Tìm x thuộc Q, biết:
a) (x-4).(x+1/3)>0
b) x+1/1974 + x+2/1973 + x+3/1974 = -3
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\(\frac{x+1}{1974}+1+\frac{x+2}{1973}+1+\frac{x+3}{1972}+1=-3+3\)
\(\Leftrightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)
\(\Leftrightarrow\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)
=> x + 1975 = 0. => x = -1975 ( vì 1/1974 + 1/1973 + 1/1972 khác 0 )
\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)
\(=>\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=-3+3\)
\(=>\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)
\(\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)
\(=>x+1975=0=>x=-1975\)
Vậy \(x=-1975\)
\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)
\(\Leftrightarrow\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=0\)
\(\Leftrightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)
\(\Leftrightarrow\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)
\(\Leftrightarrow x+1975=0\)
\(\Leftrightarrow x=-1975\)
\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)
\(\Rightarrow\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=0\)
\(\Rightarrow\frac{x+1+1974}{1974}+\frac{x+2+1973}{1973}+\frac{x+3+1972}{1972}=0\)
\(\Rightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)
\(\Rightarrow\left(x+1975\right)\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}=0\)
Mà \(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\ne0\)
\(\Rightarrow x+1975=0\)
\(\Rightarrow x=0+1975\)
\(\Rightarrow x=1975\)
Vậy \(x=1975\)
b) phần này làm tương tự phần a nha, chuyển -3 sang vế bên trái r cộng từng p.số vs 1 và sau đó nhóm tử số chung ra ngoài ^^
a: \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)