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Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
\(a,\frac{3x+2}{5x+7}=\frac{3x-1}{5x-1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x-1\right)}=\frac{3}{8};\frac{3x+2}{5x+7}=\frac{3}{8}\Leftrightarrow24x+16=15x+21\Leftrightarrow9x=5\Leftrightarrow x=\frac{5}{9}\) \(b,\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow37.7-7x=3x+39\Leftrightarrow259-7x=3x+39\Leftrightarrow220-7x=3x\Leftrightarrow10x=220\Leftrightarrow x=22\) \(c,\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{x+4}{2x+6}=\frac{\left(x+4\right)-\left(x+1\right)}{2x+6-\left(2x+1\right)}=\frac{3}{5};\frac{x+1}{2x+1}=\frac{3}{5}\Leftrightarrow5x+5=6x+3\Leftrightarrow x=2\) \(d,\frac{x-2}{x+2}=\frac{x+3}{x-4}=\frac{\left(x+3\right)-\left(x-2\right)}{\left(x-4\right)-\left(x+2\right)}=\frac{5}{-6};\frac{x-2}{x+2}=\frac{5}{-6}\Leftrightarrow6\left(2-x\right)=5x+10\Leftrightarrow2-6x=5x\Leftrightarrow x=\frac{2}{11}\) \(f,\frac{3x-5}{x}=\frac{9x}{3x+2}=\frac{9x-15}{3x}=\frac{9x-\left(9x-15\right)}{\left(3x+2\right)-3x}=\frac{15}{2};\frac{9x}{3x+2}=\frac{15}{2}\Leftrightarrow18x=45x+30\Leftrightarrow27x+30=0\Leftrightarrow x=\frac{-10}{9}\) \(e,\frac{x+2}{6}=\frac{5x-1}{5}\Leftrightarrow5\left(x+2\right)=6\left(5x-1\right)\Leftrightarrow5x+10=30x-6\Leftrightarrow10=25x-6\Leftrightarrow25x=16\Leftrightarrow x=\frac{16}{25}\)
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
a) => \(\frac{2,5}{4x}=\frac{0,5}{0,2}\) => 4x. 0,5 = 2,5 .0,2 => 2x = 0,5 => x = 0,5: 2 = 0,25
b) => \(\frac{1}{5}x:3=\frac{2}{3}:\frac{1}{4}\) => \(\frac{1}{5}x:3=\frac{8}{3}\) => \(\frac{1}{5}x=\frac{8}{3}.3=8\)=> x = 8: \(\frac{1}{5}\) = 40
c) tương tự câu a
d) => \(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)=> \(\frac{10}{1}=\frac{26}{2x-1}\) => 10.(2x -1) = 26 => 20x - 10 = 26
=> 20x = 36 => x = 36/20 = 1,8
e) Tương tự d)
f) => 7.(37 - x) = 3.(x+13)
=> 259 - 7x = 3x + 39
=> 259 - 39 = 7x + 3x
=> 220 = 10 x => x = 22