\)
\(b,\frac{2}{3}\cdot x-\frac{2}{5}x=\frac{9}{3}\)
\(\Rightarrow x\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{8}{3}\)
\(\Rightarrow x\cdot\frac{4}{15}=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}:\frac{4}{15}\)
\(\Rightarrow x=10\)
\(c,\frac{-2}{3}|x|+1\frac{1}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|+\frac{3}{2}=\frac{2}{5}\)
\(\Rightarrow\frac{-2}{3}|x|=\frac{2}{5}-\frac{3}{2}\)
\(\Rightarrow\frac{-2}{3}|x|=-\frac{11}{10}\)
\(\Rightarrow|x|=\frac{-11}{10}:\frac{-2}{3}\)
\(\Rightarrow|x|=\frac{33}{20}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{33}{20}\\x=-\frac{33}{20}\end{cases}}\)
\(d,|2x-\frac{1}{3}|+\frac{1}{6}=\frac{3}{4}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{3}{4}-\frac{1}{6}\)
\(\Rightarrow|2x-\frac{1}{3}|=\frac{7}{12}\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=\frac{7}{12}\\2x-\frac{1}{3}=-\frac{7}{12}\end{cases}\Rightarrow\orbr{\begin{cases}2x=\frac{11}{12}\\2x=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{24}\\x=-\frac{1}{8}\end{cases}}}\)
\(2.\)
\(a,\frac{1}{3}< \frac{x}{4}< \frac{2}{3}\)
\(\Rightarrow\frac{1\times4}{3\times4}< \frac{x\times3}{4\times3}< \frac{2\times4}{3\times4}\)
\(\Rightarrow\frac{4}{12}< \frac{x\times3}{12}< \frac{8}{12}\)
\(\Rightarrow4< x\times3< 8\)
\(\Rightarrow x\times3\in\left\{5;6;7\right\}\)
Ta có bảng giá trị :
| \(x\times3\) | \(5\) | \(6\) | \(7\) |
| \(x\) | \(\varnothing\) | \(2\) | \(\varnothing\) |
| \(NX\) | Loại | TM | Loại |
Vậy \(x=2\)
\(b,\frac{3}{5}< \frac{-x}{3}< \frac{4}{5}\)
\(\Rightarrow\frac{3\times3}{5\times3}< \frac{-x\times5}{3\times5}< \frac{4\times3}{5\times3}\)
\(\Rightarrow\frac{9}{15}< \frac{-x\times5}{15}< \frac{12}{15}\)
\(\Rightarrow9< -x\times5< 12\)
\(\Rightarrow-x\times5\in\left\{10;11\right\}\)
Ta có bảng giá trị :
| \(-x\times5\) | \(10\) | \(11\) |
| \(-x\) | \(2\) | \(\varnothing\) |
| \(x\) | \(-2\) | \(\varnothing\) |
| \(NX\) | TM | Loại |
Vậy \(x=-2\)
a) \(\left|x+\frac{3}{5}\right|-\frac{1}{2}=\frac{1}{2}\)
\(\left|x+\frac{3}{5}\right|=\frac{1}{2}+\frac{1}{2}\)
\(\left|x+\frac{3}{5}\right|=1\)
\(=>x+\frac{3}{5}=1\) hoặc \(x+\frac{3}{5}=-1\)
=> x = \(1-\frac{3}{5}\) x = \(-1-\frac{3}{5}\)
x = \(\frac{2}{5}\) x = \(\frac{-8}{5}\)
b) -c) - d) làm tương tự