\(3x^2-3x=\sqrt{\frac{4x+1}{12}}\)
\(ĐK:x\ge1\)
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10 tháng 7 2021
a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)
TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)
TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)
Vậy x=0,5...
d, đk \(x\ge-1\)
=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)
\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)
10 tháng 7 2021
a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow\left|3x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b) Ta có: \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)
\(\Leftrightarrow\left|x-3\right|=4-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
10 tháng 12 2021
1/ ĐKXĐ: $4x^2-4x-11\geq 0$
PT $\Leftrightarrow \sqrt{4x^2-4x-11}=2(4x^2-4x-11)-6$
$\Leftrightarrow a=2a^2-6$ (đặt $\sqrt{4x^2-4x-11}=a, a\geq 0$)
$\Leftrightarrow 2a^2-a-6=0$
$\Leftrightarrow (a-2)(2a+3)=0$
Vì $a\geq 0$ nên $a=2$
$\Leftrightarrow \sqrt{4x^2-4x-11}=2$
$\Leftrightarrow 4x^2-4x-11=4$
$\Leftrightarrow 4x^2-4x-15=0$
$\Leftrightarrow (2x-5)(2x+3)=0$
$\Rightarrow x=\frac{5}{2}$ hoặc $x=\frac{-3}{2}$ (tm)
AH
Akai Haruma
Giáo viên
10 tháng 12 2021
2/ ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{3x^2+9x+8}=\frac{1}{3}(3x^2+9x+8)-\frac{14}{3}$
$\Leftrightarrow a=\frac{1}{3}a^2-\frac{14}{3}$ (đặt $\sqrt{3x^2+9x+8}=a, a\geq 0$)
$\Leftrightarrow a^2-3a-14=0$
$\Rightarrow a=\frac{3+\sqrt{65}}{2}$ (do $a\geq 0$)
$\Leftrightarrow 3x^2+9x+8=\frac{37+3\sqrt{65}}{2}$
$\Rightarrow x=\frac{1}{2}(-3\pm \sqrt{23+2\sqrt{65}})$
22 tháng 1 2022
\(a,4x-6< 7x-12\)
\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)
\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)
\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)
\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)
\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)
\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)
\(\Leftrightarrow-16x\ge11\)
\(\Leftrightarrow x\le-\frac{11}{16}\)
\(d,-12-8x>3+2x-\left(5-7x\right)\)
\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)
\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)
\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)
\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)
BH
4
TN
20 tháng 9 2017
Đặt \(\hept{\begin{cases}a=\sqrt{4x+1}\\b=\sqrt{3x-2}\end{cases}\ge}0\) thì có:
\(\Rightarrow a^2-b^2=x+3\)\(\Rightarrow a-b=\frac{a^2-b^2}{5}\)
\(\Rightarrow a-b-\frac{\left(a-b\right)\left(a+b\right)}{5}=0\)
\(\Rightarrow\left(a-b\right)\left(1-\frac{a+b}{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=b\\a+b=5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\sqrt{4x+1}=\sqrt{3x-2}\\\sqrt{4x+1}+\sqrt{3x-2}=5\end{cases}}\)\(\Rightarrow x=2\)
NN
9
4 tháng 10 2016
ĐKXĐ: z>0
pt<=> \(\frac{x^3+3x^2\sqrt[3]{3x-2}-12x+\sqrt{x}-\sqrt{x}-8}{x}=0\)
<=> \(x^3+3x^2\sqrt[3]{3x+2}-12x-8=0\)
<=> \(3x^2\sqrt[3]{3x-2}-6x^2+x^3-6x^2+12x-8=0\)
<=> \(3x^2\left(\sqrt[3]{3x-2}-2\right)+\left(x-2\right)^3=0\)
<=> \(3x^2\cdot\frac{3x-2-8}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^3=0\)
<=> \(\left(x-2\right)\left(\frac{9x^2}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^2\right)=0\)
<=> \(x=2\)( vì cái trong ngoặc thứ 2 luôn dương vs mọi x>0)
vậy x=2
NH
0
NT
0
MT
1