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a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
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.
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