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a)\(N=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
\(=\) \(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}+\frac{\sqrt{x+2}}{\left(\sqrt{x-2}\right)\left(\sqrt{x+2}\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x-2}\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
Thay x= 25 vào biểu thức N, ta được: \(N=\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{5-2}=\frac{5}{3}\)
b) Để \(N=\frac{-1}{3}\) thì:
\(\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{-1}{3}\)
\(\Leftrightarrow3\sqrt{x}=2-\sqrt{x}\)
\(\Leftrightarrow3\sqrt{x}+\sqrt{x}=2\)
\(\Leftrightarrow4\sqrt{x}=2\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\)
\(\Rightarrow x=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)
ĐKXĐ của cả A và B : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(B=\frac{x+3\sqrt{x}}{x-25}+\frac{1}{\sqrt{x}+5}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\frac{\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x+4\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{x-\sqrt{x}+5\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-5}\)
\(M=\frac{B}{A}=\frac{\frac{\sqrt{x}-1}{\sqrt{x}-5}}{\frac{\sqrt{x}+2}{\sqrt{x}-5}}=\frac{\sqrt{x}-1}{\sqrt{x}-5}\times\frac{\sqrt{x}-5}{\sqrt{x}+2}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
ĐKXĐ của M : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)
\(M\times\left(\sqrt{x}+2\right)\ge3x-3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}\times\left(\sqrt{x}+2\right)\ge3x-3\)( ĐK : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\))
\(\Leftrightarrow\sqrt{x}-1\ge3x-3\)
\(\Leftrightarrow3x-\sqrt{x}-3+1\ge0\)
\(\Leftrightarrow3x-\sqrt{x}-2\ge0\)
\(\Leftrightarrow3x-3\sqrt{x}+2\sqrt{x}-2\ge0\)
\(\Leftrightarrow3\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}+2\right)\ge0\)
Dễ dàng nhận thấy \(3\sqrt{x}+2\ge2>0\forall x\ge0\)
\(\Rightarrow\sqrt{x}-1\ge0\)
\(\Leftrightarrow x\ge1\)
Kết hợp với điều kiện => Với 0 ≤ x ≤ 1 thì thỏa mãn đề bài
Bài 1:
a) Ta có: \(P=\frac{x}{x+2}+\frac{x+3}{x-2}+\frac{6-9x}{4-x^2}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x+3\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{6-9x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x+x^2+5x+6-6+9x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x^2+12x}{\left(x-2\right)\left(x+2\right)}\)
b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Để P=3 thì \(\frac{2x^2+12x}{\left(x-2\right)\left(x+2\right)}=3\)
\(\Leftrightarrow2x^2+12x=3\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow2x^2+12x=3\left(x^2-4\right)\)
\(\Leftrightarrow2x^2+12x=3x^2-12\)
\(\Leftrightarrow2x^2+12x-3x^2+12=0\)
\(\Leftrightarrow-x^2+12x+12=0\)
\(\Leftrightarrow x^2-12x-12=0\)
\(\Leftrightarrow x^2-12x+36-24=0\)
\(\Leftrightarrow\left(x-6\right)^2=24\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=\sqrt{24}\\x-6=-\sqrt{24}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6+2\sqrt{6}\left(nhận\right)\\x=6-2\sqrt{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: khi P=3 thì \(x\in\left\{6+2\sqrt{6};6-2\sqrt{6}\right\}\)
Bài 2:
a) Ta có: \(B=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)
\(=\frac{2a^2}{\left(a+1\right)\left(a-1\right)}+\frac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}-\frac{a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(=\frac{2a^2+a^2-a-a^2-a}{\left(a+1\right)\cdot\left(a-1\right)}=\frac{2a^2-2a}{\left(a+1\right)\left(a-1\right)}\)
\(=\frac{2a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}=\frac{2a}{a+1}\)
b) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)
Để B là số nguyên thì \(2a⋮a+1\)
\(\Leftrightarrow2a+2-2⋮a+1\)
\(\Leftrightarrow-2⋮a+1\)
\(\Leftrightarrow a+1\inƯ\left(-2\right)\)
\(\Leftrightarrow a+1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{0;-2;1;-3\right\}\)
mà \(a\notin\left\{1;-1\right\}\)
nên \(a\in\left\{0;-2;-3\right\}\)
Vậy: khi B có giá trị nguyên thì \(a\in\left\{0;-2;-3\right\}\)
Bài 3:
Ta có: \(Q=\frac{4}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)
\(=\frac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x+2}{\left(x+2\right)\left(x-2\right)}=\frac{1}{x-2}\)
Bài 4:
a) Ta có: \(P=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)
\(=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\left(\frac{\sqrt{x}+2}{\sqrt{x}-2}+\frac{3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\right)\)
\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}+2+3\sqrt{x}-6}{\sqrt{x}-2}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\cdot4\cdot\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}-2\right)^2}\)
\(=\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Để P=-4 thì \(\frac{-16x+16\sqrt{x}}{\left(\sqrt{x}-2\right)^2}=-4\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4\left(\sqrt{x}-2\right)^2\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4\left(x-4\sqrt{x}+4\right)\)
\(\Leftrightarrow-16x+16\sqrt{x}=-4x+16\sqrt{x}-16\)
\(\Leftrightarrow-16x+16\sqrt{x}+4x-16\sqrt{x}+16=0\)
\(\Leftrightarrow-12x+16=0\)
\(\Leftrightarrow-12x=-16\)
hay \(x=\frac{4}{3}\)(nhận)
Vậy: Khi P=-4 thì \(x=\frac{4}{3}\)
a) \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)
Ta được điều phải chứng minh.
