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\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)
<=> \(3\left(x^2+1\right)^2.4+3\left(x^2+x\right).4-9.4=0.4\)
<=> \(3\left(x^2+1\right)^2+12\left(x^2+x\right)-36=0\)
<=> \(3x^4+18x^2+12x-33=0\)
<=> \(3\left(x-1\right)\left(x^3+x^2+7x+11\right)=0\)
<=> \(x-1=0\)
<=> \(x=1\)
Mà vì: \(x^3+x^2+7x+11\ne0\)
=> x = 1
\(ĐKXĐ:x\ne-1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x-2-5x-5=15\)
\(\Leftrightarrow-4x=22\Leftrightarrow x=\frac{-11}{2}\)
Vậy \(S=\left\{\frac{-11}{2}\right\}\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow\frac{1\left(x-2\right)-5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{-4x-7}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow-4x-7=15\)
\(\Leftrightarrow-4x=22\)
\(\Leftrightarrow x=22:\left(-4\right)\)
\(\Leftrightarrow x=\frac{-22}{4}=\frac{-11}{2}\)
Vậy tập nghiệm \(S=\left\{\frac{-11}{2}\right\}\)
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) 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 ::
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\left(x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
=> x =-1
<=>\(\left(x+2\right)^2-\left(x-2\right)^3=-\left(x-6\right)\left(x^2-x+2\right)\)
=>\(12x\left(x-1\right)-8=4\left(3x^2-3x-2\right)\)
=>\(-\left(x-6\right)\left(x^2-x+2\right)=4\left(3x^2-3x-2\right)\)
=>x=-5;-2 hoặc 2
<=>\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=\frac{1894289\left(x+100\right)}{6370665}\)
=>\(\frac{1894289\left(x+100\right)}{6370665}=0\)(rút gọn)
=>\(\frac{1894289x}{6370665}+\frac{37885780}{1274133}=0\)
=>\(\frac{1894289\left(x+100\right)}{6370665}=0\)(giải PT này )
=>x=-100
điều kiền x # 0
đặt \(t=x+\frac{1}{x};đk:t\ge2\)=>\(x^2+\frac{1}{x^2}=t^2-2\)
Ta được phương trình mới ẩn t : \(t^2-2t-5=0\)
tự giải phương trình nhé. lấy nghiệm t>= 2