A=(\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}\) - \(\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)) : (1+\(\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\))
a) Rút gọn A
b)Tính A khi x=\(\dfrac{1}{6-2\sqrt{5}}\)
c)Tìm x ∈ Z để A ∈ Z
Giups mình với ạ
A=(\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}\) - \(\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)) : (1-\(\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\))
a) Rút gọn A
b)Tính A khi x=\(\dfrac{1}{6-2\sqrt{5}}\)
c)Tìm x ∈ Z để A ∈ Z
Giups mình với ạ
a) \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\right)\) (ĐK: \(x>0;x\ne1\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+3}-\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+3-3+\sqrt{x}}{\sqrt{x}+3}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}:\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{2\sqrt{x}}\)
\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
b) Ta có: \(x=\dfrac{1}{6-2\sqrt{5}}=\dfrac{1}{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}=\dfrac{1}{\left(\sqrt{5}-1\right)^2}=\left(\dfrac{1}{\sqrt{5}-1}\right)^2\)
Thay vào A ta có:
\(A=\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{5}-1}\right)^2}+3}{\sqrt{\left(\dfrac{1}{\sqrt{5}-1}\right)^2}}=3\sqrt{5}-2\)
c) Ta có: \(\dfrac{\sqrt{x}+3}{\sqrt{x}}=1+\dfrac{3}{\sqrt{x}}\)
\(\Rightarrow\sqrt{x}\in\left\{1;3\right\}\)
\(\Rightarrow x\in\left\{1;9\right\}\)
Bài 1: Cho A=\(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\div\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) (x≥0; x≠9)
a, Rút gọn A
b, Tính A khi \(x=7+4\sqrt{3}\)
c, Tìm x để A=\(\dfrac{3}{5}\)
d, Tìm x để A>1
e, Tìm x∈Z để A∈Z
(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3}{\sqrt{x}+3}.\)
(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{3}\).
Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)
\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)
(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)
\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).
Vậy: \(x\in\varnothing.\)
(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))
\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)
(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)
Vậy: \(x=0.\)
Cho biểu thức
A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)-\(\dfrac{3\sqrt{x}+1}{x-1}\)
a) Rút gọn A
b) Tính giá trị của A khi x = 4 - \(2\sqrt{3}\)
c) Tìm x để A = \(\dfrac{1}{2}\)
d) Tìm x để A < 1
e) Tìm x \(\in\) Z để A nhận giá trị nguyên
f) Tìm GTNN của A
A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) + \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) - \(\dfrac{3\sqrt{x}+1}{x-1}\)
a) Rút gọn A
b) Tính giá trị của A khi x = 4 - \(2\sqrt{3}\)
c) Tìm x để A = \(\dfrac{1}{2}\)
d) Tìm x để A < 1
e) Tìm x ∈ Z để A nhận giá trị nguyên
f) Tìm GTNN của A
a, ĐK: \(x\ge0,x\ne1\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+1+2\sqrt{x}+x+1-2\sqrt{x}-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b, \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
Khi đó:
\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\left(\sqrt{3}-1\right)-1}{\left(\sqrt{3}-1\right)+1}\)
\(=\dfrac{2\sqrt{3}-3}{\sqrt{3}}\)
\(=2-\sqrt{3}\)
c, \(A=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{2}\)
\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}+1\)
\(\Leftrightarrow3\sqrt{x}=3\)
\(\Leftrightarrow x=1\left(l\right)\)
Vậy không tồn tại giá trị x thỏa mãn \(A=\dfrac{1}{2}\).
Cho biểu thức A = (\(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}\) - \(\dfrac{1}{x+\sqrt{x}+1}\)) : (\(\dfrac{x-1}{x+\sqrt{x}+1}\))
a, Rút gọn biểu thức A
b, Tính A khi x = 5 + 2\(\sqrt{3}\)
c, Tìm x để |A| ≤ 1
\(a,A=\dfrac{2\sqrt{x}+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\)
29 A=\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\)
a. rút gọn A
b. Tính A với x=\(14-6\sqrt{5}\)
c. Tìm x khi A=1
\(a,A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(b,x=14-6\sqrt{5}=\sqrt{5^2}-2.3.\sqrt{5}+3^2=\left(\sqrt{5}-3\right)^2\)
\(\Rightarrow A=\dfrac{\sqrt{\left(\sqrt{5}-3\right)^2}}{14-6\sqrt{5}+\sqrt{\left(\sqrt{5}-3\right)^2}+1}\)
\(=\dfrac{\left|\sqrt{5}-3\right|}{-6\sqrt{5}+15+\left|\sqrt{5}-3\right|}\)
\(=\dfrac{3-\sqrt{5}}{-6\sqrt{5}+15+3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}}{18-7\sqrt{5}}\)
\(c,A=1\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=1\Leftrightarrow\sqrt{x}-x-\sqrt{x}-1=0\Leftrightarrow-x-1=0\Leftrightarrow x=-1\left(ktm\right)\)
Vậy khi A = 1 thì không có giá trị x thỏa mãn.
Cho biểu thức:
A = (\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)+\(\dfrac{\sqrt{x}}{\sqrt{x}-3}\)-\(\dfrac{3x+3}{x-9}\)) : (\(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}\) - 1)
a) Rút gọn A
b) Tính giá trị của A khi x = 13 - \(4\sqrt{3}\)
c) Tìm x để A < \(-\dfrac{1}{2}\)
d) Tìm x để A = \(\dfrac{-2}{3}\)
e) Tìm x \(\in\) Z để A nhận giá trị nguyên
f) Tìm GTNN của A
\(a,ĐK:x\ge0;x\ne9\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\\ b,x=13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\\ \Leftrightarrow A=\dfrac{-3}{2\sqrt{3}-1+3}=\dfrac{-3}{2\sqrt{3}+2}=\dfrac{-3\left(2\sqrt{3}-2\right)}{8}\)
\(c,A< -\dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\\ d,A=-\dfrac{2}{3}\Leftrightarrow\dfrac{3}{\sqrt{x}+3}=\dfrac{2}{3}\\ \Leftrightarrow2\sqrt{x}+6=9\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\\ e,\Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}=0\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x=0\left(tm\right)\\ f,\sqrt{x}+3\ge3\\ \Leftrightarrow A=-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{3}{3}=-1\\ A_{min}=-1\Leftrightarrow x=0\)
Cho biểu thức:
\(A=\left(1-\dfrac{\sqrt{x}}{\sqrt{x+1}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x+6}}\right)\)
a) Rút gọn A
b) Tìm x để A<0
c) Tìm giá trị nhỏ nhất của A
d) Tính giá trị nguyên của x để A nhận giá trị nguyên
Cho A = \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
a ) Rút gọn A
b) Tìm x ϵ Z để A ϵ Z
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)