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Đặt
\(A=\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}\)
( ĐKXĐ : \(x\ne2,x\ne3,x\ne4,x\ne5,x\ne6\) )
\(=\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}\)
\(=\frac{1}{x-2}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-4}+...+\frac{1}{x-5}-\frac{1}{x-6}\)
\(=\frac{1}{x-2}-\frac{1}{x-6}\)
\(=\frac{-4}{\left(x-2\right)\left(x-6\right)}\)
Để : \(A\ge0\Leftrightarrow\frac{-4}{\left(x-2\right)\left(x-6\right)}\ge0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\le0\)
TH1 : \(\hept{\begin{cases}x-2\le0\\x-6\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\le2\\x\ge6\end{cases}}\) ( vô lý )
TH2 : \(\hept{\begin{cases}x-2\ge0\\x-6\le0\end{cases}\Leftrightarrow2\le x\le6}\)kết hợp với ĐKXĐ
\(\Rightarrow2< x< 6\)
Vậy : \(2< x< 6\) thỏa mãn bất phương trình.
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
40x-20+6x+18 (lớn hơn hoặc bằng ) 84x+36 - 96+8x
rồi giải bt @@:
x (bé hơn hoặc bằng) -(29:23)
Ko có cách nào hết
\(\frac{x+5}{x+1}-\frac{x-4}{x+6}=\frac{20}{x^2+7x+6}\left(x\ne-1;x\ne-6\right)\)
\(\Leftrightarrow\frac{x+5}{x+1}-\frac{x-4}{x+6}-\frac{20}{x^2+7x+6}=0\)
\(\Leftrightarrow\frac{x+5}{x+1}-\frac{x-4}{x+6}-\frac{20}{\left(x+1\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}-\frac{\left(x-4\right)\left(x+1\right)}{\left(x+1\right)\left(x+6\right)}-\frac{20}{\left(x+1\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\frac{x^2+11x+30}{\left(x+1\right)\left(x+6\right)}-\frac{x^2-3x-4}{\left(x+1\right)\left(x+6\right)}-\frac{20}{x^2+7x+6}=0\)
\(\Leftrightarrow\frac{x^2+11x+30-x^2+3x+4-20}{\left(x+1\right)\left(x+6\right)}=0\)
\(\Leftrightarrow\frac{14x+14}{\left(x+1\right)\left(x+6\right)}=0\)
=> 14x+14=0
<=> x=-1 (ktm)
Vậy pt vô nghiệm