A=(\(\dfrac{1}{\sqrt{x}+1}\)+\(\dfrac{1}{x-1}\))chia \(\dfrac{x-2\sqrt{x}}{x-1}\)
a. Rút gọn a
b. tìm x để A bằng 1
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
a: \(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}=\dfrac{2x-3\sqrt{x}+1}{x-1}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b: Để A nguyên thì \(2\sqrt{x}+2-3⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;3\right\}\)
=>x=0 hoặc x=4
c: Để A<1 thì A-1<0
=>\(\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\)
=>căn x-2<0
=>0<=x<4
a: \(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}=\dfrac{2x-3\sqrt{x}+1}{x-1}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b: Để A nguyên thì \(2\sqrt{x}+2-3⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;3\right\}\)
=>x=0 hoặc x=4
c: Để A<1 thì A-1<0
=>\(\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\)
=>căn x-2<0
=>0<=x<4
Câu 2,cho biểu thức
A=\(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
a,Rút gọn A
b,tìm x để A=1/2
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\cdot\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Cho biểu thức:
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\right):\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
a. Rút gọn A
b. Tìm x để A = 2
a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\right):\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2x-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}=-\dfrac{1}{\sqrt{x}-1}\)
b) \(A=2\Rightarrow\dfrac{-1}{\sqrt{x}-1}=2\Rightarrow-1=2\sqrt{x}-2\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}\)
Lời giải:
ĐK: $x\geq 0; x\neq 1$
a.
\(A=\frac{\sqrt{x}(\sqrt{x}-1)-x}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{2x-\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(=\frac{-\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}:\frac{x-\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{-\sqrt{x}}{x-\sqrt{x}}=\frac{-\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}=\frac{1}{1-\sqrt{x}}\)
b.
$A=2\Leftrightarrow 1-\sqrt{x}=\frac{1}{2}$
$\Leftrightarrow \sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}$ (tm)
A=(\(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{x}{x-1}\) )\(\div\left(\dfrac{2x}{x-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) (ĐK \(x\ge0,x\ne1\) )
=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x}{x-1}\div\left(\dfrac{2x-\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}\right)\)
=\(\dfrac{x-\sqrt{x}-x}{x-1}.\dfrac{x-1}{2x-x-\sqrt{x}}=\dfrac{-\sqrt{x}}{x-1}.\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
=\(\dfrac{-1}{\sqrt{x}-1}\)
câu b
Khi x=2\(\Rightarrow A=\dfrac{-1}{\sqrt{2}-1}=\dfrac{-\left(\sqrt{2}+1\right)}{2-1}=-\left(\sqrt{2}+1\right)\)
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 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 \(A=\dfrac{1}{2}\).
\(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right).\left(\dfrac{X-\sqrt{X}}{\sqrt{X}+1}-\dfrac{X+\sqrt{X}}{\sqrt{X}-1}\right)\)
a) Rút gọn biểu thức A
b) Tìm giá trị của x để A>-6
a: \(A=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)-\sqrt{x}\left(x+2\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{2\sqrt{x}}=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
b: Để A>-6 thì \(2\sqrt{x}< 6\)
=>0<x<9
Kết hợp ĐKXĐ, ta được:
0<x<9 và x<>1
\(a,\)
\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{-4x}{x-1}\)
\(=-2\sqrt{x}\)
Vậy \(A=-2\sqrt{x}\)
\(b,\)Đề \(A\ge-6\) thì \(-2\sqrt{x}\ge-6\) \(\Leftrightarrow\sqrt{x}\le3\) \(\Leftrightarrow x\le3^2\Leftrightarrow x\le9\)
Vậy \(x\le9\) thì \(A\ge-6\)
A=\(\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{1}{x-\sqrt{x}}\)
a,rút gọn A
b,tìm x để A=5
a: ĐKXĐ: x>0 và x<>1
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{1}{x-\sqrt{x}}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{1}{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}\)
\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{1}\)
\(=\sqrt{x}+1\)
b: A=5
=>\(\sqrt{x}+1=5\)
=>\(\sqrt{x}=4\)
=>x=16(nhận)
Cho biểu thức A= \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
a, Rút gọn biểu thức A
b, Tìm các giá trị để \(\dfrac{P}{A}\left(x-1\right)=0\)
\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) \(\dfrac{P}{A}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)
Vậy \(S=\varnothing\)