Giải phương trình
\(\sqrt{2x-1}-2\sqrt{x-1}=-1\)
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LT
0
18 tháng 5 2021
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
18 tháng 5 2021
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
NT
0
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2022
ĐKXĐ: \(-\dfrac{1}{2}\le x\le\dfrac{1}{2}\)
\(\sqrt{1-2x}+\sqrt{1+2x}=2-x^2\)
\(\Leftrightarrow2+2\sqrt{1-4x^2}=\left(2-x^2\right)^2\)
Đặt \(\sqrt{1-4x^2}=t\ge0\Rightarrow x^2=\dfrac{1-t^2}{4}\)
Pt trở thành:
\(2+2t=\left(2-\dfrac{1-t^2}{4}\right)^2\)
\(\Leftrightarrow\left(t^2+7\right)^2=32\left(t+1\right)\)
\(\Leftrightarrow t^4+14t^2-32t+17=0\)
\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t+17\right)=0\)
\(\Leftrightarrow t=1\Rightarrow\sqrt{1-4x^2}=1\Rightarrow x=0\)
AH
Akai Haruma
Giáo viên
22 tháng 6 2021
Lời giải:
a. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$
$\Leftrightarrow \sqrt{x-4}+2=2$
$\Leftrightarrow \sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
b. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Rightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
c.
PT \(\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x(x-1)=0\end{matrix}\right.\Rightarrow x=1\)
HM
0
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2021
a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)
\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)
\(\Leftrightarrow x^2+2x=\dfrac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)
\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)
\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2021
b. ĐKXĐ: \(x\ge5\)
\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)
\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
MA
0
Đk: x \(\ge\)1
Ta có: \(\sqrt{2x-1}-2\sqrt{x-1}=-1\)
<=> \(\sqrt{2x-1}+1=2\sqrt{x-1}\)
<=> \(2x-1+1+2\sqrt{2x-1}=4x-4\)
<=> \(2\sqrt{2x-1}=2x-4\)
<=> \(\sqrt{2x-1}=x-2\)(x \(\ge\)2)
<=> \(2x-1=x^2-4x+4\)
<=> \(x^2-6x+5=0\)
<=> \(\left(x-1\right)\left(x-5\right)=0\)
<=> \(\orbr{\begin{cases}x=1\left(ktm\right)\\x=5\left(tm\right)\end{cases}}\)
vậy S = {5}
ĐK : x ≥ 1
\(\Leftrightarrow\left(\sqrt{2x-1}-3\right)-\left(2\sqrt{x-1}-4\right)=0\)
\(\Leftrightarrow\frac{2x-1-9}{\sqrt{2x-1}+3}-\frac{4x-4-16}{2\sqrt{x-1}+4}=0\)
\(\Leftrightarrow\frac{2\left(x-5\right)}{\sqrt{2x-1}+3}-\frac{4\left(x-5\right)}{2\sqrt{x-1}+4}=0\)
\(\Leftrightarrow2\left(x-5\right)\left(\frac{1}{\sqrt{2x-1}+3}-\frac{2}{2\sqrt{x-1}+4}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\\frac{1}{\sqrt{2x-1}+3}-\frac{2}{2\sqrt{x-1}+4}=0\left(1\right)\end{cases}}\)( t lười giải pt 1 quá ông tự làm nhé =)) )