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Hướng dẫn trả lời:
ĐKXĐ: 0 < x ≠ 1.
Đặt √x = a (a > 0 và a ≠ 1)
Ta có:
(2+√xx+2√x+1−√x−2x−1).x√x+x−√x−1√x=[2+aa2+2a+1−a−2a2−1].a3+a2−a−1a=[(2+a)(a−1)−(a−2)(a+1)(a+1)(a2−1)].(a+1)(a2−1)a=2a(a+1)(a2−1).(a+1)(a2−1)a=2
DK: x>1
\(P=\dfrac{1}{x}\left(\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2+\left(\sqrt{x+1}-\sqrt{x-1}\right)^2}{\left(\sqrt{x+1}-\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}\right)\)
\(P=\dfrac{1}{x}\left(\dfrac{x+1+x-1+2\sqrt{x^2-1}+x+1+x-1-2\sqrt{x^2-1}}{x+1-x+1}\right)\)\(P=\dfrac{1}{x}\cdot\dfrac{4x}{2}\)
\(P=2\)
\(P=\dfrac{1}{x}\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}+\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\right)=\dfrac{1}{x}\left[\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2+\left(\sqrt{x+1}-\sqrt{x-1}\right)^2}{\sqrt{\left(x+1\right)}^2-\sqrt{\left(x-1\right)^2}}\right]=\dfrac{1}{x}\left(\dfrac{x+1+2\sqrt{\left(x-1\right)\left(x+1\right)}+x-1+x+1-2\sqrt{\left(x-1\right)\left(x+1\right)}+x-1}{x+1-x+1}\right)=\dfrac{1}{x}\cdot\dfrac{4x}{2}=\dfrac{1}{x}\cdot2x=2\)
=> Giá trị của biểu thức P không phụ thuộc vào biến
\(\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}+1}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)
\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2\sqrt{x^3}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{12x+22\sqrt{x}+2\sqrt{x^3}+12}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)
\(=\frac{2\left(6x+11\sqrt{x}+\sqrt{x^3}+6\right)}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)
\(=2\) (ko phụ thuộc vào biến ) (đpcm)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
\(=\dfrac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+11}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+12x+22\sqrt{x}+13}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
biểu thức này có phụ thuộc vào biến nha bạn
\(R=\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)
\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
=\(\frac{2x\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\sqrt{x}+6x+5x+10\sqrt{x}+x+\sqrt{x}+10\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\sqrt{x}+12x+21\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
@@@@@@@@@@@ Đề sai hay mình sai??@@@@@@@@@@
ủa sao lại ko phụ thuộc ? Thấy kì kì