Cho biểu thức \(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
a) Rút gon
b) Tìm giá trị nhỏ nhất của C
c) Tìm x để \(\dfrac{2\sqrt{x}}{C}\) là số nguyên
K
Khách
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6 tháng 1 2022
a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
DA
1 tháng 1 2022
a) Điều kiện: \(x\ge0;x\ne1;x\ne\dfrac{1}{4}\)\(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt[]{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right).\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{2x\sqrt{x}+x-\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{x\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{x\sqrt{x}-2\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(E=\dfrac{x\sqrt{x}-2\sqrt{x}+x\sqrt{x}+x+\sqrt{x}}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{2x\sqrt{x}-\sqrt{x}+x}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)
\(E=\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\)
b)Vì \(x\ge0\) nên \(x+\sqrt{x}\ge0\) và \(x+\sqrt{x}+1>0\)
Do đó: \(E\ge0\). Dấu "=" xảy ra \(\Leftrightarrow x=0\)
c)\(E\ge\dfrac{6}{7}\Leftrightarrow\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\ge\dfrac{6}{7}\Leftrightarrow7x+7\sqrt{x}\ge6x+6\sqrt{x}+6\)
\(\Leftrightarrow x+\sqrt{x}-6\ge0\Leftrightarrow x-2\sqrt{x}+3\sqrt{x}-6\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\Leftrightarrow\sqrt{x}\ge2\Leftrightarrow x\ge4\)
18 tháng 10 2021
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
15 tháng 12 2022
a: \(C=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để \(C=\sqrt{x}\) thì \(x-\sqrt{x}=\sqrt{x}+1\)
=>\(x-2\sqrt{x}-1=0\)
=>\(\Leftrightarrow x=3+2\sqrt{2}\)
c: |2x-5|=3
=>2x-5=3 hoặc 2x-5=-3
=>2x=2 hoặc 2x=8
=>x=4(nhận) hoặc x=1(loại)
Khi x=4 thì \(C=\dfrac{2+1}{2-1}=3\)
30 tháng 8 2021
a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
AH
Akai Haruma
Giáo viên
5 tháng 11 2023
Lời giải:
a.
\(A=\frac{\sqrt{x}(\sqrt{x^3}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}+\frac{2(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-(2\sqrt{x}+1)+2(\sqrt{x}+1)\)
\(=\sqrt{x}(\sqrt{x}-1)-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)
b.
$A=x-\sqrt{x}+1=(x-\sqrt{x}+\frac{1}{4})+\frac{3}{4}$
$=(\sqrt{x}-\frac{1}{2})^2+\frac{3}{4}\geq 0+\frac{3}{4}=\frac{3}{4}$
$\Rightarrow A_{\min}=\frac{3}{4}$
Giá trị này đạt tại $\sqrt{x}-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{4}$
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 5 2021
ĐKXĐ: \(x>0;x\ne1\)
\(P=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
b.
\(P=x-\sqrt{x}+1=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(P_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)
MM
3 tháng 5 2021
a) đk: \(\left\{{}\begin{matrix}\sqrt{x}+1>0\\\sqrt{x}-1>0\\x>0\end{matrix}\right.=>\sqrt{x}>\pm1\)
rút gọn pt
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\) \(\dfrac{\left(x^2-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2x+\sqrt{x}\right)\left(\sqrt{x}-1\right)\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}.\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2\left(x-1\right)x\left(x+1\right)}{x\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\)