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) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)

Câu 3: đề là \(\sqrt{x+5}-\sqrt{x-2}\) hay \(\sqrt{x+5}-\sqrt{x+2}\)?

Câu 4:

ĐKXĐ: \(x\le9\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x-4}=a\\\sqrt{9-x}=b\end{matrix}\right.\) ta có hệ:

\(\left\{{}\begin{matrix}a-b=-1\\a^3+b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=a+1\\a^3+b^2=5\end{matrix}\right.\)

\(\Rightarrow a^3+\left(a+1\right)^2=5\)

\(\Leftrightarrow a^3+a^2+2a-4=0\) \(\Rightarrow a=1\)

\(\Rightarrow\sqrt[3]{x-4}=1\Rightarrow x-4=1\Rightarrow x=5\)

5.

ĐKXĐ: \(x\ge-\frac{17}{16}\)

\(\Leftrightarrow8x^2-15x-23-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left(x+1\right)\left(8x-23\right)-\left(x+1\right)\sqrt{16x+17}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8x-23=\sqrt{16x+17}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow16x+17-2\sqrt{16x+17}-63=0\)

Đặt \(\sqrt{16x+17}=t\ge0\)

\(\Rightarrow t^2-2t-63=0\Rightarrow\left[{}\begin{matrix}t=9\\t=-7\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{16x+17}=9\Leftrightarrow x=\frac{32}{3}\)

mình cần phần 3 4 5 nữa thui ạ

a/ ĐKXĐ: \(x\ge1\)

Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm

b/ \(x\ge1\)

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

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

Đặt \(\sqrt{x-1}=a\ge0\) ta được:

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

\(\Leftrightarrow\left[{}\begin{matrix}a=0\Rightarrow x=1\\\sqrt{\left(a-1\right)^2}+\sqrt{\left(a-2\right)^2}=\sqrt{a}\left(1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)

- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)

- Với \(0\le a\le1\) ta được:

\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

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

- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)

c/ ĐKXĐ: \(x\ge\frac{49}{14}\)

\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)

Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(7-\sqrt{14x-49}\ge0\)

\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)

Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)

d/ ĐKXĐ: \(x\ge\frac{1}{2}\)

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

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

TH1: \(\sqrt{2x-1}\ge3\Rightarrow x\ge5\)

\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=4\)

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

\(\Leftrightarrow x=13\)

TH2: \(2\le\sqrt{2x-1}< 3\Rightarrow\frac{5}{2}\le x< 5\)

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

\(\Leftrightarrow\sqrt{2x-1}=2\Rightarrow x=\frac{5}{2}\)

TH3: \(1\le\sqrt{2x-1}< 2\Rightarrow1\le x< \frac{5}{2}\)

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

\(\Leftrightarrow4=4\) (luôn đúng)

TH4: \(\frac{1}{2}\le x< 1\)

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

\(\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(l\right)\)

Vậy nghiệm của pt là: \(\left[{}\begin{matrix}1\le x\le\frac{5}{2}\\x=13\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

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

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

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

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

a: Đặt \(x^2-4=a\)

Pt sẽ là \(a=3\sqrt{xa}\)

\(\Rightarrow a^2=9xa\)

\(\Leftrightarrow a\left(a-9x\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)

hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)

d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)

Pt sẽ là 2a+b=ab+2

=>(b-2)(1-a)=0

=>b=2 và 1-a

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

2. \(\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}=\dfrac{7}{\sqrt{x-3}}\) (2)

\(\Leftrightarrow\dfrac{\sqrt{x^2}-16}{\sqrt{x-3}}+\sqrt{x+3}-\dfrac{7}{\sqrt{x-3}}=0\)

\(\Leftrightarrow\dfrac{\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7}{\sqrt{x-3}}=0\)

\(\Leftrightarrow\sqrt{x^2}-16+\sqrt{\left(x-3\right)\left(x+3\right)}-7=0\)

\(\Leftrightarrow\left|x\right|-16+\sqrt{x^2-9}-7=0\)

\(\Leftrightarrow\left|x\right|-23+\sqrt{x^2-9}=0\)

\(\Leftrightarrow\sqrt{x^2-9}=-\left|x\right|+23\)

\(\Leftrightarrow x^2-9=-\left(-\left|x\right|+23\right)^2\)

\(\Leftrightarrow x^2-9=-\left(-\left|x\right|\right)^2-46\cdot\left|x\right|+529\)

\(\Leftrightarrow x^2-9=\left|x\right|^2-46+\left|x\right|+529\)

\(\Leftrightarrow x^2-9=x^2-46\cdot\left|x\right|+529\)

\(\Leftrightarrow-9=-46\cdot\left|x\right|+529\)

\(\Leftrightarrow46\cdot\left|x\right|=529+9\)

\(\Leftrightarrow49\cdot\left|x\right|=538\)

\(\Leftrightarrow\left|x\right|=\dfrac{269}{23}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{269}{23}\\x=-\dfrac{269}{23}\end{matrix}\right.\)

Sau khi dùng phép thử ta nhận thấy \(x\ne-\dfrac{269}{23}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{269}{23}\right\}\)

3. sửa đề: \(\sqrt{14-x}=\sqrt{x-4}\sqrt{x-1}\) (3)

\(\Leftrightarrow\sqrt{14-x}=\sqrt{\left(x-4\right)\left(x-1\right)}\)

\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-x-4x+4}\)

\(\Leftrightarrow\sqrt{14-x}=\sqrt{x^2-5x+4}\)

\(\Leftrightarrow14-x=x^2-5x+4\)

\(\Leftrightarrow14-x-x^2+5x-4=0\)

\(\Leftrightarrow10+4x-x^2=0\)

\(\Leftrightarrow-x^2+4x+10=0\)

\(\Leftrightarrow x^2-4x-10=0\)

\(\Leftrightarrow x=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4\cdot1\cdot\left(-10\right)}}{2\cdot1}\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{16+40}}{2}\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{56}}{2}\)

\(\Leftrightarrow x=\dfrac{4\pm2\sqrt{14}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-2\sqrt{14}}{2}\\x=\dfrac{4+2\sqrt{14}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{14}\\x=2-\sqrt{14}\end{matrix}\right.\)

sau khi dùng phép thử ta nhận thấy \(x\ne2-\sqrt{14}\)

Vậy tập nghiệm phương trình (3) là \(S=\left\{2+\sqrt{14}\right\}\)

a.

ĐKXĐ: \(x\ge1\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

c.

ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)

\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)

\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)

Dấu "=" xảy ra khi và chỉ khi \(x=-1\)

d.

ĐKXĐ: \(x>1\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x-1}=b>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2-1}{a}=\dfrac{1-b^2}{b}\)

\(\Leftrightarrow a-\dfrac{1}{a}=\dfrac{1}{b}-b\)

\(\Leftrightarrow a+b-\dfrac{a+b}{ab}=0\)

\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=0\)

\(\Leftrightarrow1-\dfrac{1}{ab}=0\)

\(\Leftrightarrow ab=1\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=1\)

\(\Leftrightarrow x^3-1=1\)

\(\Leftrightarrow x=\sqrt[3]{2}\)