Cho:
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}-1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a,Rút gọn A
b,Tìm x \(\in\)Z để \(A\in Z\)
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 \(Q\in Z\)
1. A = \(\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
a) rút gọn
b) tìm x để A <-1
2. Cho A = \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x-2\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+3}{x-1}-\dfrac{2}{\sqrt{x}+1}\right)\)
a) Rút gọn
b) tìm x \(\in\) Z để A \(\in\) Z
1. a) \(A=\left(\dfrac{\sqrt{x}-1+x-\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}\right).\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)ĐK x\(\ne\)0,1
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(x-1\right)}=\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
b) A<-1 <=> \(\dfrac{2\sqrt{x}}{x-\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}+1< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x-\sqrt{x}}{x-\sqrt{x}}< 0\)\(\Leftrightarrow\dfrac{x+\sqrt{x}}{x-\sqrt{x}}< 0\)
\(\Leftrightarrow x-\sqrt{x}< 0\) (vì \(x+\sqrt{x}>0\left(\forall x>0\right)\))
\(\Leftrightarrow x< \sqrt{x}\Leftrightarrow x^2< x\Leftrightarrow x^2-x< 0\)
\(\Leftrightarrow x\in\left(0;1\right)\Leftrightarrow0< x< 1\)
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: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
Cho biểu thức:
\(P=\left(1+\dfrac{4}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\left(\dfrac{x+2\sqrt{x}}{x-1}\right)\) Với \(x>0;x\ne1\)
a, Rút gọn biểu thức.
b, Tìm \(x\in Z\) để P nhận giá trị nguyên.
a) \(P=\dfrac{x-1+4\left(\sqrt{x}+1\right)+1}{x-1}.\dfrac{x-1}{x+2\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+4}{x+2\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
b) \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}=1+\dfrac{2}{\sqrt{x}}\in Z\)
Do \(\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{1;4\right\}\)
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 sau: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
1, Rút gọn P
2, Tính giá trị nhỏ nhất của P
3, Tìm \(x\in Z\) sao cho \(Q=\dfrac{2\sqrt{x}}{P}\in Z\)
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
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
Bài 1: Cho A = \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
a) Rút gọn A
b) Tìm x để \(\left|A\right|>A\)
Bài 2: Cho B = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)
a) Rút gọn B
b) Tìm tất cả các giá trị của x sao cho B<0
6.A=\(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
a) Rút gọn A
b)Tìm a ϵ Z để biểu thức A nhận giá trị nguyên
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
Cho hai bt A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)và B=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\)
a) rút gọn B
b)tìm x thuộc Z để C= A(B-2) có giá trị nguyên
a) \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\left(đk:x\ge0,x\ne4\right)\)
\(=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)
c) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{-2}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0\right\}\)
\(\Rightarrow x\in\left\{0;1;9;16\right\}\)