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`8(x-3)(x+1)=8x^2 +11`
`<=>8(x^2 +x-3x-3)-8x^2 -11=0`
`<=>8x^2 +8x-24x-24-8x^2 -11=0`
`<=>-16x-35=0`
`<=>-16x=35`
`<=>x=-35/16`
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(x\ne0;x\ne2\right)\\ < =>\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)
suy ra
`x^2 +2x-2=x-2`
`<=>x^2 +2x-x-2+2=0`
`<=>x^2 +x=0`
`<=>x(x+1)=0`
\(< =>\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ < =>x=-1\)
\(a,8\left(x-3\right)\left(x+1\right)=8x^2+11\\ \Leftrightarrow\left(8x-24\right)\left(x+1\right)=8x^2+11\\ \Leftrightarrow8x^2-24x+8x-24-8x^2-11=0\\ \Leftrightarrow-16x-35=0\\ \Leftrightarrow x=\dfrac{-35}{16}\)
Vậy \(x=-\dfrac{35}{16}\)
\(b,đkxđ:x\ne2;x\ne0\)
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}-\dfrac{1}{x}=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=0\\ \Leftrightarrow x^2+2x-2-x+2=0\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
Vậy \(x=-1\)
@ducminh
a. ĐKXĐ: \(x\ne2\).
\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
⇔\(\dfrac{1}{x-2}+\dfrac{3x-6}{x-2}=\dfrac{3-x}{x-2}\)
⇔\(1+3x-6=3-x\)
⇔\(4x-8=0\)
⇔\(x=2\) (không thỏa mãn)
-Vậy S=∅.
b. ĐKXĐ: \(x\ne-1\)
\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\)
⇔\(\dfrac{5x}{2\left(x+1\right)}+1=-\dfrac{6}{x+1}\)
⇔\(\dfrac{5x}{2\left(x+1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)}=-\dfrac{12}{2\left(x+1\right)}\)
⇔\(5x+2\left(x+1\right)=-12\)
⇔\(5x+2x+2+12=0\)
⇔\(7x+14=0\)
⇔\(x=-2\) (thỏa mãn).
-Vậy \(S=\left\{-2\right\}\)
a, \(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3.\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\\ \Leftrightarrow1+3x-6=3-x\)
\(\Leftrightarrow3x+x=3-1+6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=\dfrac{8}{4}=2\\ Vậy.S=\left\{2\right\}\)
b, \(\Leftrightarrow\)\(\dfrac{5x}{2x+2}+\dfrac{2x+2}{2x+2}=\dfrac{-6.2}{2.\left(x+1\right)}\)
\(\Leftrightarrow5x+2x+2=-12\\ \Leftrightarrow7x=-12-2\\ \Leftrightarrow7x=-14\\ \Leftrightarrow x=-\dfrac{14}{7}=-2\\ Vậy.S=\left\{-2\right\}\)
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).
a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)
\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)
\(\Leftrightarrow4-2x-6x-12\le3x-51\)
\(\Leftrightarrow-11x\le-43\)
\(\Leftrightarrow x\ge\dfrac{43}{11}\)
Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)
\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)
\(\Leftrightarrow0x\le-10\) (vô lý)
Vậy \(S=\varnothing\)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}=3\)
b) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow-x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
a, đk : x khác 5;-6
\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm)
b, đk : x khác 1;3
\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)
pt vô nghiệm
a, đk : x khác 5;-6
x2+12x+36+x2−10x+25=2x2+23x+61x2+12x+36+x2−10x+25=2x2+23x+61
⇔2x+61=23x+61⇔21x=0⇔x=0⇔2x+61=23x+61⇔21x=0⇔x=0(tm)
b, đk : x khác 1;3
x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)
pt vô nghiệm
1:
a: =>28x-8=9x+3
=>19x=11
=>x=11/19
b: =>(3x-1)(x-1)=(2x+1)(x+1)
=>3x^2-4x+1=2x^2+3x+1
=>x^2-7x=0
=>x=0 hoặc x=7
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}`
\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)
`<=> 20(2x-1) +15(3x-2) =12(4x-3)`
`<=> 40x - 20 + 45x - 30 = 48x - 36`
`<=> 85x -50 = 48x - 36`
`<=> 85x-48x = -36+50`
`<=> 37x =14`
`<=> x= 14/37`
Vậy phương trình có nghiệm `x=14/37`
__
\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)
\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
`=> 5x + 15 + 4x -12=x-6`
`<=> 9x + 3=x-6`
`<=> 9x-x=-6-3`
`<=> 8x = -9`
`<=>x=-9/8(tm)`
Vậy phương trình có nghiệm `x=-9/8`
` @ yngoc`
