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1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
a)\(\frac{1}{x-1}\)-\(\frac{3x2}{x3-1}\)=\(\frac{2x}{x2+x+1}\)
<=> \(\frac{1}{x-1}\)-\(\frac{3x2}{\left(x-1\right)\left(x2+x+1\right)}\)=\(\frac{2x}{x2+x+1}\) ĐKXĐ: x khác 1
<=> x2+x+1 - 3x2 = 2x(x-1)
<=>x2+x+1 - 3x2 = 2x2-2x
<=>x2-3x-1=0( đoạn này làm nhanh nhé)
<=>x2-2*\(\frac{3}{2}\)x +\(\frac{9}{4}\)-\(\frac{9}{4}\)-1=0
<=>(x-\(\frac{3}{2}\))2-\(\frac{13}{4}\)=0
<=>(x-\(\frac{3-\sqrt{13}}{2}\))(x-\(\frac{3+\sqrt{13}}{2}\))=0
\(\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}\)
b) pt... đkxđ x khác 1;2;3
<=> 3(x-3) +2(x-2)=x-1
<=> 3x-9 +2x-4 = x-1
<=> 4x= 12
<=> x=3 ( ko thỏa đk)
vậy pt vô nghiệm
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(=>x^2+x+1-3x^2=2x\left(x-1\right)\)
\(=>-2x^2+x+1=2x^2-2x\)
\(=>-4x^2+3x+1=0\)
\(=>\left(x-1\right)\left(x+\frac{1}{4}\right)=0\)'
\(=>\orbr{\begin{cases}x-1=0\\x+\frac{1}{4}\end{cases}=>\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)
a) ĐKXĐ: \(x\ne-1;x\ne2\)
Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(x-2-5x-5+15=0\)
⇔\(-4x+8=0\)
⇔\(-4x=-8\)
⇔\(x=\frac{-8}{-4}=2\)(loại)
Vậy: x không có giá trị
b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
⇔\(x-3-10x+15=0\)
⇔\(-9x+12=0\)
⇔\(-9x=-12\)
⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ:\(x\ne3;x\ne1\)
Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2\left(x-3\right)}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)
⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)
⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)
⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)
⇔6x-18-8x+8=0
⇔-2x-10=0
⇔-2(x+5)=0
Vì 2≠0 nên x+5=0
hay x=-5
Vậy: x=-5
(2x+9)/(x+1)(x+8)-(2x+15)/(x+8)(x+7)+(2x+10)/(x+7)(x+3)=4/3
(x+1+x+8)/(x+1)(x+8)-(x+8+x+7)/(x+8)(x+7)+(x+7+x+3)/(x+7)(x+3)=4/3
1/(x+8)+1/(x+1)-1/(x+7)-1/(x+8)+1/(x+7)+1/(x+3)=4/3
1/(x+1)+1/(x+3)=4/3
(x+3+x+1)/(x+3)(x+1)=4/3
(2x+4)/(x+3)(x+1)=4/3
=>(2x+4).3=(x+3)(x+1).4
6(x+2)=4(x+3)(x+1)
3(x+2)=2(x+3)(x+1)
3x+6=2(x^2+4x+3)
3x+6=2x^2+8x+6
2x^2+8x+6-3x-6=0
2x^2+5x=0
x(2x+5)=0
=> x=0 hoac 2x+5=0
=> x=0 hoac x=-5/2
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
\(\frac{1}{x+3}+\frac{8}{\left(x-1\right)\left(x-3\right)}=\frac{x+3}{x^2-2x=3}\)
\(\Leftrightarrow\frac{1}{x+3}+\frac{8}{\left(x-1\right)\left(x-3\right)}=\frac{x+3}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+1\right)+8\left(x+3\right)\left(x+1\right)=\left(x+3\right)^2\left(x-1\right)\)
\(\Leftrightarrow x^3+5x^2+31x+27=x^3+5x^2+3x-9\)
\(\Leftrightarrow5x^2+31x+27=5x^2+3x-9\)
\(\Leftrightarrow31x+27=3x-9\)
\(\Leftrightarrow31x+27-3x=-9\)
\(\Leftrightarrow28x+27=-9\)
\(\Leftrightarrow28x=-36\)
\(\Leftrightarrow x=\frac{-36}{28}=-\frac{9}{7}\)
Vậy \(x=-\frac{9}{7}\)