Cho 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) Tìm x để A nguyên
c) Tìm x để A<1
cho biểu thức A= \(\left(\dfrac{1}{x-4}+\dfrac{1}{\sqrt{x}+2}\right)\):\(\dfrac{\sqrt{x}-1}{x+2\sqrt{x}}\)
a) nêu đkxđ và rút gọn
b) tìm giá trị nguyên của x để A có giá trị nguyên
c) tìm x để A<0
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
\(A=\left(\dfrac{1}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{1+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\inƯ\left(2\right)\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{3;1;4;0\right\}\)
=>\(x\in\left\{9;1;16;0\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{9;16\right\}\)
c: A<0
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
Kết hợp ĐKXĐ, ta được: 0<x<4 và 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
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 \(A=\dfrac{1}{2}\).
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=\(\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
a rút gọn a
b tìm x để a=7
c tính giá trị của a khi x=2(2+\(\sqrt{3}\))
d tìm x để a<1
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1$
a)
\(A=\frac{x+\sqrt{x}+1}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{(\sqrt{x}-1)(x+1)}\right]\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+1)}=\frac{x+\sqrt{x}+1}{x+1}.\frac{(\sqrt{x}-1)(x+1)}{(\sqrt{x}-1)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b)
\(A=7\Leftrightarrow x+\sqrt{x}+1=7(\sqrt{x}-1)\)
\(\Leftrightarrow x-6\sqrt{x}+8=0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x=4\\ x=16\end{matrix}\right.\) (đều thỏa mãn)
c)
\(x=2(2+\sqrt{3})=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\frac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}}=\frac{6+3\sqrt{3}}{\sqrt{3}}=3+2\sqrt{3}\)
d)
\(A< 1\Leftrightarrow \frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{x-2\sqrt{x}+2}{\sqrt{x}-1}<0\)
\(\Leftrightarrow \frac{(\sqrt{x}-1)^2+1}{\sqrt{x}-1}<0\Leftrightarrow \sqrt{x}-1< 0\Leftrightarrow 0\leq x< 1\)
1) Cho biểu thức A= (\(\dfrac{3\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{1}{\sqrt{x}+1}\)- 3) . \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) với x≥0 và x≠1
a) rút gọn A
b) tìm x để A<0
\(a,A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(đk:x\ge0;x\ne1\right)\)
\(=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{3x+2\sqrt{x}+1-3x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2}{\sqrt{x}-1}\)
\(---\)
\(b,A< 0\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Kết hợp với điều kiện của \(x\), ta được:
\(0\le x< 1\)
Vậy: ...
\(Toru\)
a) \(A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2}{\sqrt{x}-1}\)
b) \(A< 0\) khi
\(\dfrac{2}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Kết hợp với đk:
\(0\le x< 1\)
Cho \(A=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)vớix\ge0,x\ne1\)
a) Rút gọn A
b) Tìm x để A=3
c) Tìm x ϵ N để A≤ 5
a: \(A=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
b: A=3
=>căn x-1=3
=>căn x=4
=>x=16
c: A<=5
=>căn x-1<=5
=>căn x<=6
=>0<=x<=36
=>\(x\in\left\{0;2;3;4;...;36\right\}\)
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
Cho \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\) với \(x\ge0;x\ne9\)
a. Rút gọn A
b. Tìm x để \(A>\dfrac{1}{3}\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
b: Để \(A>\dfrac{1}{3}\) thì \(A-\dfrac{1}{3}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)
a) \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\left(đk:x\ge0,x\ne0\right)\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\)
b) \(A>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\)
\(\Leftrightarrow6>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)