Question

cho biểu thức: A= \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{_{\left(1-x^2\right)}}{2}\)
a) Rút gọn A nếu \(x\ge0,x\ne1\)
b) Tìm x để A dương 
c) Tìm giá trị lớn nhất của A

Answer 1

E mới 7 - 8 thui !!! nhưng e sẽ cố giúp

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

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

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

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

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

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

b )

ĐKXĐ : \(x\ge0\)

Vì \(\sqrt{x}+1>0\forall x\) Để \(A=\frac{\sqrt{x}\left(x+1\right)}{\sqrt{x}+1}>0\) \(\Leftrightarrow\sqrt{x}\left(x+1\right)>0\)

\(\Rightarrow\hept{\begin{cases}\sqrt{x}\ne0\\x+1>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x>-1\end{cases}}}\) Mà theo đxxd thì \(x\ge0\) nên \(x>0\)

Vậy với \(x>0\) thì \(A>0\)

c ) Lớp 7 chưa bt làm :((

Answer 2

E ghi rõ nèk

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

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

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

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

Answer 3

\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)

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

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

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

\(=\frac{-2\sqrt{x}\left(\sqrt{x}+1\right).\left(x-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2.2}\)

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

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

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

Answer 4

đức hùng cố lên