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1/ \(2\left(x-5\right)=\left(-x-5\right)\)
\(\Leftrightarrow2x-10=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)
==========
2/ \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy: \(S=\left\{7\right\}\)
==========
3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1=x-19\)
\(\Leftrightarrow0x=0\)
Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\)
===========
4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2=10-15x\)
\(\Leftrightarrow14x=1\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)
==========
5/ \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x=7x+1\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy: \(S=\left\{-3\right\}\)
[---]
Chúc bạn học tốt.
1. \(2\left(x-5\right)=-x-5\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
Vậy \(S=\left\{\dfrac{5}{3}\right\}\)
2. \(2\left(x+3\right)-3\left(x-1\right)=2\)
\(\Leftrightarrow2x+6-3x+3=2\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)
\(\Leftrightarrow4x-20-3x+1-x+19=0\)
\(\Leftrightarrow0x=0\)
Vậy \(S=\left\{x\in R\right\}\)
4. \(7-\left(x-2\right)=5\left(2-3x\right)\)
\(\Leftrightarrow7-x+2-10+15x=0\)
\(\Leftrightarrow14x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{14}\)
Vậy \(S=\left\{\dfrac{1}{14}\right\}\)
4. \(2x-\left(5-3x\right)=7x+1\)
\(\Leftrightarrow2x-5+3x-7x-1=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
\(\text{a) 5(2x-3)-4(5x-7)=19-2(x+11)}\)
\(10x-15-20x+28=19-2x-22\)
\(10x-20x+2x=19-22-28+15\)
\(-8x=-16\)
\(\Rightarrow x=2\)
\(\text{b) 4(x+3)-7x+17=8(5x-1)+166}\)
\(4x+12-7x+17=40x-8+166\)
\(4x-7x-40x=-8+166-17-12\)
\(-43x=129\)
\(x=-3\)
\(\text{c) 17-14(x+1)=13-4(x+1)-5(x-3)}\)
\(17-14x+14=13-4x-4-5x+15\)
\(-14x+4x+5x=13-4+15-14-17\)
\(-5x=-7\)
\(x=\frac{7}{5}\)
\(\text{d) 5x+3,5+(3x-4)=7x-3(x-0,5)}\)
\(5x+3,5+3x-4=7x-3x+1,5\)
\(5x+3x-7x+3x=1,5-3,5\)
\(x=-2\)
\(\text{e) 7(4x+3)-4(x-1)=15(x+0,75)+7}\)
\(28x+21-4x+4=15x+11,25+7\)
\(28x-4x-15x=11,25+7-4-21\)
\(9x=\frac{-27}{4}\)
\(x=\frac{-3}{4}\)
\(\text{f) 3x+2,42+o,8x=3,38-0,2x}\)
\(3x+0,8x+0,2x=3,38-2,42\)
\(4x=\frac{24}{25}\)
\(x=\frac{6}{25}\)
chúc bạn học tốt !!
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
1/ \(7x-5=13-5x\)
\(\Leftrightarrow12x=18\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)
==========
2/ \(19+3x=5-18x\)
\(\Leftrightarrow21x=-14\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)
==========
3/ \(x^2+2x-4=-12+3x+x^2\)
\(\Leftrightarrow-x=-8\)
\(\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
===========
4/ \(-\left(x+5\right)=3\left(x-5\right)\)
\(\Leftrightarrow-x-5=3x-15\)
\(\Leftrightarrow-4x=-10\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)
==========
5/ \(3\left(x+4\right)=\left(-x+4\right)\)
\(\Leftrightarrow3x+12=-x+4\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Vậy: \(S=\left\{-2\right\}\)
[----------]
1. \(7x-5=13-5x\) \(\Leftrightarrow12x=18\Leftrightarrow x=\dfrac{3}{2}\)
2. \(19+3x=5-18x\Leftrightarrow21x=-14\Leftrightarrow x=-\dfrac{2}{3}\)
3. \(x^2+2x-4=-12+3x+x^2\Leftrightarrow-x=-8\Leftrightarrow x=8\)
4. \(-\left(x+5\right)=3\left(x-5\right)\Leftrightarrow-x-5=3x-15\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)
5. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)