\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\sqrt{2x-1}-2x\sqrt{x+3}\\ \Leftrightarrow\left(2x-2\right)-\left(2\sqrt{2x^2+5x-3}-4\right)=\left(x\sqrt{2x-1}-x\right)-\left(2x\sqrt{x+3}-4x\right)-3x+3\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(2x^2+5x-7\right)}{\sqrt{2x^2+5x-3}+4}=\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}-\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}-3\left(x-1\right)\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(x-1\right)\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x\left(x-1\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge\dfrac{1}{2}\Leftrightarrow-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}>-\dfrac{2\cdot8}{4}=-4\)

\(-\dfrac{2x}{\sqrt{2x-1}+2}>-\dfrac{1}{2};\dfrac{2x}{\sqrt{x+3}+4x}>0\)

Do đó \(\left(1\right)>2-4-\dfrac{1}{2}+3=\dfrac{1}{2}>0\) nên (1) vô nghiệm

Vậy PT có nghiệm duy nhất \(x=1\)

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

\(\Leftrightarrow5x^4+2x-2\sqrt{2x+1}+2=0\)

\(\Leftrightarrow5x^4+\left(2x+1-2\sqrt{2x+1}+1\right)=0\)

\(\Leftrightarrow5x^4+\left(\sqrt{2x+1}-1\right)^2=0\)

có \(\hept{\begin{cases}5x^4\ge0\\\left(\sqrt{2x+1}-1\right)^2\ge0\end{cases}}\)mà \(5x^4+\left(\sqrt{2x+1}-1\right)^2=0\Rightarrow\hept{\begin{cases}5x^4=0\\\left(\sqrt{2x+1}-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^4=0\\\sqrt{2x+1}=1\end{cases}\Leftrightarrow x=0}\)

vạy x=0 là nghiệm của phương trình

Cre: Đàm Hải Ngọc

cái này dùng liên hợp dễ hơn 

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

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

\(< =>x\left(5x^3+2\right)-x.\frac{4}{\sqrt{2x+1}+1}=0\)

\(< =>x\left(5x^3+2-\frac{4}{\sqrt{2x+1}+1}\right)=0< =>x=0\)

giờ dùng đk đánh giá cái ngoặc to vô nghiệm là ok

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)

\(\Leftrightarrow cos^2x+2cosx+1=0\)

\(\Leftrightarrow\left(cosx+1\right)^2=0\)

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

điều kiện xác đinh \(x\ge-\frac{1}{2}\)

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

\(\Leftrightarrow5x^4+2x-2\sqrt{2x+1}+2=0\Leftrightarrow5x^4+2x+1-2\sqrt{2x+1}+1=0\)

\(\Leftrightarrow5x^4+\left(\sqrt{2x+1}-1\right)^2=0=>\orbr{\begin{cases}5x^4=0\\\sqrt{2x+1}-1=0\end{cases}\Leftrightarrow x=0\left(nhận\right)}\)

zậy \(S=\left\{0\right\}\)

ĐK: \(x\ge\frac{-1}{2}\). PT đã cho có thể viết lại thành 

\(5x^4+\left(\sqrt{2x+1}-1\right)^2=0\)

Do \(5x^4\ge0,\left(\sqrt{2x+1}-1\right)^2\ge0\)nên PT trên chỉ thỏa mãn khi \(\hept{\begin{cases}5x^4=0\\\left(\sqrt{2x+1}-1\right)^2=0\end{cases}}\)

Giải hệ này ta được x=0

Vậy PT đã cho có nghiệm duy nhất x=0

a.

ĐKXĐ: \(x^2+2x-1\ge0\)

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

Đặt \(\sqrt{x^2+2x-1}=t\ge0\)

\(\Rightarrow t^2+2\left(x-1\right)t-4x=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-5=0\\3x^2-2x+1=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=-1\pm\sqrt{6}\)

b.

ĐKXĐ: \(x\ge\dfrac{1}{5}\)

\(2x^2+x-3+2x-\sqrt{5x-1}+2-\sqrt[3]{9-x}=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\dfrac{\left(x-1\right)\left(4x-1\right)}{2x+\sqrt[]{5x-1}}+\dfrac{x-1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3+\dfrac{4x-1}{2x+\sqrt[]{5x-1}}+\dfrac{1}{4+2\sqrt[3]{9-x}+\sqrt[3]{\left(9-x\right)^2}}\right)=0\)

\(\Leftrightarrow x=1\) (ngoặc đằng sau luôn dương)