A=\((\frac{\sqrt{x}+1}{x-2\sqrt{x}}-\frac{1}{\sqrt{x}-2}).(x-3\sqrt{x}+2)\) (với x>0; x \(\ne\)4)
a,Rút gọn A
b, Tìm x để A <\(\frac{1}{2}\)
c,Tìm các gt nguyên của x để A có gt nguyên
(làm ơn giúp mk vs huhu..)
K
Khách
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Những câu hỏi liên quan
3 tháng 9 2016
a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}=4x-\sqrt{8}+x=5x-\sqrt{8}\)
3 tháng 8 2017
Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:
\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)
=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)
=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)
Y
11 tháng 8 2019
2.
a)
\(\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(2-\frac{2\sqrt{a}-a}{\sqrt{a}-2}\right)\\ =\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(2+\frac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\right)\\ =\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)\\ =2^2-\left(\sqrt{a}\right)^2\\ =4-a\)
b)
\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{\sqrt{x}+1}{x}\\ =\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\frac{x}{\sqrt{x}+1}\\ =\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\cdot\frac{x}{\sqrt{x}+1}\\ =\frac{x-1}{\sqrt{x}}\cdot\frac{x}{\sqrt{x}+1}\\ =\sqrt{x}\left(\sqrt{x}-1\right)\\ =x-\sqrt{x}\)
c)
\(\left(\frac{1-x\sqrt{x}}{1-x}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\\ =\left(\frac{1-\sqrt{x^3}}{1-x}+\sqrt{x}\right)\cdot\frac{\left(1-\sqrt{x}\right)^2}{\left(1-x\right)^2}\\ =\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\sqrt{x}\right)\cdot\frac{\left(1-\sqrt{x}\right)^2}{\left[\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\right]^2}\\ =\left(\frac{1+\sqrt{x}+x+\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\\ =\frac{2x+2\sqrt{x}+1}{1+\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{2x+2\sqrt{x}+1}{\left(1+\sqrt{x}\right)^3}\)
Y
11 tháng 8 2019
1. (Ko viết lại đề nha :v)
a)
\(A=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\\ =\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\\ =\left(\frac{x+2\sqrt{x}-\sqrt{x}-2-x-\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\\ =\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-1\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{2}{x-1}\)
b) Để A đạt giá trị nguyên thì \(2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)\)
\(\Leftrightarrow x-1\in\left\{-1;1;-2;2\right\}\\ \Leftrightarrow x\in\left\{0;2;-1;3\right\}\)
Vậy......
KT
1 tháng 8 2018
\(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\)
\(=\left(\frac{3x-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}+2\right)\)
\(=\frac{3x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}+2\right)\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=3\sqrt{x}\)
T
13 tháng 9 2019
ĐK: \(x\ge-7\)
PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)
\(\Leftrightarrow x=9\)
P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((
a.
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\left(x-\sqrt{x}-2\sqrt{x}+2\right)\\ =\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\right]\\ =\frac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\right]\\ =\frac{\sqrt{x}-1}{\sqrt{x}}\)
b.
\(A=\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\\ \Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{1}{2}< 0\\ \Leftrightarrow\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}}{2\sqrt{x}}< 0\\ \Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}}< 0\\ \Leftrightarrow\sqrt{x}-2< 0\\ \Leftrightarrow x< 4\)
Vậy với 0<x<4 thì A < \(\frac{1}{2}\)
c. Ta có \(A=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)
Để A đạt giá trị nguyên thì \(1⋮\sqrt{x}\Leftrightarrow\sqrt{x}\inƯ\left(1\right)\)
Mà \(\sqrt{x}>0\forall x>0\Rightarrow x=1\)
Vậy với x=1 thì A đạt giá trị nguyên