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Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)ĐKXĐ : \(x>1\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)
\(A=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)
\(A=\frac{x+2}{\sqrt{x}}\)
\(a)\)\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)
\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\right)+\left(\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}\right)\)
\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\)
\(b)\) Áp dụng Cosi với hai số dương ta có :
\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}=\frac{2}{\sqrt{x}}\)
\(\Leftrightarrow\)\(x=2\)
Vậy GTNN của \(A\) là \(2\sqrt{2}\) khi \(x=2\)
Chúc bạn học tốt ~
PS : mới lớp 8 ko chắc nhé :v
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
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...........
\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\left(đkxđ\Leftrightarrow x\ge0\right).\)
\(=\frac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)\)
\(=x+\sqrt{x}-2\sqrt{x}-1=x-\sqrt{x}-1\)
\(P=x-\sqrt{x}-1=\sqrt{x}^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1\)
\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\)
\(\Rightarrow P_{min}=-\frac{5}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)