\(\dfrac{x}{\sqrt{4x-1}}+\dfrac{\sqrt{4x-1}}{x}=2\)
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11 tháng 4 2023
1: ĐKXĐ: x>1/2
=>\(\dfrac{x}{\sqrt{2x-1}}+\dfrac{x}{\sqrt[4]{4x-3}}=2\)
x^2-2x+1>=0
=>x^2>=2x-1
=>\(\dfrac{x}{\sqrt{2x-1}}>=1\)
Dấu = xảy ra khi x=1
(x^2-2x+1)(x^2+2x+3)>=0
=>x^4-4x+3>=0
=>x^4>=4x-3
=>\(\dfrac{x}{\sqrt[4]{4x-3}}>=1\)
=>VT>=2
Dấu = xảy ra khi x=1
2: 4x-1=x+x+2x-1
5x-2=x+2x-1+2x-1
\(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}\right)\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)>=9\)
=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{9}{\sqrt{x}+\sqrt{x}+\sqrt{2x-1}}\)
\(\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)^2< =3\left(4x-1\right)\)
=>\(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}< =\sqrt{3\left(4x-1\right)}\)
=>\(\dfrac{2}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{4x-1}}\)
Tương tự, ta cũng có: \(\dfrac{1}{\sqrt{x}}+\dfrac{2}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{5x-2}}\)
=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\sqrt{3}\left(\dfrac{1}{\sqrt{4x-1}}+\dfrac{1}{\sqrt{5x-2}}\right)\)
Dấu = xảy ra khi x=1
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
8 tháng 8 2021
\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(A=1-\dfrac{2\left(2\sqrt{x}-1\right)-5\sqrt{x}+\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}+1\right)^2}\)
\(A=1-\dfrac{4\sqrt{x}-2-5\sqrt{x}+2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(A=1-\dfrac{2\sqrt{x}+1}{2\sqrt{x}-1}=\dfrac{2\sqrt{x}-1-2\sqrt{x}-1}{2\sqrt{x}-1}=\dfrac{-2}{2\sqrt{x}-1}\)
Tick hộ nha
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
7 tháng 10 2021
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
8 tháng 10 2021
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
13 tháng 3 2021
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt{8x^2+5x+2}}=\dfrac{1+\sqrt{1+\dfrac{2}{x^2}}}{\sqrt{8+\dfrac{5}{x}+\dfrac{2}{x^2}}}=\dfrac{1+\sqrt{1}}{\sqrt{8}}=\dfrac{\sqrt{2}}{2}\).
13 tháng 3 2021
Thiếu \(\lim\limits_{x\rightarrow-\infty}\) ở sau dấu bằng thứ nhất nha
Q
20 tháng 6 2017
\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(=1-\dfrac{2\left(4x-1\right)-\left(1-2\sqrt{x}\right)-5\sqrt{x}\cdot\left(1+2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)-\left(1-2\sqrt{x}\right)\cdot\left(4x-1\right)}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{4x-4x\sqrt{x}-1+\sqrt{x}}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{4x\cdot\left(1-\sqrt{x}\right)-\left(1-\sqrt{x}\right)}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{\left(4x-1\right)\cdot\left(1-\sqrt{x}\right)}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{1-\sqrt{x}}{\left(1+2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{-\left(\sqrt{x}-1\right)}{\left(1+2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{-1}{\left(1-2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\left(4x+4\sqrt{x}+1\right)\)
\(=1+\dfrac{1}{1-4x}\cdot\left(4x+4\sqrt{x}+1\right)\)
\(=1+\dfrac{4x+4\sqrt{x}+1}{1-4x}\)
\(=\dfrac{1-4x+4x+4\sqrt{x}+1}{1-4x}\)
\(=\dfrac{2+4\sqrt{x}}{1-4x}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 8 2021
ĐKXĐ: \(x>\dfrac{1}{4}\)
Đặt \(\dfrac{x}{\sqrt{4x-1}}=t>0\)
\(\Rightarrow t+\dfrac{1}{t}=2\Rightarrow t^2-2t+1=0\)
\(\Rightarrow t=1\Rightarrow x=\sqrt{4x-1}\)
\(\Rightarrow x^2-4x+1=0\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
\(ĐKXĐ:x>\dfrac{1}{4}\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{x}{\sqrt{4x-1}}+\dfrac{\sqrt{4x-1}}{x}\ge2\sqrt{\dfrac{x}{\sqrt{4x-1}}.\dfrac{\sqrt{4x-1}}{x}}=2\)
\("="\Leftrightarrow\dfrac{x}{\sqrt{4x-1}}=\dfrac{\sqrt{4x-1}}{x}\Leftrightarrow x^2=4x-1\)
\(\Leftrightarrow x^2-4x+4-3=0\Leftrightarrow\left(x-2\right)^2-3=0\)
\(\Leftrightarrow\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2-\sqrt{3}\left(KTM\right)\\x=2+\sqrt{3}\left(TM\right)\end{matrix}\right.\)
KL.....
\(\dfrac{x}{\sqrt{4x-1}}+\dfrac{\sqrt{4x-1}}{x}=2\Leftrightarrow\dfrac{x^2+4x-1}{x\sqrt{4x-1}}=2\)