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

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

\(\sqrt{5x+1}-\sqrt{1-4x}=3\sqrt{x}\)

Đk:\(0\le x\le\frac{1}{4}\)

\(pt\Leftrightarrow\sqrt{5x+1}-\left(2x+1\right)-\sqrt{1-4x}-\left(4x-1\right)=3\sqrt{x}-6x\)

\(\Leftrightarrow\frac{5x+1-\left(2x+1\right)^2}{\sqrt{5x+1}+\left(2x+1\right)}-\frac{1-4x-\left(4x-1\right)^2}{\sqrt{1-4x}+\left(4x-1\right)}=\frac{9x-36x^2}{3\sqrt{x}+6x}\)

\(\Leftrightarrow\frac{5x+1-4x^2-4x-1}{\sqrt{5x+1}+\left(2x+1\right)}-\frac{1-4x-16x^2+8x-1^2}{\sqrt{1-4x}+\left(4x-1\right)}-\frac{9x-36x^2}{3\sqrt{x}+6x}=0\)

\(\Leftrightarrow\frac{-x\left(4x-1\right)}{\sqrt{5x+1}+\left(2x+1\right)}-\frac{-4x\left(4x-1\right)}{\sqrt{1-4x}+\left(4x-1\right)}-\frac{-9x\left(4x-1\right)}{3\sqrt{x}+6x}=0\)

\(\Leftrightarrow x\left(4x-1\right)\left(\frac{-1}{\sqrt{5x+1}+\left(2x+1\right)}-\frac{-4}{\sqrt{1-4x}+\left(4x-1\right)}-\frac{-9}{3\sqrt{x}+6x}\right)=0\)

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

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

=>\(\sqrt{\left(x-2\right)^2}=7\)

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

=>\(\sqrt{x+3}=2\)

=>x+3=4

=>x=1(nhận)

ĐKXĐ: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)

b) ĐKXĐ: \(-1\le x\le3\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\\x\ne3\end{matrix}\right.\).

d) ĐKXĐ: \(x< \dfrac{3}{5}\).

1.Ta co:

\(\text{ }\sqrt{5x^2+10x+9}=\sqrt{5\left(x+1\right)^2+4}\ge2\)

\(\sqrt{2x^2+4x+3}=\sqrt{2\left(x+1\right)^2+1}\ge1\)

\(\Rightarrow A=\sqrt{5x^2+10x+9}+\sqrt{2x^2+4x+3}\ge2+1=3\)

Dau '=' xay ra khi \(x=-1\)

Vay \(A_{min}=3\)khi \(x=-1\)

2c.

\(DK:x\ge\frac{1}{2}\)

\(\Leftrightarrow\text{ }2x+1+\sqrt{2x-1}=0\)

\(\Leftrightarrow2x-1+\sqrt{2x-1}+2=0\)

\(\Leftrightarrow\left(\sqrt{2x-1}+\frac{1}{2}\right)^2+\frac{7}{4}=0\)

Ma \(\left(\sqrt{2x-1}+\frac{1}{2}\right)^2+\frac{7}{4}>0\)

Vay PT vo nghiem

\(\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+2x+1-5x-1\right)\left(x+\sqrt{4x-3}\right)}{\left(x^2-4x+3\right)\left(x+1+\sqrt{5x+1}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{x\left(x-3\right)\left(x+\sqrt{4x-3}\right)}{\left(x-1\right)\left(x-3\right)\left(x+1+\sqrt{5x+1}\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{x\left(x+\sqrt{4x-3}\right)}{\left(x-1\right)\left(x+1+\sqrt{5x+1}\right)}=\dfrac{9}{8}\)

\(\sqrt{4\left(1-x\right)^2}-6=0\) 

<=> \(\left|2\left(1-x\right)\right|=6\)

TH1: x \(\ge\)1 Khi đó pt trở thành:

\(2\left(x-1\right)=6\)

<=> x - 1 = 3

<=> x = 4 (tm)

TH2: x < 1, khi đó pt trở thành:

