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\(R=\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)
\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
=\(\frac{2x\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\sqrt{x}+6x+5x+10\sqrt{x}+x+\sqrt{x}+10\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\sqrt{x}+12x+21\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
@@@@@@@@@@@ Đề sai hay mình sai??@@@@@@@@@@
\(\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}+1}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)
\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2\sqrt{x^3}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{12x+22\sqrt{x}+2\sqrt{x^3}+12}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)
\(=\frac{2\left(6x+11\sqrt{x}+\sqrt{x^3}+6\right)}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)
\(=2\) (ko phụ thuộc vào biến ) (đpcm)
\(=\dfrac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+11}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+12x+22\sqrt{x}+13}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
biểu thức này có phụ thuộc vào biến nha bạn
Bài 2:
a: ĐKXĐ: 2/3x-1/5>=0
=>2/3x>=1/5
hay x>=3/10
b: ĐKXĐ: \(\dfrac{x+1}{2x-3}>=0\)
=>2x-3>0 hoặc x+1<=0
=>x>3/2 hoặc x<=-1
c: ĐKXĐ: \(\left\{{}\begin{matrix}3x-5>=0\\x-4>0\end{matrix}\right.\Leftrightarrow x>4\)
a: \(A=\dfrac{5\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-5\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{5\sqrt{x}-11-3x-7\sqrt{x}+6-2x+5\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+3\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
b: Q-2/3
\(=\dfrac{-15x+9\sqrt{x}-24-2\left(x+2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-17x+5\sqrt{x}-18}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}< 0\)
=>Q<2/3