1) \(ĐK:x\in R\)

2) \(ĐK:x< 0\)

3) \(ĐK:x\in\varnothing\)

4) \(=\sqrt{\left(x+1\right)^2+2}\) 

\(ĐK:x\in R\)

5) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)

1) xài qui nạp để cm \(\sqrt{1^3+2^3+...+x^3}=1+2+3+...+x=\frac{x\left(x+1\right)}{2}\)

2) a) Vô nghiệm vì ĐKXĐ không tm

b) auto do 

2.

A=\(\sqrt{\sqrt{\left(\sqrt{16}-\sqrt{12}\right)^2}}-\sqrt{\sqrt{\left(\sqrt{16}+\sqrt{12}\right)^2}}\)

\(=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

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

\(=\sqrt{3}-1-\left(\sqrt{3}+1\right)\)

\(=\sqrt{3}-1-\sqrt{3}-1\)

\(=-2\)

B= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+\sqrt{1}\right)^2}}}\)

\(=\sqrt{5-2\sqrt{2+\sqrt{8}+1}}\)

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

\(=\sqrt{5-2\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}}\)

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

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

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

\(=\sqrt{2}-1\)

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)

Đạt được khi x = 9

2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)

\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)

\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)

Không có GTLN nhé

3/ Điều kiện xác định bạn tự làm nhé

\(\frac{16}{\sqrt{x}+3}=\frac{-8\sqrt{x}+5}{3\sqrt{x}+1}\)

\(\Leftrightarrow8x+67\sqrt{x}+1=0\)

Tới đây thì bạn xem như phương trình bậc 2 là giải tiếp được. Nhớ đối chiếu điều kiện để loại nghiệm

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

Lời giải:

a)

PT \(\Leftrightarrow \sqrt{(3x-1)^2}=\sqrt{(x+4)^2}\)

\(\Leftrightarrow |3x-1|=|x+4|\)

\(\Rightarrow \left[\begin{matrix} 3x-1=x+4\\ 3x-1=-(x+4)\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2.5\\ x=-0.75\end{matrix}\right.\)

Vậy........

b) ĐK: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=5\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=5\)

\(\Leftrightarrow |\sqrt{x-1}+2|+|\sqrt{x-1}-3|=5\)

Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:

\(|\sqrt{x-1}+2|+|\sqrt{x-1}-3|=|\sqrt{x-1}+2|+|3-\sqrt{x-1}|\geq |\sqrt{x-1}+2+3-\sqrt{x-1}|=5\)

Dấu "=" xảy ra khi \((\sqrt{x-1}+2)(3-\sqrt{x-1})\geq 0\)

\(\Leftrightarrow -2\leq \sqrt{x-1}\leq 3\)

\(\Leftrightarrow 1\leq x\leq 10\)

Vậy.........

a, \(\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-2\right)^2}=3\)

\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)

b, \(\sqrt{x^2-10x+25}=x+3\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+3\)

\(\Leftrightarrow x-5=x+3\Leftrightarrow0\ne8\)( vô nghiệm ) 

câu c nữa bạn!!!!!!!!!!

a) Đk: \(\forall x\in R\)

a) \(\sqrt{x^2-4x+4}=3\) <=> \(\sqrt{\left(x-2\right)^2}=3\) <=> \(\left|x-2\right|=3\)

<=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

Vậy S = {5; -1}

b) Đk: \(\forall x\in R\)

Ta có: \(\sqrt{x^2-10x+25}=x+3\)

<=> \(\sqrt{\left(x-5\right)^2}=x+3\)

<=> \(\left|x-5\right|=x+3\)

<=> \(\orbr{\begin{cases}x-5=x+3\\5-x=x+3\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=8\left(vl\right)\\2=2x\end{cases}}\) <=> x = 1

Vậy S = {1}

c)Đk: x \(\ge\)0

 \(\sqrt{x+1+2\sqrt{x}}-\sqrt{x+16-8\sqrt{x}}=3\)

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

<=> \(\left|\sqrt{x}+1\right|-\left|\sqrt{x}-4\right|=3\)

Do \(x\ge0\) => \(\sqrt{x}+1>0\)

<=> \(\sqrt{x}+1-\left|\sqrt{x}-4\right|=3\)

<=> \(\sqrt{x}-2=\left|\sqrt{x}-4\right|\)

<=> \(\orbr{\begin{cases}\sqrt{x}-2=\sqrt{x}-4\left(đk:x\ge16\right)\\\sqrt{x}-2=4-\sqrt{x}\left(đk:0\le x\le16\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=-2\left(vl\right)\\2\sqrt{x}=6\end{cases}}\) <=> \(x=9\)

Vậy S = {9}

a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)

\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)

Pt trở thành:

\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)

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

\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)

\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)

b.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)

Pt trở thành:

\(t=t^2-4-16\Leftrightarrow...\)