cho biểu thức A=\(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
a, Rút gọn biểu thức A
b, tìm tất cả các giá trị nguyên của biêu thức A
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Có: \(\sqrt{a}\sqrt{b}=\sqrt{âb}\)
\(\Rightarrow\sqrt{20}=\sqrt{4}\sqrt{5}=2\sqrt{5}\)
\(\sqrt{a+c}-\sqrt{a}< \sqrt{b+c}-\sqrt{b}\)
\(\Leftrightarrow\sqrt{a+c}+\sqrt{b}< \sqrt{b+c}+\sqrt{a}\)
\(\Leftrightarrow\left(\sqrt{a+c}+\sqrt{b}\right)^2< \left(\sqrt{b+c}+\sqrt{a}\right)^2\)
\(\Leftrightarrow a+b+c+2\sqrt{ab+bc}< a+b+c+2\sqrt{ab+ac}\)
\(\Leftrightarrow2\sqrt{ab+bc}< 2\sqrt{ab+ac}\Leftrightarrow\sqrt{ab+bc}< \sqrt{ab+ac}\)(đúng vs a>b) .Vậy bđt cần cm đúng
1) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
Xin lỗi xin lỗi :v
1)\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
= \(\sqrt{7}.\left(3\sqrt{7}-2\sqrt{14}\right)+14\sqrt{2}\)
= 21 - \(14\sqrt{2}+14\sqrt{2}\)
= 21
2) \(\left(\sqrt{8}-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{18}-\sqrt{8}+\sqrt{5}\right)\)
= \(\left(2\sqrt{2}-\sqrt{2}-\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{5}-2\sqrt{2}\right)\)
= \(\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)\)
=\(\left(\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2\)
= -3
\(\(A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)\)
\(\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)\)
\(\(=\frac{\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)\)
\(\(=\frac{x+2\sqrt{x}-\sqrt{x}-2-\left(x+\sqrt{x}-2\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)\)
\(=\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}:\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
Vậy \(A=\frac{2}{x-1}vs\left(x\ge0;x\ne1\right)\)
_Ko chắc , đag bận nên còn phần b , tí mk giải nối_
_Minh ngụy_
\(ĐK:x\ge0;x\ne1\)
\(a,A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+2\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)\sqrt{x}}\)
\(=\frac{2}{x-1}\)
Vậy với \(x\ge0;x\ne1\)thì \(A=\frac{2}{x-1}\)
\(b,\)Ta có:\(A=\frac{2}{x-1}\)
Để A nhận giá trị nguyên \(\Leftrightarrow2⋮x-1\)
Vì \(x\in Z\Rightarrow x-1\inƯ_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)
Ta có bảng sau:
Vậy để A nhận giá trị nguyên \(x\in\left\{2;0;3\right\}\)