Rút gọn Q=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\)
Cho Q = \(\dfrac{3x+9\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
Rút gọn biểu thức Q
\(Q=\left(\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\right)+\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\)
Rút gọn Q với x>0, x≠0, x≠9
\(Q=\dfrac{\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-1-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{-6\sqrt{x}+1+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ Q=\dfrac{2\sqrt{x}+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
cho biểu thức Q=\(\left(\dfrac{1}{\sqrt{X}-1}-\dfrac{1}{\sqrt{X}}\right):\left(\dfrac{\sqrt{X}+1}{\sqrt{X}-2}-\dfrac{\sqrt{X}+2}{\sqrt{X-1}}\right)\)
a rút gọn Q
b tìm x để Q>0
b) Q > 0
⇔ \(\dfrac{\sqrt{\text{x}}-2}{3\sqrt{\text{x}}}\) > 0
Do \(\text{3}\sqrt{\text{x}}>0\) ∀x⩾0
⇒ \(\sqrt{\text{x}}-2>0\)
⇔ \(\sqrt{\text{x}}>2\)
⇔ x > 4
Vậy x > 4 thì Q > 0
Cho Q=\(\left(\dfrac{x-1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-1}{x-1}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)^2\)
a;Rút gọn Q với x≥0;x≠1
b;Tìm x để Q<1
a) Ta có: \(Q=\left(\dfrac{x-1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-1}{x-1}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)^2\)
\(=\dfrac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}:\dfrac{\left(x-\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\left(x-\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}}{\left(x-\sqrt{x}+1\right)^2}\)
Câu 1: \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)
a,Rút gọn Q
b, Tìm \(x\in Z\) để \(Q\in Z\)
1.
\(a,Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{x+5}{x-\sqrt{x}-2}\)
\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(x+7\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ Q=\dfrac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\)
\(b,Q\in Z\Leftrightarrow\dfrac{-\sqrt{x}-6}{\sqrt{x}-2}\in Z\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-2\right)-8}{\sqrt{x}-2}\in Z\\ \Leftrightarrow-1-\dfrac{8}{\sqrt{x}-2}\in Z\)
Mà \(-1\in Z\Leftrightarrow\dfrac{8}{\sqrt{x}-2}\in Z\)
\(\Leftrightarrow8⋮\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{-8,-4,-2,-1,1,2,4,8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;1;3;4;6;10\right\}\)
Mà \(x\in Z\) và \(\sqrt{x}\ge0\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;1;4\right\}\\ \Leftrightarrow x\in\left\{0;1;4\right\}\)
Vậy \(x\in\left\{0;1;4\right\}\) thì \(Q\in Z\)
(\(\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3x+3}{x-9}\)):(\(\dfrac{2\sqrt{x}-2}{\sqrt{x}+3}-1\))
a) Rút gọn biểu thức
b) Tìm x để Q<\(\dfrac{-1}{2}\)
c) Tìm min Q
\(a,=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}+3}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}\)
a: \(=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
RÚT GỌN BIỂU THỨC:
18) \(Q = \dfrac{x + 2\sqrt{x} - 10}{x - \sqrt{x} - 6} - \dfrac{\sqrt{x} - 2}{\sqrt{x} - 3} - \dfrac{1}{\sqrt{x} - 2}\)
\(Q=\dfrac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+2}\) (ĐK: \(x\ge0;x\ne9\))
\(Q=\dfrac{x+2\sqrt{x}-10}{x+2\sqrt{x}-3\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+2}\)
\(Q=\dfrac{x+2\sqrt{x}-10}{\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+2}\)
\(Q=\dfrac{x+2\sqrt{x}-10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\dfrac{x+2\sqrt{x}-10-x+4-\sqrt{x}+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(Q=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(Q=\dfrac{1}{\sqrt{x}+2}\)
Rút gọn:
\(A=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a) Rút gọn Q
b) Tìm số nguyên x để Q có giá trị nguyên
a) \(ĐK:x>0,x\ne1\)\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2}{x-1}\)
b) \(P=\dfrac{2}{x-1}\in Z\)
\(\Rightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp với đk
\(\Rightarrow x\in\left\{0;2;3\right\}\)