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Bài `1:`
`h)(3/4x-1)(5/3x+2)=0`
`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`
______________
Bài `2:`
`b)3x-15=2x(x-5)`
`<=>3(x-5)-2x(x-5)=0`
`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`
`d)x(x+6)-7x-42=0`
`<=>x(x+6)-7(x+6)=0`
`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`
`f)x^3-2x^2-(x-2)=0`
`<=>x^2(x-2)-(x-2)=0`
`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`
`h)(3x-1)(6x+1)=(x+7)(3x-1)`
`<=>18x^2+3x-6x-1=3x^2-x+21x-7`
`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`
`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`
`j)(2x-5)^2-(x+2)^2=0`
`<=>(2x-5-x-2)(2x-5+x+2)=0`
`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`
`w)x^2-x-12=0`
`<=>x^2-4x+3x-12=0`
`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`
`m)(1-x)(5x+3)=(3x-7)(x-1)`
`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`
`<=>(1-x)(5x+3+3x-7)=0`
`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`
`p)(2x-1)^2-4=0`
`<=>(2x-1-2)(2x-1+2)=0`
`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`
`r)(2x-1)^2=49`
`<=>(2x-1-7)(2x-1+7)=0`
`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`
`t)(5x-3)^2-(4x-7)^2=0`
`<=>(5x-3-4x+7)(5x-3+4x-7)=0`
`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`
`u)x^2-10x+16=0`
`<=>x^2-8x-2x+16=0`
`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`
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
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
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 \(S=\left\{-3\right\}\)
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. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
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 \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
