5. \(y=\dfrac{-3x}{x+2}\)

xác định khi: \(x+2\ne0\Leftrightarrow x\ne-2\)

vậy D= (\(-\infty;+\infty\))\{-2}

6. \(y=\sqrt{-2x-3}\)

xác định khi: \(-2x-3\ge0\Leftrightarrow x\le\dfrac{-3}{2}\)

vậy D= (\(-\infty;\dfrac{-3}{2}\)]

7. \(y=\dfrac{3-x}{\sqrt{x-4}}\)

xác định khi: x-4 >0 <=> x>4

vậy D= (\(4;+\infty\))

8. \(y=\dfrac{2x-5}{\left(3-x\right)\sqrt{5-x}}\)

xác định khi: \(\left\{{}\begin{matrix}3-x\ne0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x< 5\end{matrix}\right.\)

vậy D= (\(-\infty;5\))\ {3}

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

xác định khi: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{-1}{2}\le x\le\dfrac{4}{3}\)

vậy D= [\(\dfrac{-1}{2};\dfrac{4}{3}\)]

1. \(y=\dfrac{3x-2}{x^2-4x+3}\)

xác định khi : \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)

vậy tập xác định là: D = \(\left(-\infty;+\infty\right)\backslash\left\{3;1\right\}\)

2.\(y=2\sqrt{5-4x}\)

xác định khi \(5-4x\ge0\Leftrightarrow x\le\dfrac{5}{4}\)

vậy D= (\(-\infty;\dfrac{5}{4}\)]

3. \(y=\dfrac{2}{\sqrt{x+3}}+\sqrt{5-2x}\)

xác định khi: \(\left\{{}\begin{matrix}x+3>0\\5-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow-3< x\le\dfrac{5}{2}\)

vậy D= (\(-3;\dfrac{5}{2}\)]

4.\(\sqrt{9-x}+\dfrac{1}{\sqrt{x+2}-2}\)

xác định khi: \(\left\{{}\begin{matrix}9-x\ge0\\x+2\ge0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x\ge-2\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le9\\x\ne2\end{matrix}\right.\)

Vậy D= [\(-2;9\)]\{2}

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

=>\(-2\sqrt{3x}=-4\)

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

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

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

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

=>\(\sqrt{1-3x}=2\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

a) \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(=\frac{\sqrt{2}.\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)

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

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

\(=\frac{\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

ok  mk giải dk tối qua rồi , dù s cx thanks

a) \(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{\left(2x+3\right)\left(x+1\right)}-16\)

Đặt \(t=\sqrt{2x+3}+\sqrt{x+1}\left(t\ge0\right)\)

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

\(\Rightarrow2\sqrt{\left(2x+3\right)\left(x+1\right)}=t^2-3x-4\)

Pt <=> \(t=3x+t^2-3x-4-16\)

\(\Leftrightarrow t^2-t-20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5\)

\(\Leftrightarrow3x+4+2\sqrt{\left(2x+3\right)\left(x+1\right)}=25\)

\(\Leftrightarrow2\sqrt{\left(2x+3\right)\left(x+1\right)}=21-3x\)

\(\Leftrightarrow x^2-146x+429=0\)

...

Câu b giải tương tự

Bạn đọc rồi kiểm tra hộ mình ạ

\(\sqrt{25t^2-9}=2\sqrt{5t-3}\left(t\ge\dfrac{3}{5}\right)\)hoặc\(t\le-\dfrac{3}{5}\))

\(=\sqrt{\left(5t-3\right)\left(5t+3\right)}-2\sqrt{5t-3}=0\)

\(< =>\sqrt{5t-3}\left(\sqrt{5t+3}-2\right)=0\)

\(=>\left[{}\begin{matrix}\sqrt{5t-3}=0\\\sqrt{5t+3}-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=0,6\left(TM\right)\\t=0,2\left(loai\right)\end{matrix}\right.\)

vậy t=0,6

\(\sqrt{-2x^2+6}=x-1\)(\(-\sqrt{3}\le x\le\sqrt{3}\) \(\))

\(=>-2x^2+6=x^2-2x+1\)

\(< =>-3x^2+2x+5=0\)

\(\Delta=\left(2\right)^2-4.5.\left(-3\right)=64>0\)

\(=>\left[{}\begin{matrix}x1=\dfrac{-2+\sqrt{64}}{2\left(-3\right)}=-1\left(loai\right)\\x2=\dfrac{-2-\sqrt{64}}{2\left(-3\right)}=\dfrac{5}{3}\left(TM\right)\end{matrix}\right.\)vậy x=5/3

a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)

\(\left(\sqrt{x-1}+\sqrt{x+3}\right)^2+\sqrt{x-1}+\sqrt{x-3}-6=0\)

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)

\(\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-\left(\sqrt{3x-2}+\sqrt{x-1}\right)-6=0\)

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

a/ ĐKXĐ: $x\geq 1$

Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:

$a+b+2ab=6-(a^2+b^2)$

$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$

$\Leftrightarrow (a+b)^2+(a+b)-6=0$

$\Leftrightarrow (a+b-2)(a+b+3)=0$

Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$

$\Leftrightarrow a+b=2$

Mà $b^2-a^2=(x+3)-(x-1)=4$

$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$

$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$

$\Leftrightarrow x=1$ (tm)

a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)

\(\Leftrightarrow2x+9=5-4x\)

\(\Leftrightarrow6x=-4\)

hay \(x=-\dfrac{2}{3}\left(nhận\right)\)

b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)

\(\Leftrightarrow2x-1=x-1\)

hay x=0(loại)

c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)

\(\Leftrightarrow x^2+3x=x\)

\(\Leftrightarrow x\left(x+2\right)=0\)

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

a. \(\sqrt{2x+9}=\sqrt{5-4x}\)

<=> 2x + 9 = 5 - 4x 

<=> 2x + 4x = 5 - 9

<=> 6x = -4

<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)