ĐKXĐ: \(\left\{{}\begin{matrix}5x-2>0\\3-2x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{2}{5}\\x\le\dfrac{3}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{5}< x\le\dfrac{3}{2}\)

ĐK: `{(3x+4>=0),(1+2x>=0),(x+3>=0):}<=> {(x>=-4/3),(x>=-1/2),(x>=-3):} <=> x>=-1/2`

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

a: ĐKXĐ: \(x\ge1\)

b: ĐKXĐ: \(x< 0\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

1) ĐKXĐ: \(\left\{{}\begin{matrix}2x+11\ge0\\x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

2) ĐKXĐ: \(\left\{{}\begin{matrix}-5x\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x< 0\)

3) ĐKXĐ: \(7x^2+1\ge0\left(đúng\forall x\right)\Leftrightarrow x\in R\)

4) ĐKXĐ: \(x^2-14x+33\ge0\Leftrightarrow\left(x-11\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-11\ge0\\x-3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-11\le0\\x-3\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge11\\x\le3\end{matrix}\right.\)

5) ĐKXĐ: 

+) \(-x^2+6x+16\ge0\)

\(\Leftrightarrow-\left(x^2-6x+9\right)+25\ge0\)

\(\Leftrightarrow\left(x-3\right)^2\le25\Leftrightarrow-5\le x-3\le5\)

\(\Leftrightarrow-2\le x\le8\)

+) \(3x^2\ne0\Leftrightarrow x\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}-2\le x\le8\\x\ne0\end{matrix}\right.\)

a/ ĐKXĐ: \(-\frac{1}{2}\le x\le4\)

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

\(\Leftrightarrow4-x=3x+2+2\sqrt{2x^2+3x+1}\)

\(\Leftrightarrow1-2x=\sqrt{2x^2+3x+1}\) (\(x\le\frac{1}{2}\))

\(\Leftrightarrow4x^2-4x+1=2x^2+3x+1\)

\(\Leftrightarrow2x^2-7x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{7}{2}\left(l\right)\end{matrix}\right.\)

Bài này liên hợp cũng được

b/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{5x+1}^2-\sqrt{5x+1}\left(\sqrt{14x+7}-\sqrt{2x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\Rightarrow x=-\frac{1}{5}\\\sqrt{5x+1}-\sqrt{14x+7}+\sqrt{2x+3}=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{5x+1}+\sqrt{2x+3}=\sqrt{14x+7}\)

\(\Leftrightarrow7x+4+2\sqrt{10x^2+17x+3}=14x+7\)

\(\Leftrightarrow2\sqrt{10x^2+17x+3}=7x+3\)

\(\Leftrightarrow4\left(10x^2+17x+3\right)=\left(7x+3\right)^2\)

\(\Leftrightarrow...\)

c/ ĐKXĐ: \(x\ge\frac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-2x}=a\\\sqrt{2x-1}=b\end{matrix}\right.\) ta được:

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

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

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

\(\Rightarrow\left[{}\begin{matrix}2-2x=0\\2-2x=1\\2-2x=-8\end{matrix}\right.\)

d/ ĐKXĐ: \(x\le\frac{5}{4}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{5-4x}=a\\\sqrt[3]{x+7}=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\a^2+4b^3=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3-b\\a^2+4b^3=33\end{matrix}\right.\)

\(\Leftrightarrow\left(3-b\right)^2+4b^3=33\)

\(\Leftrightarrow4b^3+b^2-6b-24=0\)

\(\Leftrightarrow\left(b-2\right)\left(4b^2+9b+12\right)=0\)

\(\Rightarrow b=2\Rightarrow\sqrt[3]{x+7}=2\Rightarrow x=1\)

       \(\sqrt{\frac{x-4}{2x+3}}=2\)

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

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

\(\Leftrightarrow x-4=4\left(2x+3\right)\)

\(\Leftrightarrow x-4=8x+12\)

\(\Leftrightarrow x-8x=12+4\)

\(\Leftrightarrow-7x=16\)

\(\Leftrightarrow x=\frac{16}{-7}=\frac{-16}{7}\)

Vậy tập nghiệm của pt là \(S=\left\{-\frac{16}{7}\right\}\)

Nếu bạn thiếu số 2 bên cạnh $\sqrt{2x^2+5x+3}$ thì có thể tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/tim-x-sao-cho-sqrt2x3sqrtx13x2sqrt2x25x3-16.235781793134

Đỗ Thanh Hải: uh ha, mình đã sửa lại rồi.