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a) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{3x}+\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4}{12x}+\dfrac{6}{12x}=\dfrac{3x}{12x}\)
Suy ra: \(3x=10\)
\(\Leftrightarrow x=\dfrac{10}{3}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{10}{3}\right\}\)
b) ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)
\(\Leftrightarrow\dfrac{3x}{8x^2}-\dfrac{4x}{8x^2}=\dfrac{8}{8x^2}\)
Suy ra: \(3x-4x=8\)
\(\Leftrightarrow-x=8\)
hay x=-8(thỏa ĐK)
Vậy: S={-8}
c)ĐKXĐ: \(x\ne0\)
Ta có: \(\dfrac{1}{2x}+\dfrac{3}{4x}=\dfrac{5}{2x^2}\)
\(\Leftrightarrow\dfrac{2x}{4x^2}+\dfrac{3x}{4x^2}=\dfrac{10}{4x^2}\)
Suy ra: 2x+3x=10
\(\Leftrightarrow5x=10\)
hay x=2(thỏa ĐK)
Vậy: S={2}
d, \(\dfrac{2a}{x+a}=1\) (x \(\ne\) -a)
\(\Leftrightarrow\) \(\dfrac{2a}{x+a}-\dfrac{x+a}{x+a}=0\)
\(\Leftrightarrow\) \(\dfrac{a-x}{x+a}=0\)
\(\Leftrightarrow\) a - x = 0 (x + a \(\ne\) 0)
\(\Leftrightarrow\) x = a (TM)
Vậy S = {a}
Chúc bn học tốt!
b) Đặt \(x^2+2x+3=a\)(a>0)
Ta có: \(\dfrac{x^2+2x+7}{\left(x+1\right)^2+2}=x^2+2x+4\)
\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+1+2}=x^2+2x+4\)
\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+3}=x^2+2x+4\)
\(\Leftrightarrow\dfrac{a+4}{a}=a+1\)
\(\Leftrightarrow a^2+a=a+4\)
\(\Leftrightarrow a^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(nhận\right)\\a=-2\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2+2x+3=2\)
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Vậy: S={-1}
ĐKXĐ của cả 2 pt trên đều là `x in RR`
`a,1/(x^2-2x+2)+2/(x^2-2x+3)=6/(x^2-2x+4)`
Đặt `a=x^+2x+3(a>=2)` ta có:
`1/(a-1)+2/a=6/(a+1)`
`<=>a(a+1)+2(a-1)(a+1)=6a(a-1)`
`<=>a^2+a+2(a^2-1)=6a^2-6a`
`<=>a^2+a+2a^2-2=6a^2-6a`
`<=>3a^2-5a+2=0`
`<=>3a^2-3a-2a+2=0`
`<=>3a(a-1)-2(a-1)=0`
`<=>(a-1)(3a-2)=0`
`a>=2=>a-1>=1>0`
`a>=2=>3a-2>=4>0`
Vậy pt vô nghiệm
`(x^2+2x+7)/((x+1)^2+2)=x^2+2x+4`
`<=>(x^2+2x+7)=(x^2+2x+4)(x^2+2x+3)`
Đặt `a=x^2+2x+3(a>=2)`
`pt<=>a+4=a(a+1)`
`<=>a^2+a=a+4`
`<=>a^2=4`
`<=>a=2` do `a>=2`
`<=>(x+1)^2+2=2`
`<=>(x+1)^2=0`
`<=>x=-1`
Vậy `S={-1}`
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
a: =>(x-2)(2x+5)=0
=>x-2=0 hoặc 2x+5=0
=>x=2 hoặc x=-5/2
c: \(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\)
=>\(\dfrac{2x^2+2x-x^2+x}{x^2-1}=1\)
=>x^2+3x=x^2-1
=>3x=-1
=>x=-1/3
\(a,\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{2;\dfrac{5}{2}\right\}\)
\(c,\Leftrightarrow2x.\left(x+1\right)-x.\left(x-1\right)=\left(x-1\right)\left(x+1\right)\) ( ĐKXĐ: \(x\ne-1;x\ne1\) )
\(\Leftrightarrow2x^2+2x-x^2+x=x^2-1\\ \Leftrightarrow x^2-x^2+3x=-1\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\) ( nhận )
Vậy phương trình có tập nghiệm S = \(\left\{-\dfrac{1}{3}\right\}\)
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)
giúp mk câu k nhé đề bài như trên
b: \(\Leftrightarrow4x+8-9=4x-4\)
=>-1=-4(loại)
d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)
=>8x=3x-6+2x+2=5x-4
=>3x=-4
=>x=-4/3
f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)
=>3x+6+8x-12=2x-24
=>11x-6=2x-24
=>9x=-18
=>x=-2
1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
=>-4x<12
hay x>-3
2: \(\Leftrightarrow6+2x+2>2x-1-12\)
=>8>-13(đúng)
4: \(\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)
=>x-3<0
hay x<3
6: =>(x+4)(x-1)<=0
=>-4<=x<=1
a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)
=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)
=>2x-2018<0
=>x<2019
b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)
=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)
=>\(x+97< 0\)
=>x<-97