Tìm GTLN của biểu thức:
a. \(A=\dfrac{1}{x-\sqrt{x}+1}\)
b. \(B=\dfrac{2x-2\sqrt{x}+5}{x-\sqrt{x}+2}\)
K
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NT
0
10 tháng 7 2021
a) ĐKXĐ: \(\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x< -2\end{matrix}\right.\)
b) ĐKXĐ: \(-\sqrt{2}\le x\le\sqrt{2}\)
c) ĐKXĐ: \(x\ge1\)
11 tháng 10 2021
a: Ta có: \(A=\left(1-\dfrac{2\sqrt{x}-2}{x-1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{x-1}:\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}\)
13 tháng 8 2021
a,\(A=2\sqrt{x^2+x+\dfrac{1}{2}}=2\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{1}{4}}=2\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)
\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) dấu"=" xảy ra<=>x=-1/2
\(B=\sqrt{2\left(x^2-2x+\dfrac{5}{2}\right)}=\sqrt{2\left[x^2-2x+1+\dfrac{3}{2}\right]}\)
\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\) dấu"=" xảy ra<=>x=1
\(C=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\ge\dfrac{-2}{-\sqrt{2}}=\sqrt{2}\) dấu"=" xảy ra<=>x=1
\(D=x-2\sqrt{x+2}\ge-2\) dấu"=" xảy ra<=>x=-2
HP
22 tháng 10 2021
a.
A = \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right).\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{\left(x-2+\sqrt{x}\right)\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x+\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-\left(x\sqrt{x}+2x+x+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{x\sqrt{x}-x-2\sqrt{x}+2+x-x\sqrt{x}-2x-x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-3x-4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\left(3x+4\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{-\sqrt{x}\left(3\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}+2\sqrt{x}-2}\)
A = \(\dfrac{3\sqrt{x}+4}{x-\sqrt{x}-2}\)
AH
Akai Haruma
Giáo viên
1 tháng 4 2021
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1; x\neq 25$
a)
\(A=\frac{4\sqrt{x}}{\sqrt{x}-5}:\left[\frac{(\sqrt{x}-2)(\sqrt{x}+2)+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2}+\frac{5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\right]\)
\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\)
\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{4\sqrt{x}}{\sqrt{x}-5}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{4(\sqrt{x}+2)}{\sqrt{x}-5}\)
b) Tại $x=81$ thì $\sqrt{x}=9$.
Khi đó: $A=\frac{4(9+2)}{9-5}=11$
c) $A< 4\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}-5}< 1$
$\Leftrightarrow \frac{7}{\sqrt{x}-5}< 0\Leftrightarrow \sqrt{x}-5< 0$
$\Leftrightarrow 0\leq x< 25$. Kết hợp với ĐKXĐ suy ra: $0\leq x< 25; x\neq 1$
19 tháng 1 2023
a: \(B=\dfrac{1}{\sqrt{x}+1}\)
\(B-1=\dfrac{\sqrt{x}+1-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}>=0\)
=>B>=1
b: \(P=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(P\cdot\sqrt{x}+2x-\sqrt{x}=3x-2\sqrt{x-4}+3\)
=>\(x+\sqrt{x}+1+2x-\sqrt{x}=3x+3-2\sqrt{x-4}\)
=>\(-2\sqrt{x-4}+3=1\)
=>x-4=1
=>x=5