CHO M=\(\frac{1}{\sqrt{X}}\)+\(\frac{\sqrt{X}}{\sqrt{X}+1}\)
N=\(\frac{\sqrt{X}}{X+\sqrt{X}}\)
a, RÚT GỌN S=M:N
b, TÌM GIÁ TRỊ NHỎ NHẤT CỦA S
CÁC BN ƠI GIÚP MK LÀM BÀI NÀY VS NHA!
MK CẢM ƠN CÁC BN TRƯỚC NHA!
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NN
18 tháng 10 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
\(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{x-9}\)
\(=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b) Ta có: \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\left|\sqrt{3}-1\right|}\)
\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)( thỏa mãn ĐKXĐ )
Thay \(x=1\)vào M ta được:
\(M=\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{1-3}=\frac{-3}{2}\)
c) \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)
Vì \(x\inℕ\)\(\Rightarrow\)Để M là số tự nhiên thì \(\frac{9}{\sqrt{x}-3}\inℕ\)
\(\Rightarrow9⋮\left(\sqrt{x}-3\right)\)\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\)(1)
Vì \(x\ge0\)\(\Rightarrow\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3\ge-3\)(2)
Từ (1) và (2) \(\Rightarrow\sqrt{x}-3\in\left\{-3;-1;1;3;9\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\)\(\Rightarrow x\in\left\{0;4;16;36;144\right\}\)( thỏa mãn ĐKXĐ )
Thử lại với \(x=4\)ta thấy M không là số tự nhiên
Vậy \(x\in\left\{0;16;36;144\right\}\)
S
22 tháng 8 2021
a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)
b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)
=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)
check giùm mik
20 tháng 4 2020
ĐK: x > 0
a) Rút gọn M
M = \(\frac{\sqrt{x}}{x+\sqrt{x}}:\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
= \(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\left(\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(\frac{1}{M}=\frac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2+1=3\)
=> M \(\le\)1/3
=> GTLN của M =1/ 3 khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=1\) thỏa mãn
Vậy max M = 1/3 tại x = 1
21 tháng 7 2017
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne-1\\x\ne1\end{cases}}\)
Ta có \(P=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)-1\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)-1\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}-1\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1=\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}-1}\)
b. Ta có \(P-\sqrt{x}=\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=\frac{x+2-x+\sqrt{x}}{\sqrt{x}-1}=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Để \(P-\sqrt{x}\in Z\Rightarrow\sqrt{x}-1\inƯ\left(3\right)\Rightarrow\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)
| \(\sqrt{x}-1\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(\sqrt{x}\) | -2 | 0 | 2 | 4 |
| x | 0 | 4 | 16 | |
| (l) | (n) | (n) | (n) |
Vậy \(x\in\left\{0;4;16\right\}\)thì \(P-\sqrt{x}\in Z\)
6 tháng 7 2017
a. ĐK \(x\ge0\)và \(x\ne1\)
A =\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\cdot\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}+1+\sqrt{x}-x-1+\sqrt{x}}\)
\(=\frac{x+1}{4\sqrt{x}}\)
b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)
c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)
Vậy A >1/2
