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\(\frac{x\left(3-x\right)}{x+1}\left(x+\frac{3-x}{x+1}\right)=2\)
\(\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\left(\frac{x^2+x+3-x}{x+1}\right)=2\)
\(\Leftrightarrow\frac{x\left(3-x\right)}{x+1}.\frac{x^2+3}{x+1}=2\)
\(\Leftrightarrow\frac{x\left(3-x\right)}{x+1}.\frac{3x+3+x^2-3x}{x+1}=2\)
\(\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\left(1+\frac{x^2-3x}{x+1}\right)=2\)
Đặt \(a=\frac{x\left(3-x\right)}{x+1}\)
\(\Leftrightarrow a\left(1+a=2\right)\)
\frac{x\left(3-x\right)}{x+1}\left(x+\frac{3-x}{x+1}\right)=2x+1x(3−x)(x+x+13−x)=2
\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\left(\frac{x^2+x+3-x}{x+1}\right)=2⇔x+1x(3−x)(x+1x2+x+3−x)=2
\Leftrightarrow\frac{x\left(3-x\right)}{x+1}.\frac{x^2+3}{x+1}=2⇔x+1x(3−x).x+1x2+3=2
\Leftrightarrow\frac{x\left(3-x\right)}{x+1}.\frac{3x+3+x^2-3x}{x+1}=2⇔x+1x(3−x).x+13x+3+x2−3x=2
\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\left(1+\frac{x^2-3x}{x+1}\right)=2⇔x+1x(3−x)(1+x+1x2−3x)=2
Đặt a=\frac{x\left(3-x\right)}{x+1}a=x+1x(3−x)
\Leftrightarrow a\left(1+a=2\right)⇔a(1+a=2)
c, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
- Ta có : \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
=> \(\frac{12\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}=\frac{8\left(x-1\right)}{2\left(x-3\right)\left(x-1\right)}\)
=> \(12\left(x-3\right)-8\left(x-1\right)=8\left(x-1\right)\)
=> \(12x-36-8x+8-8x+8=0\)
=> \(-4x-20=0\)
=> \(x=-5\) ( TM )
Vậy phương trình trên có tập nghiệm là \(S=\left\{-5\right\}\)
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
Ta có : \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
=> \(\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
=> \(x-3=5\left(2x-3\right)\)
=> \(x-3-10x+15=0\)
=> \(-9x=-12\)
=> \(x=\frac{4}{3}\) ( TM )
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{4}{3}\right\}\)
\(a,\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\frac{2-x}{\left(x+1\right)\left(2-x\right)}+\frac{5x+5}{\left(2-x\right)\left(x+1\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow7+4x=15\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow Ptvn\)
\(b,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{10x-15}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-3-10x+15=0\)
\(\Leftrightarrow-9x+12=0\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow\frac{4}{3}\)
\(c,\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{6x-18}{\left(x-1\right)\left(x-3\right)}-\frac{4x-4}{\left(x-1\right)\left(x-3\right)}=\frac{4x-4}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow6x-18-4x+4=4x-4\)
\(\Leftrightarrow2x-14=4x-4\)
\(\Leftrightarrow-2x=10\)
\(\Leftrightarrow x=-5\)
\(d,\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\) \(Đkxđ:\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{3x-9}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow3x-9+2x-4=x-1\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow Ptvn\)
Vậy .................................
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
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
a) ĐK: \(x\ne-1\)
\(x.\frac{3-x}{x+1}\left(x+\frac{3-x}{x+1}\right)=2\)
\(\Leftrightarrow\frac{x^2\left(3-x\right)}{x+1}+\frac{x\left(3-x\right)^2}{\left(x+1\right)^2}-2=0\)
\(\Leftrightarrow\frac{\left(3x^2-x^3\right)\left(x+1\right)+x\left(9-6x+x^2\right)-2\left(x^2+2x+1\right)}{\left(x+1\right)^2}=0\)
\(\Leftrightarrow\left(3x^2-x^3\right)\left(x+1\right)+x\left(9-6x+x^2\right)-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow-x^4+3x^3-5x^2+5x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^3+2x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(-x^2+x-1\right)=0\)
Do \(-x^2+x-1\ne0\forall x\) nên \(x-1=0\Leftrightarrow x=1\)
b) Tương tự.