Giải phương trình
c)\(\frac{1}{x-1}\) + \(\frac{2}{x^2+x+1}\) = \(\frac{3x^2}{x^3-1}\)
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\(a,2x\left(x+5\right)=x+5\)
\(2x^2+10x=x+5\)
\(2x^2+10x-x-5=0\)
\(2x^2+9x-5=0\)
\(2x^2+x-10x-5=0\)
\(x\left(2x+1\right)-5\left(2x+1\right)=0\)
\(\left(x-5\right)\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\2x=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-\frac{1}{2}\end{cases}}}\)
\(pt\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x^2+3x-1}{x^3-1}=\frac{3x^2}{x^3-1}\)
\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow3x-1=2x^2\Leftrightarrow2x^2-3x+1=0\Leftrightarrow x^2-\frac{3}{2}x+\frac{1}{2}=0\)
đến đây là pt bậc 2
\(\left(2x+1\right)^2-3\left(x-1\right)^2-\left(x+1\right)\left(x-1\right)\)
\(=\left(2.\left(-\frac{1}{2}\right)+1\right)^2-3\left(-\frac{1}{2}-1\right)^2-\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)\)
\(=-3\left(-\frac{9}{4}\right)-\frac{1}{2}.\left(-\frac{3}{2}\right)\)
\(=\frac{27}{4}+\frac{3}{4}=\frac{31}{4}\)
còn đâu tự lm nha !
\(x\left(x-y\right)+y\left(y-x\right)=x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)=\left(x-y\right)^2=\left(124-24\right)^2=100^2=10000\)
\(\left(x-3\right)^2-\left(x+1\right)^3+12x\left(x-1\right)=\frac{49}{4}-\frac{1}{8}+\frac{\left(-6\right).\left(-3\right)}{2}\)
\(=\frac{97}{8}+9=\frac{169}{8}\)
X(X-Y)+Y(Y-X)=X2 -XY +Y2 -XY=(X-Y)2 =(124-24)2 =1002 =10000
(x-3)2 -(x+1)3 +12x(x-1)=x2 -6x+9-x3 -3x2 -3x-1+12x2 -12x=-x3 +10x2 -9x+8
cho n thuộc N, n>1
CMR : 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/n+n < 3/4
giúp mik vs mik đang cần gập ạ 😚😚😚
(3x+5)(2x-1)+(4x-1)(3x+2)
=6x2-3x+10x-5+12x2+8x-3x-2
=18x2+12x-7
#hoctot :)
\(\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)
\(=6x^2-3+10x-5+12x^2+8x-3x-2\)
\(=18x^2+12x-7\)
giúp mik vs
ĐKXĐ:\(x\ne1\)
\(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)
\(\Leftrightarrow\frac{x^2+x+1+2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow x^2+3x-1=3x^2\)\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(KTMĐK\right)\\x=\frac{1}{2}\left(TMĐK\right)\end{cases}}}\)
Vậy nghiệm của pt là \(x=\frac{1}{2}\)