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\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
`e)3/(3x)-3/12=4/5-(7/x-2)`
`<=>1/x-1/4=4/5-7/x+2`
`<=>8/x=1/4+4/5+2=61/20`
`<=>1/x=61/160`
`<=>x=160/61`
`f)1/(x-1)+(-2)/3(3/4-6/5)=5/(2-2x)`
`<=>1/(x-1)+5/(2x-2)=2/3(3/4-6/5)=-3/10`
`<=>7/(2x-1)=-3/10`
`<=>2x-1=-70/3`
`<=>2x=-67/3`
`<=>x=-67/6`
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
a)Ta có: \(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}x\right)\)
\(\Leftrightarrow\dfrac{x}{2}-\dfrac{3x-13}{5}=\dfrac{-7}{5}-\dfrac{7x}{10}\)
\(\Leftrightarrow\dfrac{5x}{10}-\dfrac{2\left(3x-13\right)}{10}=\dfrac{-14}{10}-\dfrac{7x}{10}\)
\(\Leftrightarrow5x-6x+26=-14-7x\)
\(\Leftrightarrow-x+26+14+7x=0\)
\(\Leftrightarrow6x=-40\)
hay \(x=-\dfrac{20}{3}\)
d) Ta có: \(\dfrac{2x-3}{2}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}+\dfrac{-9}{6}=\dfrac{5-3x}{6}-\dfrac{2}{6}\)
\(\Leftrightarrow6x-9-9=5-3x-2\)
\(\Leftrightarrow6x-18-3+3x=0\)
\(\Leftrightarrow x=\dfrac{7}{3}\)
Bài 1:
a.
$|x+\frac{7}{4}|=\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)
b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$
$|2x+1|=\frac{11}{15}$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)
c.
$3x(x+\frac{2}{3})=0$
\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)
d.
$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$
$\Leftrightarrow x=\frac{2}{5}$
Nguyễn Quý Trung:
\(x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
Bạn bớt 2 vế đi 1/3 thì \(x=\dfrac{2}{5}\)
\(a.ĐK:x\ne3;1\)
\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)
\(\Leftrightarrow7x-21=7x^2-28x+21\)
\(\Leftrightarrow7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
b.\(ĐK:x\ne2;4\)
\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(4-x\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)
\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)
\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)
=>(x-3)(7x-14)=0
=>x=3(loại) hoặc x=2(nhận)
b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)
\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)
\(\Leftrightarrow2x^2-4x=0\)
=>2x(x-2)=0
=>x=0(nhận) hoặc x=2(loại)