Tìm x để bt nguyê
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HP
18 tháng 12 2021
a, ĐK: \(x\ge0,x\ne1\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+1+2\sqrt{x}+x+1-2\sqrt{x}-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
HP
31 tháng 8 2021
a, ĐK: \(x\ge0;x\ne9\)
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=-\dfrac{3}{\sqrt{x}-3}\)
HP
31 tháng 8 2021
b, \(P>0\Leftrightarrow-\dfrac{3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3>0\)
\(\Leftrightarrow x>9\)
c, \(P=-\dfrac{3}{\sqrt{x}-3}\in Z\)
\(\Leftrightarrow\sqrt{x}-3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)
\(\Leftrightarrow x\in\left\{0;4;16;36\right\}\)
AH
Akai Haruma
Giáo viên
12 tháng 1 2022
Lời giải:
ĐKXĐ: $x\geq 0$
Áp dụng BĐT Cô-si: $x+1\geq 2\sqrt{x}$
$\Rightarrow \frac{3\sqrt{x}}{x+1}\leq \frac{3\sqrt{x}}{2\sqrt{x}}=\frac{3}{2}$
Vậy gtln của bt là $\frac{3}{2}$ khi $x=1$
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 5 2021
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(x+2\sqrt{x}\right).x.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}=\dfrac{x}{\sqrt{x}-1}\)
b.
\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)
c.
Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)
Ta có:
\(\sqrt{A}=\sqrt{\dfrac{x}{\sqrt{x}-1}}=\sqrt{\dfrac{x}{\sqrt{x}-1}-4+4}=\sqrt{\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge\sqrt{4}=2\)
Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)
S
1
19 tháng 7 2023
\(A=\dfrac{x+2}{\sqrt{x}}.\dfrac{2\sqrt{x}}{\sqrt{x}+1}\) \(ĐK:x\ge0\)
\(=\dfrac{2\left(x+2\right)}{\sqrt{x}+1}\)
\(A=4\Leftrightarrow\dfrac{2\left(x+2\right)}{\sqrt{x}+1}=4\Leftrightarrow2\left(x+2\right)=4\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow2x+4=4\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)
CC
1
24 tháng 8 2021
Để biểu thức nguyên thì \(3\sqrt{x}+1⋮2\sqrt{x}-1\)
\(\Leftrightarrow6\sqrt{x}+2⋮2\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}-1\in\left\{-1;1;5\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{0;2;6\right\}\)
hay \(x\in\left\{0;1;9\right\}\)
VE
0
27 tháng 10 2021
1) \(A=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2x-2\sqrt{x}\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(A=\dfrac{2\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{5}{4}\)
c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Rightarrow2\sqrt{x}-1< \sqrt{x}+1\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
27 tháng 10 2021
\(1,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ 2,x=9\Leftrightarrow A=\dfrac{6-1}{3+1}=\dfrac{5}{4}\\ 3,A< 1\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\Leftrightarrow\sqrt{x}-2< 0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)
23 tháng 8 2021
Bài 1:
Để biểu thức nhận giá trị nguyên thì \(3\sqrt{x}+1⋮2\sqrt{x}-1\)
\(\Leftrightarrow6\sqrt{x}+2⋮2\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}-1\in\left\{1;-1;5\right\}\)
\(\Leftrightarrow2\sqrt{x}\in\left\{2;0;6\right\}\)
hay \(x\in\left\{4;0;36\right\}\)
