Bạn xem lại đề câu b và c nhé !

a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)

\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ

\(\Rightarrow x\ge2\) thỏa mãn đề.

d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)

Pt tương đương :

\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )

e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)

\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)

Phương trình (1) tương đương :

\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )

1) đặt đk rùi bình phương 2 vế là ok

2) \(pt\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+2}}{x-x-2}+\frac{\sqrt{x+2}-\sqrt{x+4}}{x+2-x-4}+\frac{\sqrt{x+4}-\sqrt{x+6}}{x+4-x-6}=\frac{\sqrt{10}}{2}-1\)(ĐKXĐ : \(x\ge0\))

<=> \(\frac{\sqrt{x}-\sqrt{x+6}}{-2}=\frac{\sqrt{10}}{2}-1\)

<=> \(\frac{\sqrt{x+6}-\sqrt{x}}{2}=\frac{\sqrt{10}-2}{2}\)

<=> \(\sqrt{x+6}-\sqrt{x}=\sqrt{10}-2\)

<=> \(\sqrt{x+6}+2=\sqrt{10}+\sqrt{x}\)

đến đây bình phương 2 vế rùi giải bình thường nhé 

hello bạn

1.a) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)

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

\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\sqrt{x+2}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x+2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy x=2 hoặc x=-1

\(\dfrac{-17}{15}\)

b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)

\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)

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

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

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

Pt trong ngoặc VN suy ra x=2

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

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

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

\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)

pt trong căn vô nghiệm

suy ra x=1; x=-1

c)\(8x^2-13x+7=1+\frac{1}{x}\sqrt[3]{3x^2-2}\)

\(\Leftrightarrow8x^2-13x+7-2=\frac{1}{x}\sqrt[3]{3x^2-2}-1\)

\(\Leftrightarrow\left(x-1\right)\left(8x-5\right)-\frac{\frac{3x^2-2}{x^3}-1}{\frac{1}{x}\sqrt[3]{3x^2-2}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(8x-5\right)-\frac{\frac{-\left(x-1\right)\left(x^2-2x-2\right)}{x^3}}{\frac{1}{x}\sqrt[3]{3x^2-2}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\left(8x-5\right)-\frac{\frac{-\left(x^2-2x-2\right)}{x^3}}{\frac{1}{x}\sqrt[3]{3x^2-2}+1}\right)=0\)

SUy ra x=1 và 1 nghiệm lẻ nx trong ngoặc bn tự làm :V