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a. ĐKXĐ: x\(\ne1\) x, \(\ne-1\)
b. \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
=\(\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
=\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\dfrac{4}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1^2}{2}=2\left(\sqrt{x}-1\right)=2\sqrt{x}-2\)
c. khi x=0,16 thì G=\(2\sqrt{x}-2=2\sqrt{0,16}-2=2.0,4-2=0,8-2=-1,2\)
ĐKXĐ: \(x\ge0;x\ne1\)
\(G=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}.\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}=-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x\)
\(x=0,16\Rightarrow G=\sqrt{0,16}-0,16=\)
\(G=\frac{1}{4}-x+\sqrt{x}-\frac{1}{4}=\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\)
\(\Rightarrow G_{max}=\frac{1}{4}\) khi \(\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
G nguyên khi \(\sqrt{x}\) nguyên \(\Rightarrow x=k^2\) với \(k\in Z\)
Vậy với mọi x có dạng \(x=k^2\) thì G nguyên
\(G=\sqrt{x}\left(1-\sqrt{x}\right)\)
Với \(0< x< 1\Rightarrow0< \sqrt{x}< 1\Rightarrow1-\sqrt{x}>0\)
\(\Rightarrow G=\sqrt{x}\left(1-\sqrt{x}\right)>0\Rightarrow G\) dương
Để \(G< 0\Rightarrow1-\sqrt{x}< 0\Rightarrow x>1\)
Câu 2:
a: ĐKXĐ: x>=0; x<>1
b: \(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=-\dfrac{2\sqrt{x}}{2}\cdot\left(\sqrt{x}-1\right)=-\sqrt{x}\left(\sqrt{x}-1\right)\)
c: Thay x=4/25 vào G, ta được:
\(G=-\dfrac{2}{5}\cdot\left(\dfrac{2}{5}-1\right)=\dfrac{-2}{5}\cdot\dfrac{-3}{5}=\dfrac{6}{25}\)
điều kiện xác định : \(x\ge0;x\ne1\)
a) ta có : \(G=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
\(\Leftrightarrow G=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(x-1\right)^2}{2}\)
\(\Leftrightarrow G=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(x-1\right)^2}{2}\) \(\Leftrightarrow G=\left(\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(x-1\right)^2}{2}=\sqrt{x}-x\)
b) thay \(x=0,16\) vào \(G\) ta có : \(G=\sqrt{0,16}-0,16=0,24\)
c) ta có : \(G=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge\dfrac{-1}{4}\)
\(\Rightarrow G_{max}=\dfrac{-1}{4}\) khi \(\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)
d) ta có : \(G=\sqrt{x}-x\) \(\Rightarrow\) để \(G\in Z\) \(\Rightarrow x=a^2\ne1\)
e) ta có : \(G>0\Leftrightarrow\sqrt{x}-x>0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\1-\sqrt{x}< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0< x< 1\\x\in\varnothing\end{matrix}\right.\) \(\Rightarrow\left(đpcm\right)\)
f) để \(G< 0\Leftrightarrow\sqrt{x}-x< 0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\1-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>1\\x\in\varnothing\end{matrix}\right.\) vậy \(x>1\)
bạn có thể làm chi tiết dòng thứ tư phần rút gọn đc ko ?
Đầu tiên bạn rút gọn biểu thức G,mik phân tích được:
G=x - 3 \(\sqrt{x}+2\)
(do ko có thời gian nên mik ko giải thick đâu nha.khi nào rảnh mik giải thích cho nếu bạn muốn)
Ta có: G= \(x-3\sqrt{x}+2\)
4G= \(4x-12\sqrt{x}+8\)
= \(\left(2x-3\right)^2-1\)
vì 0 <x<1 nên 0<2x<1 =>-3<2x-3<-2
=>3>(2x-3)2>2
=>2>(2x-3)2>1
Vậy G luôn dương khi 0<x<1.
mik nhầm dòng thứ 2 dưới lên nha bạn sửa thành
2>(2x-3)2-1>1
xin lỗi nhiều nha.
a: ĐKXĐ: x>=0; x<>1
b: \(G=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{2}\cdot\left(\sqrt{x}-1\right)\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
c: Thay x=0,16 vào G, ta được:
\(H=-0,4\cdot\left(0,4-1\right)=-0,4\cdot0,3=-0,12\)