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19 tháng 6 2021
Đk:\(x\ne0;x\ge-\dfrac{1}{3}\)
Pt \(\Leftrightarrow12x^2-3x-1=4x\sqrt{3x+1}\)
\(\Leftrightarrow16x^2=4x^2+4x\sqrt{3x+1}+3x+1\)
\(\Leftrightarrow16x^2=\left(2x+\sqrt{3x+1}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=2x+\sqrt{3x+1}\\4x=-\left(2x+\sqrt{3x+1}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{3x+1}\left(1\right)\\6x=-\sqrt{3x+1}\left(2\right)\end{matrix}\right.\)
TH1 \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=3x+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-1\right)\left(4x+1\right)=0\end{matrix}\right.\)\(\Rightarrow x=1\) (thỏa)
TH2\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\36x^2=3x+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{24}\\x=\dfrac{1-\sqrt{17}}{24}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=\dfrac{1-\sqrt{17}}{24}\)(tm)
Vậy...
AH
Akai Haruma
Giáo viên
19 tháng 6 2021
Lời giải:
ĐKXĐ: $x\ge \frac{-1}{3}; x\neq 0$
PT \(\Leftrightarrow 3(x-1)+\frac{x-1}{4x}=\sqrt{3x+1}-2\)
\(\Leftrightarrow 3(x-1)+\frac{x-1}{4x}=\frac{3(x-1)}{\sqrt{3x+1}+2}\)
\(\Leftrightarrow (x-1)(3+\frac{1}{4x}-\frac{3}{\sqrt{3x+1}+2})=0\)
Nếu $x-1=0\Leftrightarrow x=1$ (tm)
Nếu $3+\frac{1}{4x}-\frac{3}{\sqrt{3x+1}+2}=0$
$\Leftrightarrow 12x\sqrt{3x+1}+12x+\sqrt{3x+1}+2=0$
$\Leftrightarrow \sqrt{3x+1}(12x+1)=-(12x+2)$
Từ đây suy ra $x\leq \frac{-1}{6}$
Bình phương 2 vế:
$(3x+1)(12x+1)^2=[(12x+1)+1]^2$
$\Leftrightarrow 3x(12x+1)^2=2(12x+1)+1$
$\Leftrightarrow 144x^3+24x^2-7x-1=0$
$\Leftrightarrow (4x+1)(36x^2-3x-1)=0$
Vì $x\leq \frac{-1}{6}$ nên $x=\frac{1-\sqrt{17}}{24}$
17 tháng 3 2021
\(\left\{{}\begin{matrix}\dfrac{6}{3x-2}-2\sqrt{1-y}=1\\\dfrac{2}{3x-2}+\sqrt{1-y}=2\end{matrix}\right.\) (x \(\ne\) \(\dfrac{2}{3}\); y \(\le\) 1)
Đặt \(\dfrac{1}{3x-2}=a\); \(\sqrt{1-y}=b\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6a-2b=1\\2a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6a-2b=1\\4a+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}10a=5\\4a+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\4\cdot\dfrac{1}{2}+2b=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{1}{3x-2}=2\\\sqrt{1-y}=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2\left(3x-2\right)=1\\1-y=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{5}{6}\\y=0\end{matrix}\right.\) (TM)
Vậy ...
Chúc bn học tốt!
KC
1
AH
Akai Haruma
Giáo viên
16 tháng 5 2021
Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$
PT $\Leftrightarrow \frac{x}{\sqrt{x+2}}=\sqrt{3x+1}-\sqrt{x+1}$
$\Leftrightarrow \frac{x}{\sqrt{x+2}}=\frac{2x}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow x\left(\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}\right)=0$
Xét các TH:
TH1: $x=0$ (thỏa mãn)
TH2: $\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}$
$\Leftrightarrow \sqrt{3x+1}+\sqrt{x+1}=2\sqrt{x+2}$
$\Rightarrow 4x+2+2\sqrt{(3x+1)(x+1)}=4(x+2)$
$\Leftrightarrow \sqrt{(3x+1)(x+1)}=3$
$\Rightarrow (3x+1)(x+1)=9$
$\Leftrightarrow 3x^2+4x-8=0$
$\Rightarrow x=\frac{-2\pm 2\sqrt{7}}{3}$
Kết hợp với ĐKXĐ suy ra $x=\frac{-2+2\sqrt{7}}{3}$
Vậy............
CC
1
31 tháng 5 2018
Đặt \(\sqrt{3x+1}=a\)
\(\Rightarrow\frac{a^2-1}{\sqrt{a^2+9}}=a-1\)
\(\Leftrightarrow\left(a-1\right)\left(\frac{a+1}{\sqrt{a^2+9}}-1\right)=0\)
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 )
25 tháng 11 2023
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
