a.

ĐKXĐ: \(x\ge-1\)

\(7+12\sqrt{x+1}=x+4\sqrt{x^2+3x+2}\)

\(\Leftrightarrow4\sqrt{\left(x+1\right)\left(x+2\right)}-12\sqrt{x+1}+x-7=0\)

\(\Leftrightarrow4\sqrt{x+1}\left(\sqrt{x+2}-3\right)+x-7=0\)

\(\Leftrightarrow4\sqrt{x+1}\left(\dfrac{x-7}{\sqrt{x+2}+3}\right)+x-7=0\)

\(\Leftrightarrow\left(x-7\right)\left(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1\right)=0\)

\(\Leftrightarrow x-7=0\) (do \(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1>0;\forall x\ge-1\))

\(\Rightarrow x=7\)

b.

ĐKXĐ: \(x\ne-\dfrac{1}{3}\)

\(\Rightarrow3x^2+3x+2=\left(3x+1\right)\sqrt{x^2+x+2}\)

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

Đặt \(\sqrt{x^2+x+2}=t\)

\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)

\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1+x-1}{2}=2x\\t=\dfrac{3x+1-\left(x-1\right)}{2}=x+1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4x^2\left(x\ge0\right)\\x^2+x+2=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\\\end{matrix}\right.\)

a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)

\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)

đặt\(x^2+x+1=t\left(t>0\right)\)

\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)

bình phương 2 vế pt trở thành:

\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)

\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m

vậy pt vô nghiệm

a/ ĐKXĐ: ...

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)

\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)

\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))

\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)

\(\Leftrightarrow11a^2+6a-25=0\)

Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó

b/

Đặt \(x^2+x+1=a>0\)

\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)

\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)

\(\Leftrightarrow\sqrt{a^2+3a}=2\)

\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x^2+x+1=1\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3x^2-5x+7}=3-\sqrt{3x^2-7x+2}\)

\(\Rightarrow3x^2-5x+7=3x^2-7x+11-6\sqrt{3x^2-7x+2}\)

\(\Leftrightarrow3\sqrt{3x^2-7x+2}=2-x\) (\(x\le2\))

\(\Leftrightarrow9\left(3x^2-7x+2\right)=x^2-4x+4\)

\(\Leftrightarrow26x^2-59x+14=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{26}\end{matrix}\right.\)

Do biến đổi ko tương đương nên cần thay lại nghiệm vào pt ban đầu kiểm tra

d/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{x^2+3x+2}+\sqrt{x^2+6x+5}=\sqrt{2x^2+9x+7}\)

\(\Leftrightarrow2x^2+9x+7+2\sqrt{\left(x^2+3x+2\right)\left(x^2+6x+5\right)}=2x^2+9x+7\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\left(l\right)\\x=-5\end{matrix}\right.\)

b) đặt \(\sqrt{3x+1}=a\)(\(a\ge0\))

\(PT\Leftrightarrow\dfrac{a^2-1}{\sqrt{a^2+9}}+1=a\)

\(\Leftrightarrow\left(a-1\right)\left(1-\dfrac{a+1}{\sqrt{a^2+9}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+1=\sqrt{a^2+9}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)(tm)

c) bunyalovsky:

\(VT^2\le2\left(7-x+x-5\right)=4\)

\(\Leftrightarrow VT\le2\)

\(VF=\left(x-6\right)^2+2\ge2\)

Dấu = xảy ra khi x=6

a,đk -1<x<7

x+1+2 căn 7-x-2 căn x+1=căn (x+1)(7-x)

c.

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

Đặt \(\sqrt{x^2+3}=t>0\)

\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)

\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1-x+1}{2}=x+1\\t=\dfrac{3x+1+x-1}{2}=2x\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\left(x\ge-1\right)\\x^2+3=4x^2\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

a.

Đề bài ko chính xác, pt này ko giải được

b.

ĐKXĐ: \(x\ge-\dfrac{7}{2}\)

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

Đặt \(\sqrt{2x+7}=t\ge0\)

\(\Rightarrow t^2-\left(2x+7\right)t+x^2+7x=0\)

\(\Delta=\left(2x+7\right)^2-4\left(x^2+7x\right)=49\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+7-7}{2}=x\\t=\dfrac{2x+7+7}{2}=x+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+7}=x\left(x\ge0\right)\\\sqrt{2x+7}=x+7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-7=0\left(x\ge0\right)\\x^2+12x+42=0\left(vn\right)\end{matrix}\right.\)

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