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1: Ta có: \(\dfrac{3}{x+2}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)
Suy ra: \(3x-6-x+1=2x+4\)
\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)
2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)
Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)
\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)
\(\Leftrightarrow5x=16\)
hay \(x=\dfrac{16}{5}\)
\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow x\left(x+8\right)=105\)
\(\Leftrightarrow x^2+8x-105=0\)
\(\Leftrightarrow x^2-7x+15x-105=0\)
\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
`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, 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
d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow-x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
đkxđ \(x\ne0;2\)
\(\dfrac{2}{x^2-2x}+\dfrac{1}{x}=\dfrac{x+2}{x-2}\\ \Leftrightarrow\dfrac{2+\left(x-2\right)-\left(x+2\right)x}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{-x^2-x}{x\left(x-2\right)}=0\\ \Leftrightarrow-x^2-x=0\\ \Leftrightarrow-x\left(x+1\right)=0\)
mà \(x\ne0\Rightarrow x+1=0\\ \Leftrightarrow x=-1\)