Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé.

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

ĐKXĐ: 0<=x<=3/2

\(-3x^2 + x + 3 + (\sqrt{3x + 2} - 4)\sqrt{3x - 2x^2} + (x - 1)\sqrt{3x + 2} = 0 \quad (1)\)

=>\(-3x^2 + x + 3 + \sqrt{3x + 2} \cdot \sqrt{3x - 2x^2} - 4\sqrt{3x - 2x^2} + (x - 1)\sqrt{3x + 2} = 0\)

Đặt \(u=\sqrt{3x + 2}\ge0;v=\sqrt{3x - 2x^2}\ge0\)

Phương trình sẽ trở thành: \(-3x^2 + x + 3 + uv - 4v + (x - 1)u = 0\)

=>\(v^2-(x-1)(x+3)+uv-4v+(x-1)u=0\)

\(\iff\Big[v^2+uv-(x+3)v-v\Big]+(x-1)u-(x-1)(x+3)=0\)

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

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

TH1: \(x - 1 + \sqrt{3x - 2x^2} = 0\)

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

=>1-x>=0 và \(3x-2x^2=\left(1-x\right)^2\)

=>x<=1 và \(3x-2x^2-x^2+2x-1=0\)

=>x<=1 và \(-3x^2+5x-1=0\)

=>\(x=\frac{5-\sqrt{13}}{6}\)

TH2: \(\sqrt{3x + 2} + \sqrt{3x - 2x^2} - 3x - 1 = 0\)

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

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

=>\(6x+2-2x^2+2\sqrt{-6x^3 + 5x^2 + 6x}=9x^2+6x+1\)

=>\(2\sqrt{-6x^3 + 5x^2 + 6x}=11x^2-1\)

=>\(4(-6x^3 + 5x^2 + 6x) = (11x^2 - 1)^2\) và \(11x^2-1\ge0\)

=>\(-24x^3+20x^2+24x=121x^4-22x^2+1\) và \(11x^2\ge1\)

=>\(121x^4+24x^3-42x^2-24x+1=0\) và \(x^2\ge\frac{1}{11}\)

=>x≃0,6092(nhận)

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

ĐKXĐ:

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

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

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

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

1.

ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)

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

\(\Leftrightarrow x^2-4x-3-2\sqrt{\left(x^2-x\right)\left(x+1\right)}=0\)

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

\(\Rightarrow a^2-3b^2-2ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)

\(\Leftrightarrow a=3b\)

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

\(\Leftrightarrow x^2-x=9\left(x+1\right)\)

\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)

2.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:

\(x^3+3\left(x^2-4a^2\right)a=0\)

\(\Leftrightarrow x^3+3ax^2-4a^3=0\)

\(\Leftrightarrow\left(x-a\right)\left(x+2a\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-4x-4=0\end{matrix}\right.\)

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

a, ĐK: \(x\le-1,x\ge3\)

\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)

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

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

\(\Leftrightarrow x^2-2x-3=1\)

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

\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)

b, ĐK: \(-2\le x\le2\)

Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)

Khi đó phương trình tương đương:

\(3t-t^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)

Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm

Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)

=>x>=-1 và 3x^2+5x-13=x^2+2x+1

=>x>=-1 và 2x^2+3x-14=0

=>x>=-1 và 2x^2+7x-4x-14=0

=>x>=-1 và (2x+7)(x-2)=0

=>x=2

\(\sqrt{3x^2+5x-13}=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3x^2+5x-13=\left(x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\3x^2+5x-13=x^2+2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\2x^2+3x-14=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=2\left(\text{nhận}\right)\\x=-\dfrac{7}{2}\left(\text{loại}\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy: \(S=\left\{2\right\}\)