Bạn viết lại cái biểu thức được không?

a, Ta có : \(x=81\Rightarrow\sqrt{x}=9\)

Thay \(\sqrt{x}=9\)vào biểu thức A ta được : 

\(A=\frac{2}{9+1}=\frac{2}{10}=\frac{1}{5}\)

b, Ta có : \(P=\frac{B}{A}\)hay\(P=\frac{\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}}{\frac{2}{\sqrt{x}+1}}\)

\(=\frac{1+\sqrt{x}}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{2}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

c, Ta có \(\frac{1}{2}=\frac{\sqrt{x}}{2\sqrt{x}}\)mà \(\sqrt{x}< \sqrt{x}+1\)

nên \(P>\frac{1}{2}\)

a) \(A=\frac{2}{\sqrt{x}+1}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}\)

b) \(B=\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}+1}\)

\(=\frac{1+\sqrt{x}}{\left(1+\sqrt{x}\right)\sqrt{x}}=\frac{1}{\sqrt{x}}\)

\(\Rightarrow P=\frac{B}{A}=\frac{1}{\sqrt{x}}\div\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{2\sqrt{x}}\)

c) Ta có: \(P=\frac{\sqrt{x}+1}{2\sqrt{x}}=\frac{1}{2}+\frac{1}{\sqrt{x}}+\frac{1}{2}+0=\frac{1}{2}\)

=> P>1/2

a)Thay $x=81(\mathrm{t}\mathrm{m}dk)$ vào biều thức $A$, ta được \mathrm{A}=\frac{2}{\sqrt{81}+1}=\frac{2}{9+1}=\frac{1}{5}A=81​+12​=9+12​=51​
Vậy $x=81$ thì \mathrm{A}=\frac{1}{5}A=51​

b) B=\frac{1}{\sqrt{x}(\sqrt{x}+1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}B=x​(x​+1)1​+x​(x​+1)x​​
B=\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}B=x​(x​+1)x​+1​
B=\frac{1}{\sqrt{x}}B=x​1​
P=\frac{1}{\sqrt{x}} \div \frac{2}{\sqrt{x}+1}P=x​1​÷x​+12​
P=\frac{1}{\sqrt{x}} \cdot \frac{\sqrt{x}+1}{2}P=x​1​⋅2x​+1​
P=\frac{\sqrt{x}+1}{2 \sqrt{x}}P=2x​x​+1​

c) Ta có P-\frac{1}{2}=\frac{\sqrt{x}+1}{2 \sqrt{x}}-\frac{1}{2}=\ldots=\frac{1}{2 \sqrt{x}}P−21​=2x​x​+1​−21​=…=2x​1​
Ta có $x>0$ nên \frac{1}{2 \sqrt{x}}>0 \Rightarrow P-\frac{1}{2}>0 \Rightarrow P>\frac{1}{2}2x​1​>0⇒P−21​>0⇒P>21​
 

Câu 1: D

Câu 2: A

\(A\ge2\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}=2\left(B\right)\)

Chọn B

a) \(M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

b) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\)

c) \(M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{\sqrt{3-2\sqrt{2}}-1}{\sqrt{3-2\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{\sqrt{\left(\sqrt{2}-1\right)^2}}=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}\)

\(a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=1-\dfrac{1}{\sqrt{a}}< 1\\ c,a=3-2\sqrt{2}\Leftrightarrow\sqrt{a}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\\ \Leftrightarrow M=\dfrac{\sqrt{2}-1-1}{\sqrt{2}-1}=\dfrac{\sqrt{2}-2}{\sqrt{2}-1}=\dfrac{-\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=-\sqrt{2}\)

\(a,=\sqrt{6,25-0,49}=\sqrt{5,76}=2,4\)

\(b,=\sqrt{2,5-0,49}=\sqrt{2,01}\)

\(c,\sqrt{5,76}=2,4\)

Bài 1: Bạn đã post 1 lần

Bài 2:

\(C=\sqrt{(x-3)-2\sqrt{x-3}+1}-\sqrt{(x-3)-4\sqrt{x-3}+4}\)

\(=\sqrt{(\sqrt{x-3}-1)^2}-\sqrt{(\sqrt{x-3}-2)^2}\)

\(=|\sqrt{x-3}-1|-|\sqrt{x-3}-2|\)

Áp dụng BĐT dạng $|a|-|b|\leq |a-b|(*)$ thì:

$C\leq |\sqrt{x-3}-1-(\sqrt{x-3}-2)|$ hay $C\leq 1$

Vậy $C_{\max}=1$

Mặt khác, vẫn áp dụng BĐT $(*)$:

\(|\sqrt{x-3}-1|=|(\sqrt{x-3}-2-(-1)|\geq |\sqrt{x-3}-2|-|-1|\)

\(=|\sqrt{x-3}-2|-1\Rightarrow C\geq -1\)

Vậy $C_{\min}=-1$

a: \(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{a-1}\cdot\dfrac{1}{a\sqrt{a}}\)

\(=\dfrac{4\sqrt{a}\left(a-1+1\right)}{a-1}\cdot\dfrac{1}{a\sqrt{a}}=\dfrac{4}{a-1}\)

b: Khi a=2căn 2+1 thì \(A=\dfrac{4}{2\sqrt{2}+1-1}=\sqrt{2}\)

a)6,6
b)3,5

a) \(\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1.0,09}=\sqrt{\frac{1089}{25}}=\frac{33}{5}\)

b) \(\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{3,5.2,5.7.\frac{1}{5}}=\sqrt{\frac{49}{4}}=\frac{7}{2}\)

\(a,\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1.0,09}=\sqrt{43,56}=6,6\)

\(b,\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{3,5.2,5.7.\frac{1}{5}}=\sqrt{12,25}=3,5\)

a: \(E=1+1=2\)

b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)

\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)

d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)