1) ĐKXĐ: \(x\ge-2\)

\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)

\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

2) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

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

3) \(pt\Leftrightarrow\sqrt{\left(5x+2\right)^2}=1\Leftrightarrow\left|5x+2\right|=1\)

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

4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)

\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)

5) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)

\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)

Vậy \(S=\varnothing\)

a: \(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a) \(\sqrt{\left(2x-3\right)^2}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)

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

b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)

\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)

\(\Leftrightarrow5\sqrt{x+2}=20\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

a. \(\sqrt{\left(2x-3\right)^2}=7\)

<=> \(\left|2x-3\right|=7\)

<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)

b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\)  ĐK: \(x\ge-2\)

<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)

<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)

<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)

<=> \(5\sqrt{x+2}=20\)

<=> \(\sqrt{x+2}=4\)

<=> \(\left(\sqrt{x+2}\right)^2=4^2\)

<=> \(\left|x+2\right|=16\)

<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)

c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)             ĐK: \(x\ge3\)

<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)

<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)

<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)

a) I2x-3I=7

        2x-3=7     =>x=5

        2x-3=-7     =>x=-2

b) \(8\sqrt{3x}-5\sqrt{3x}+2\sqrt{3x}=20\)

   5\(\sqrt{3x}=20\)

   3x=16  =>x=16/3

c)  vì câu c dài nên mình chỉ cho đáp án thôi là 0,3,6

vì \(\sqrt{ }\) của 1 số luôn dương nên 3,6 thỏa mãn 

1) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x=25\Leftrightarrow x=\dfrac{25}{2}\left(tm\right)\)

2) \(=\sqrt{\dfrac{\dfrac{1}{4}}{9}}=\dfrac{\dfrac{1}{2}}{3}=\dfrac{1}{6}\)

3) \(=\sqrt{225a^2}=15a\left(do.a\ge0\right)\)

4) \(=2y^2.\dfrac{x^2}{2\left|y\right|}=\left[{}\begin{matrix}x^2y\left(y>0\right)\\-x^2y\left(y< 0\right)\end{matrix}\right.\)

Bài 1:

a)\(Q=2x-\sqrt{x^2+2x+1}=2x-\sqrt{\left(x+1\right)^2}=2x-\left|x+1\right|\)

b)Tại x=7 thay vào Q ta được:

\(Q=2.7-\left|7+1\right|=14-8=6\)

Bài 2:

\(\sqrt{x^2-6x}+7x=13\)\(\Leftrightarrow\sqrt{x^2-6x}=13-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-7x\ge0\\x^2-6x=\left(13-7x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\0=48x^2-85x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{13}{7}\\\Delta=\left(-85\right)^2-4.48.169=-25223< 0\end{matrix}\right.\)

\(\Rightarrow x\in\varnothing\)

Vậy pt vô nghiệm.

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

\(=\left|x+1\right|+\left|x-1\right|=\left|x+1\right|+\left|1-x\right|\)

Áp dụng bđt \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có:

\(A\ge\left|x+1+1-x\right|=2\)

Vậy GTNN của A là 2 khi \(-1\le x\le1\)

Ta có 

\(A=\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}\)

\(\Rightarrow A=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-1\right)^2}\)

\(\Rightarrow A=\left|x+1\right|+\left|x-1\right|\)

\(\Rightarrow A=\left|x+1\right|+\left|1-x\right|\)

Vì \(\begin{cases}\left|x+1\right|\ge x+1\\\left|1-x\right|\ge1-x\end{cases}\)\(\Rightarrow\left|x+1\right|+\left|1-x\right|\ge x+1+1-x\)

\(\Rightarrow\left|x+1\right|+\left|1-x\right|\ge2\)

Dấu " = " xảy ra khi \(\begin{cases}x+1\ge0\\1-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge-1\\x\le1\end{cases}\)

Vậy MIN_A=2 khi \(-1\le x\le1\)

\(A=\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}\)

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

 -Nêú \(x\ge1\)thì \(\sqrt{\left(x+1\right)^2}=x+1\)và\(\sqrt{\left(x-1\right)^2}=x-1\)

Ta có:\(A=x+1+x-1=2x\ge2\)

Dấu "=" xảy ra khi x=1

-Nếu\(1>x\ge-1\)thì \(\sqrt{\left(x+1\right)^2}=x+1\)và\(\sqrt{\left(x-1\right)^2}=1-x\)

Ta có:\(A=x+1+1-x=2\)

-Nếu x<-1 thì \(\sqrt{\left(x+1\right)^2}=-x-1\)và\(\sqrt{\left(x-1\right)^2}=1-x\)

Ta có:\(A=-x-1+1-x=-2x\ge2\)

Dấu "=" xảy ra khi x=-1

Vậy GTNN của A là 2 tại x=1 hoặc x=-1

Ta có  x – 2√x + 3 = (√x – 1)² + 2.  Mà (√x – 1)² ≥ 0 với mọi x ≥ 0 ⇒ (√x – 1)² + 2 ≥ 2 với mọi x ≥ 0

⇒ \(A=\frac{1}{\left(\sqrt{X}-1\right)^2+2}\le\frac{1}{2}\)

Vậy GTLN của A = 1/2  ⇔ √x = 1 ⇔ x =1

ủNG HỘ MK NHAK

P=A*B

\(=\dfrac{x-7}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{x-7}{\sqrt{x}+2}\)

P nguyên

=>x-4-3 chia hết cho căn x+2

=>căn x+2 thuộc Ư(-3)

=>căn x+2=3

=>x=1