Xét a>0 thì 6^a-2^a+4 chia hết cho 2 trong khi 43^a không chia hết cho 2 >>>loại.

Xét a=0,VT=4;VP=1>>>loại.

>>>Không có n(xem lại đề nha,chắc bạn ghi nhầm đề rồi.

Hép mi nha

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

ĐK:\(x\ge\frac{1}{2}\)

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

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

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

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

Suy ra x=1 và pt trong ngoặc chuyển vế bình phương lên đưuọc \(x=-\sqrt{2}+2\)

2)\(\left(x+1\right)\sqrt{x^2-2x+3}=x^2+1\) (bình phương luôn cũng được nhưng cơ bản là mình ko thích :| )

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

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

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

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

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

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

\(\Leftrightarrow x^2-2x+3=x^2-2x+1\Leftrightarrow3=1\) (loại)

\(\Rightarrow x^2-2x-1=0\Rightarrow x=\frac{2\pm\sqrt{8}}{2}\)

1/ (x + 1)(x - √x - 6)

cm =2 dung ko ban 

ta co \(\frac{A}{\sqrt{2}}\) \(=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

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

                            \(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{9-3}\)

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

SUY RA A=\(\sqrt{2}\left(DPCM\right)\)

1.

\(\text{ĐK: }x\ge\frac{1}{2}\)

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

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

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

\(\Leftrightarrow x=1\text{ }\left(do\text{ }....................................................>0\right)\)

cảm ơn nhìu nkoa b!!!