2(1 - x) = 6

<=> 1 - x = 3

<=> x = -2(tm)

vậy S= {4; -2}

Trả lời:

\(\sqrt{4\left(1-x\right)^2}-6=0\)

\(\Leftrightarrow2.\left|1-x\right|=6\)

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

\(\Leftrightarrow\orbr{\begin{cases}1-x=3\\1-x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)

Vậy \(x=\left\{-2,4\right\}\)

\(\sqrt{4x^2+4x+1}=x+2\)\(\left(x\ge-2\right)\)

\(\Leftrightarrow4x^2+4x+1=\left(x+2\right)^2\)

\(\Leftrightarrow4x^2+4x+1=x^2+4x+4\)

\(\Leftrightarrow3x^2=3\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\left(TM\right)\\x=-1\left(TM\right)\end{cases}}\)

Vậy \(x=\left\{1,-1\right\}\)

\(\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{20-12\sqrt{5}+9}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\sqrt{3}-2\sqrt{5}+3}}\)

\(\sqrt{4x^2+4x+1}=x+2\) (Đk: x > = -2)

<=> \(\sqrt{\left(2x+1\right)^2}=x+2\)

<=>\(\left|2x+1\right|=x+2\)

<=> \(\orbr{\begin{cases}2x+1=x+2\left(đk:x\ge-\frac{1}{2}\right)\\-2x-1=x+2\left(đk:x\le-\frac{1}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\left(tm\right)\\x=-1\left(tm\right)\end{cases}}\)

Vậy S = {1; -1}

a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)

 PT <=> 2x - 1 = 5

<=> x = 3 ( TM )

Vậy ...

b, ĐKXĐ : \(x\ge5\)

PT <=> x - 5 = 9

<=> x = 14 ( TM )

Vậy ...

c, PT <=> \(\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy ...

d, PT<=> \(\left|x-3\right|=3-x\)

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

Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)

e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)

PT <=> 2x + 5 = 1 - x

<=> 3x = -4

<=> \(x=-\dfrac{4}{3}\left(TM\right)\)

Vậy ...

f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

PT <=> \(x^2-x=3-x\)

\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )

Vậy ...

a) \(\sqrt{2x-1}=\sqrt{5}\)          (x \(\ge\dfrac{1}{2}\))

<=> 2x - 1 = 5

<=> x = 3 (tmđk)

Vậy S = \(\left\{3\right\}\)

b) \(\sqrt{x-5}=3\)           (x\(\ge5\))

<=> x - 5 = 9

<=> x = 4 (ko tmđk)

Vậy x \(\in\varnothing\)

c) \(\sqrt{4x^2+4x+1}=6\)          (x \(\in R\))

<=> \(\sqrt{\left(2x+1\right)^2}=6\)

<=> |2x + 1| = 6

<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)

Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)

Điều kiện:`5-x>=0`

`<=>x<=5`

Bình phượng hai vế ta có:

`(2x-1)^2<=(5-x)^2`

`<=>(3x-6)(x+4)<=0`

`<=>(x-2)(x+4)<=0`

Để 1 tích <=0 thì 2 số trái dấu mà `x-2<x+4` 

`<=>` \(\begin{cases}x-2 \le 0\\x+4 \ge 0\\\end{cases}\)

`<=>-4<=x<=2`

kết hợp đk:`-4<=x<=2`

a) ĐKXĐ:

$\begin{cases}1-2x\ge 0\\3-4x\ge 0\end{cases}\\\Leftrightarrow \begin{cases}2x\le 1\\4x\le 3\end{cases}\\\Leftrightarrow \begin{cases}x\le \dfrac{1}{2}\\x\le \dfrac{3}{4}\end{cases}\\\Leftrightarrow x\le \dfrac{1}{2}$

b) ĐKXĐ:

$\begin{cases}1+x\ge 0\\-4x\ge 0\end{cases}\\\Leftrightarrow \begin{cases}x\ge -1\\x\le 0\end{cases}\\\Leftrightarrow-1\le x\le 0$