Mau la \(\sqrt{X - 3} \) that sao

sử dụng phương pháp miền giá trị

bạn nói rõ hơn được không?

a. ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

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

b. Ta có \(B-5=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}-5=\frac{2x-3\sqrt{x}+2}{\sqrt{x}}=\frac{2\left(x-2.\sqrt{x}.\frac{3}{4}+\frac{9}{16}\right)-\frac{9}{8}+2}{\sqrt{x}}\)

\(=\frac{2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}}{\sqrt{x}}\)

Ta thấy \(\hept{\begin{cases}2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}>0\\\sqrt{x}>0\forall x>0\end{cases}\Rightarrow B-5>0\Rightarrow B>5}\)

Vậy \(B>5\)

B3: \(\sqrt{x^4-4x^3+2x^2+4x+1}=3x-1\)

\(pt\Leftrightarrow x^4-4x^3+2x^2+4x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow x^4-4x^3+2x^2+4x+1=9x^2-6x+1\)

\(\Leftrightarrow x^4-4x^3-7x^2+10x=0\)

\(\Leftrightarrow x\left(x^3-4x^2-7x+10\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\) (thỏa mãn (mấy cái kia loại hết))

Rút gọn : \(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\) (ĐKXĐ : \(0\le x\ne1\) )

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

Vậy B < 1

\(ĐKXĐ:x\ge0;x\ne1;0\)

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

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

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

\(A=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}\)

\(A=2\sqrt{x}+\frac{2}{\sqrt{x}}+2\)

a/d bđt cauchy 

\(2\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{2.2}=2.2=4\)

\(A\ge4+2=6\)

\(< =>A>5\)

dấu "=" xảy ra khi x=1

Ta có: \(P=\frac{\sqrt{x}-4}{\sqrt{x}}\times\frac{x+\sqrt{x}+1}{\sqrt{x}-4}\)

 \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)\(\left(ĐK:x>0\right)\)

Ta lấy \(P-2=\frac{x+\sqrt{x}+1}{\sqrt{x}}-2\)

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

                       \(=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

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

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

Vì \(x>0\Rightarrow\sqrt{x}>0\)

 \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

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

\(\Rightarrow P-2>0\)

\(\Rightarrow P>2\)

Học tốt 

a) \(ĐKXĐ:\) \(x\ne1,x>0\)

\(P=1:\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\right)\)

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

\(=1:\left[\frac{x+2+x-1-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\)

\(=1:\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Vậy \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\left(x\ne1,x>0\right)\)

b) Xét hiệu \(P-3=\frac{x+\sqrt{x}+1}{\sqrt{x}}-3\)

\(=\frac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\forall x>0,x\ne1\)

Do đó : \(P>3\)