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Bài 1: 

a: Ta có: \(\sqrt{3x^2}=\sqrt{12}\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(\sqrt{\left(x-2\right)^2}=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

a: \(f\left(-3\right)=3\cdot9=27\)

\(f\left(2\sqrt{2}\right)=3\cdot8=24\)

\(f\left(1-2\sqrt{3}\right)=3\cdot\left(13-4\sqrt{3}\right)=39-12\sqrt{3}\)

b: Ta có: \(f\left(a\right)=12+6\sqrt{3}=\left(3+\sqrt{3}\right)^2=3\left(\sqrt{3}+1\right)^2\)

nên \(3x^2=3\left(\sqrt{3}+1\right)^2\)

hay \(x\in\left\{\sqrt{3}+1;-\sqrt{3}-1\right\}\)

AH
Akai Haruma
Giáo viên
26 tháng 2 2022

c.

$f(b)\geq 6b+12$

$\Leftrightarrow 3b^2\geq 6b+12$

$\Leftrightarrow b^2\geq 2b+4$

$\Leftrightarrow b^2-2b-4\geq 0$

$\Leftrightarrow (b-1-\sqrt{5})(b-1+\sqrt{5})\geq 0$

$\Leftrightarrow b\geq 1+\sqrt{5}$ hoặc $b\leq 1-\sqrt{5}$

AH
Akai Haruma
Giáo viên
29 tháng 11 2023

Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$

PT \(\Leftrightarrow 3(\sqrt{3x^2+1}-2)+2(\sqrt{3x+1}-2)+2(x-1)=0\)

\(\Leftrightarrow 3.\frac{3(x^2-1)}{\sqrt{3x^2+1}+2}+2.\frac{3(x-1)}{\sqrt{3x+1}+2}+2(x-1)=0\\ \Leftrightarrow (x-1)\left[\frac{3(x+1)}{\sqrt{3x^2+1}+2}+\frac{2}{\sqrt{3x+1}+2}+2\right]=0\)

Dễ thấy với $x\geq \frac{-1}{3}$ thì biểu thức trong ngoặc vuông luôn dương.

$\Rightarrow x-1=0$

$\Leftrightarrow x=1$ (tm)

 

a, \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)

\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+\frac{24\sqrt{x-1}}{8}=-17\)

\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Rightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)

\(\Rightarrow\sqrt{x-1}.-1=-17\)

\(\Rightarrow\sqrt{x-1}=17\)

\(\Rightarrow x-1=289\)

\(\Rightarrow x=290\)

b, \(3x-7\sqrt{x}+4=0\)

\(\Rightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)

\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)

\(\Rightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\3\sqrt{x}-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}\sqrt{x}=1\\3\sqrt{x}=4\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{16}{9}\end{cases}}}\)

c, \(-5x+7\sqrt{x}+12=0\)

\(\Rightarrow-5x-5\sqrt{x}+12\sqrt{x}+12=0\)

\(\Rightarrow-5\sqrt{x}\left(\sqrt{x}+1\right)+12\left(x+1\right)=0\)

\(\Rightarrow\left(\sqrt{x}+1\right)\left(-5\sqrt{x}+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\-5\sqrt{x}+12=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1VN\\-5\sqrt{x}=-12\end{cases}}\Rightarrow\orbr{\begin{cases}\\\sqrt{x}=\frac{12}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}\\x=\frac{144}{25}\end{cases}}}\)

9 tháng 7 2019

1) ĐK: \(x-1\ge0\Leftrightarrow x\ge1\)

pt \(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}.3\sqrt{x-1}+\frac{24}{8}\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=17^2=289\Leftrightarrow x=290\left(tm\right)\)

b) \(3x-7\sqrt{x}+4=0\)

ĐK: \(x\ge0\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\Leftrightarrow t^2=x\)

Ta có phương trình ẩn t: 

\(3t^2-7t+4=0\)( giải đen ta)

\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=\frac{4}{3}\end{cases}}\)

Với t=1 ta có: \(\sqrt{x}=1\Leftrightarrow x=1\) (tm)

Với t=4/3 ta có: \(\sqrt{x}=\frac{4}{3}\Leftrightarrow x=\frac{16}{9}\) (tm)

Câu c em làm tương tự  câu b nhé!

4 tháng 10 2021

c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

4 tháng 10 2021

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

1 tháng 7 2019

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)

\(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

1 tháng 7 2019

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

bạn làm nốt pần này nhá

24 tháng 11 2021

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)