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

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

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

Cái ngoặc to yên tâm là vô nghiệm từ đó...

P/s: em chi có mỗi cách này thôi, ko biết có đúng không nữa..

Anh ko hiểu từ dòng đầu qua cái p/s ở dòng 2 luôn @@

Đặt \(\sqrt{x^2+1}=t>\left|x\right|\ge x\) (dễ dàng chứng minh.

PT \(\Leftrightarrow\) \(t^2-\left(x+3\right)t+3x=0\)

\(\Leftrightarrow\left(t-3\right)\left(t-x\right)=0\)

\(\Leftrightarrow t=3\Leftrightarrow\orbr{\begin{cases}x=2\sqrt{2}\\x=-2\sqrt{2}\end{cases}}\)

True?Em tính liên hợp nhưng thôi:v Mệt lắm

d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)

\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)

\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)

\(\Leftrightarrow x^3=6-5x\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow x=1\)

c/

\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)

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

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

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

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

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

\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)

\(=3-1=2\)

\(B=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)

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

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

B=4x-3x+2x+x+10

\(\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}=\sqrt{2}\left(\sqrt{5}+1\right)\sqrt{3-\sqrt{5}}=\left(\sqrt{5}+1\right)\sqrt{6-2\sqrt{5}}=\left(\sqrt{5}+1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\)\(=\left(\sqrt{5}+1\right)|\sqrt{5}-1|=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=\sqrt{5}^2-1^2=4\)