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Áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rút gọn rồi quy đồng làm nốt
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne2\end{cases}}\)
\(P=\left(1-\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\right).\left(1+\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right)\)
\(=\left(1-\sqrt{x}\right).\left(1+\sqrt{x}\right)=1-x\)
b. \(P\ge0\Rightarrow1-x\ge0\Rightarrow x\le1\)
Vậy với \(x\le1\)thì P có giá trị không âm
a) \(\left|x-2000\right|+\left|x-2002\right|=\left|x-2000\right|+\left|2002-x\right|\)
\(\ge\left|x-2000+2002-x\right|=2\) (1)
Dấu "=" \(\Leftrightarrow\left(x-2000\right)\left(2002-x\right)\ge0\)
\(\Leftrightarrow2000\le x\le2002\)
+ \(\left|x-2001\right|\ge0\forall x\). "=" \(\Leftrightarrow x=2001\) (2)
Từ (1) và (2) suy ra \(A\ge2\)
Dấu "=" \(\Leftrightarrow x=2001\)
b) \(B=\left|x-8\right|+\left|x-9\right|+\left|x-10\right|+\left|x+11\right|\)
+ \(\left|x-10\right|+\left|x+11\right|=\left|x+11\right|+\left|10-x\right|\)
\(\ge\left|x+11+10-x\right|=21\) (3)
Dấu "=" \(\Leftrightarrow\left(x+11\right)\left(10-x\right)\ge0\Leftrightarrow-11\le x\le10\)
+ \(\left|x-8\right|+\left|x-9\right|\ge\left|x-8+9-x\right|=1\) (4)
"=" \(\Leftrightarrow\left(x-8\right)\left(9-x\right)\ge0\Leftrightarrow8\le x\le9\)
Từ (3) và (4) suy ra \(B\ge22\)
"=" \(\Leftrightarrow8\le x\le9\)
a) ĐKXD : \(x\ge0;x\ne1\)
b)\(A=\left(1+\frac{\sqrt{x}}{x+1}\right):\frac{x\sqrt{x}-1}{\sqrt{x}-1}\)
\(A=\frac{\left(x+1+\sqrt{x}\right).\left(\sqrt{x}-1\right)}{\left(x+1\right).\left(x\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x^3}-1}{\left(x+1\right).\left(\sqrt{x^3}-1\right)}\)
\(A=\frac{1}{x+1}\)
c) \(A=\frac{1}{5}\Rightarrow\frac{1}{x+1}=\frac{1}{5}\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)
b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.
\(\text{ĐKXĐ: }x+1\ne0\text{ và }x-2001\ne0\)
\(\Leftrightarrow x\ne-1\text{ và }x\ne2001\)
\(\frac{\left(x^2-2000x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x^2+x-2001x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}\)
\(=\frac{\left[x.\left(x+1\right)-2001\left(x+1\right)\right].2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x-2001\right)\left(x+1\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{2001}{2002}\)