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1.
a. ĐKXĐ : x lớn hơn hoặc bằng 1/2
b. A\(\sqrt{2}\)= \(\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
= \(\sqrt{2x-1+1+2\sqrt{2x-1}}-\sqrt{2x-1+1-2\sqrt{2x-1}}\)
=\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
= \(\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)
Nếu \(x\ge1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(\sqrt{2x-1}-1\right)=2\)
\(\Rightarrow A=2\)
Nếu 1/2 \(\le x< 1thìA\sqrt{2}=\sqrt{2x-1}+1-\left(1-\sqrt{2x-1}\right)=2\sqrt{2x-1}\)
Do đó : A= \(\sqrt{4x-2}\)
Vậy ............
2.
a. \(x\ge2\)hoặc x<0
b. A= \(2\sqrt{x^2-2x}\)
c. A<2 \(\Leftrightarrow\)\(2\sqrt{x^2-2x}< 2\Leftrightarrow\sqrt{x^2-2x}< 1\Leftrightarrow x^2-2x< 1\Leftrightarrow\left(x-1\right)^2< 2\)
\(-\sqrt{2}< x-1< \sqrt{2}\Leftrightarrow1-\sqrt{2}< x< 1+\sqrt{2}\)
Kết hợp vs đk câu a , ta đc : \(1-\sqrt{2}< x< 0và2\le x< 1+\sqrt{2}\)
Vậy...........
A=\(\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
=\(\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x-2}}\)
Vậy A=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)vs x\(\ge0;x\ne4\)
C=\(\left(\frac{1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{1+x}{\sqrt{x}}\)
Vậy C=\(\frac{1+x}{\sqrt{x}}\)vs x>0
\(ĐKXĐ:x\ge0;x\ne1;0\)
\(A=\frac{2x+2}{\sqrt{x}}+\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\frac{2x+2}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(A=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(A=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}\)
\(A=2\sqrt{x}+\frac{2}{\sqrt{x}}+2\)
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
\(A=\left(\frac{2+\sqrt{x}}{x-1}+\frac{2}{\sqrt{x}+1}\right)\div\frac{3}{x+\sqrt{x}}\)
a) ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(=\left(\frac{2+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{3}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{2+\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{3}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{3}\)
\(=\frac{x}{\sqrt{x}-1}\)
b) Xét biểu thức\(\frac{x}{\sqrt{x}-1}+4\left(\sqrt{x}-1\right)\)
Vì x > 1 nên áp dụng bất đẳng thức Cauchy ta có :
\(\frac{x}{\sqrt{x}-1}+4\left(\sqrt{x}-1\right)\ge2\sqrt{\frac{x}{\sqrt{x}-1}\cdot4\left(\sqrt{x}-1\right)}=2\sqrt{4x}=4\sqrt{x}\)
=> \(\frac{x}{\sqrt{x}-1}+4\left(\sqrt{x}-1\right)\ge4\sqrt{x}\)
=> \(\frac{x}{\sqrt{x}-1}+4\sqrt{x}-4\ge4\sqrt{x}\)
=> \(\frac{x}{\sqrt{x}-1}\ge4\)
Đẳng thức xảy ra khi x = 4 ( tm )
=> MinA = 4 <=> x = 4