Cho biểu thức P=x+2x+3 - 5x2+x−6 +\(\frac{1}{2-x}\)
Tìm x để P=\(\frac{-3}{4}\)
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\(x^2-2xy+2y^2-x+8=\left(x-y-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2}\)
Dấu "=" xảy ra khi \(x=1,y=\frac{1}{2}\)
Bài làm
\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
\(=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}-\frac{1}{x-2}\)
\(=\frac{x+2}{x+3}-\frac{5}{x\left(x+3\right)-2\left(x+3\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) x2 - 9 = 0 <=> ( x - 3 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}x=3\left(nhan\right)\\x=-3\left(loai\right)\end{cases}}\)
x = 3 => \(P=\frac{3-4}{3-2}=-1\)
c) \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P đạt giá trị nguyên => \(\frac{2}{x-2}\)nguyên
=> \(2⋮x-2\)
=> \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x-2 | 1 | -1 | 2 | -2 |
x | 3 | 1 | 4 | 0 |
Vậy ...
a, \(x\ne-1;3\)
b, Ta có : \(P=\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=1\Leftrightarrow\frac{3x}{2\left(x-3\right)}=1\)
\(\Leftrightarrow3x=2x-6\Leftrightarrow x=-6\)
\(x^2-\frac{x^4}{x^2+1}-1\)
\(=\frac{x^2\left(x^2+1\right)}{x^2+1}-\frac{x^4}{x^2+1}-1\)
\(=\frac{x^4+x^2-x^4}{x^2+1}-1\)
\(=\frac{x^2}{x^2+1}-1\)
\(=\frac{x^2}{x^2+1}-\frac{x^2+1}{x^2+1}\)
\(=\frac{1}{x^2+1}\)
Bài làm
\(x^2-\frac{x^4}{x^2+1}-1\)
\(=\frac{x^2\left(x^2+1\right)}{x^2+1}-\frac{x^4}{x^2+1}-\frac{x^2+1}{x^2+1}\)
\(=\frac{x^4+x^2}{x^2+1}-\frac{x^4}{x^2+1}-\frac{x^2+1}{x^2+1}\)
\(=\frac{x^4+x^2-x^4-x^2-1}{x^2+1}\)
\(=\frac{-1}{x^2+1}\)