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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
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(=>x^2+x+1-3x^2=2x\left(x-1\right)\)
\(=>-2x^2+x+1=2x^2-2x\)
\(=>-4x^2+3x+1=0\)
\(=>\left(x-1\right)\left(x+\frac{1}{4}\right)=0\)'
\(=>\orbr{\begin{cases}x-1=0\\x+\frac{1}{4}\end{cases}=>\orbr{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}}\)
2. \(\frac{1}{x-1}-\frac{7}{x-2}=\frac{1}{\left(x-1\right)\left(2-x\right)}\) (ĐKXĐ:\(x\ne1,x\ne2\))
\(\Leftrightarrow\frac{1}{x-1}+\frac{7}{2-x}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Leftrightarrow\frac{2-x+7\left(x-1\right)}{\left(x-1\right)\left(2-x\right)}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Rightarrow2-x+7\left(x-1\right)=1\)
\(\Leftrightarrow2-x+7x-7=1\)
\(\Leftrightarrow-x+7x=1-2+7\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)
Vậy phương trình trên vô nghiệm
ko phan tich duoc nha bn
chuc bn hoc gioi
happy new year
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne\pm2\\x\ne0\end{matrix}\right.\)
Ta có : \(\frac{x-4}{x\left(x+2\right)}-\frac{1}{x\left(x-2\right)}=-\frac{2}{\left(x+2\right)\left(x-2\right)}\)
=> \(\frac{\left(x-4\right)\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}-\frac{x+2}{x\left(x-2\right)\left(x+2\right)}=-\frac{2x}{x\left(x+2\right)\left(x-2\right)}\)
=> \(\left(x-4\right)\left(x-2\right)-x-2=-2x\)
=> \(x^2-4x-2x+8-x-2=-2x\)
=> \(x^2-5x+6=0\)
=> \(\left(x-2\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=2\\x=3\left(TM\right)\end{matrix}\right.\)
=> x = 3 .
Vậy phương trình trên có tập nghiệm là \(S=\left\{3\right\}\)
b, ĐKXĐ : \(x\ne0,-3,-6,-9,-12\)
Ta có : \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+12\right)}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{1}{x}-\frac{1}{x+12}=\frac{1}{16}\)
=> \(\frac{x+12}{x\left(x+12\right)}-\frac{x}{x\left(x+12\right)}=\frac{1}{16}\)
=> \(x\left(x+12\right)=192\)
=> \(x^2+12x-192=0\)
=> \(x^2+2x.6+36-228=0\)
=> \(\left(x+6\right)^2=288\)
=> \(\left[{}\begin{matrix}x=\sqrt{288}-6\\x=-\sqrt{288}-6\end{matrix}\right.\) ( TM )
Vậy phương trình có tập nghiệm là \(S=\left\{\pm\sqrt{288}-6\right\}\)
thì sao
ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x-1}-\frac{7}{x-2}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 7(x - 1)(2 - x) = x - 2
<=> 2x - x2 - 4 + 2x - 14x + 7x2 + 14 - 7x = x - 2
<=> -17x + 6x2 + 10 = x - 2
<=> 17x - 6x2 - 10 + x - 2 = 0
<=> 18x - 6x2 - 12 = 0
<=> 6(3x - x2 - 2) = 0
<=> -6(x2 + 3x - 2) = 0
<=> x2 - 3x + 2 = 0
<=> (x - 1)(x - 2) = 0
<=> x - 1 = 0 hoặc x - 2 = 0
<=> x = 1 (ktm) hoặc x = 2 (ktm)
=> pt vo nghiem