\(\left(\sqrt{2x+5}-\left(x+1\right)\right)^2+\left(\sqrt{3\left(x+1\right)}-\sqrt{x+7}\right)^2=0.\\ \)
Đến đây chắc biết phải làm gì =))
 

1.

\(3sin^22x-2sin2x.cos2x-4cos^22x=2\)

\(\Leftrightarrow-\dfrac{3}{2}\left(1-2sin^22x\right)-2sin2x.cos2x-2\left(2cos^22x-1\right)=\dfrac{5}{2}\)

\(\Leftrightarrow sin4x+\dfrac{7}{2}cos4x=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{53}}{2}\left(\dfrac{2}{\sqrt{53}}sin4x+\dfrac{7}{\sqrt{53}}cos4x\right)=-\dfrac{5}{2}\)

\(\Leftrightarrow sin\left(4x+arccos\dfrac{2}{\sqrt{53}}\right)=-\dfrac{5}{\sqrt{53}}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+arccos\dfrac{2}{\sqrt{53}}=arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\\4x+arccos\dfrac{2}{\sqrt{53}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}+\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}-\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)

2.

\(2\sqrt{3}cos^2x+6sinx.cosx=3+\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1\right)+6sinx.cosx=3\)

\(\Leftrightarrow\sqrt{3}cos2x+3sin2x=3\)

\(\Leftrightarrow2\sqrt{3}\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)=3\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)

3.

\(3cos^24x+5sin^24x=2-2\sqrt{3}sin4x.cos4x\)

\(\Leftrightarrow4cos^24x+4sin^24x-cos^24x+sin^24x=2-2\sqrt{3}sin4x.cos4x\)

\(\Leftrightarrow4-cos8x=2-\sqrt{3}sin8x\)

\(\Leftrightarrow cos8x-\sqrt{3}sin8x=2\)

\(\Leftrightarrow\dfrac{1}{2}cos8x-\dfrac{\sqrt{3}}{2}sin8x=1\)

\(\Leftrightarrow cos\left(8x+\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow8x+\dfrac{\pi}{3}=k2\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{24}+\dfrac{k\pi}{4}\)

ĐKXĐ: \(x\geq -2\).

Nhận thấy x = -2 không là nghiệm của pt.

Xét x khác -2.

\(PT\Leftrightarrow\sqrt[3]{x^3+8}-\left(2x+4\right)=\dfrac{24x-18}{x^2-2x-7}-6\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2-6x-4\right)}{\sqrt[3]{x^3+8}+x+2}=\dfrac{-6\left(x^2-6x-4\right)}{x^2-2x-7}\)

\(\Leftrightarrow\dfrac{x+2}{\sqrt[3]{x^3+8}+x+2}=\dfrac{-6}{x^2-2x-7}\left(1\right)\) hoặc x² - 6x - 4 = 0.

\(\left(1\right)\Rightarrow\left(x+2\right)\left(x^2-2x-1\right)=-6\sqrt[3]{x^3+8}\)

+) Nếu x \(\geq 7\) thì \(\left(x+2\right)\left(x^2-2x-1\right)>0\ge-6\sqrt{x^3+8}\) (loại)

+) Nếu \(x\le7\) thì \(\left(x+2\right)\left(x^2-2x-1\right)\ge-2\left(x+2\right)>-6\sqrt[3]{3\left(x+2\right)}\ge-6\sqrt[3]{x^3+8}\) (loại)

Do đó (1) vô nghiệm.

Do đó \(x^2-6x-4=0\Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{13}\left(TMĐK\right)\\x=3-\sqrt{13}\left(loại\right)\end{matrix}\right.\)

Vậy...

Viết nốt đi bạn ơi!! 

Viết tiếp đi.Không có kết quả là bao nhiêu thì làm sao giải được???

đặt S=vế trái

ta có:S=\(\sqrt{3\left(x^2-6x+9\right)+1}+\sqrt{4\left(x^2-6x+9\right)+9}\)

S=\(\sqrt{3\left(x-3\right)^2+1}+\sqrt{4\left(x-3\right)^2+9}\)

ta thấy:\(\sqrt{3\left(x-3\right)^2+1}\ge\sqrt{1}=1\);\(\sqrt{4\left(x-3\right)^2+9}\ge\sqrt{9}=3\)

→S\(\ge\)4; xét vế phải :\(-5-x^2+6x=4-\left(x-3\right)^2\)\(\le\)4

vậy pt xảy ra khi x-3=0↔x=3

(đề là -5 -x²+6x thì khả nghi hơn)

\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)