Cho A=(\(\dfrac{x+2}{x\sqrt{x}-1}\)+\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)+\(\dfrac{1}{1-\sqrt{x}}\)):\(\dfrac{\sqrt{x}-1}{2}\)
a, Rút gọn A=\(\dfrac{2}{x+\sqrt{x}+1}\)
b,CM: A>0 và A≤2
Cho A= \(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}\)và B= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\)
a) rút gọn B
b) Cho x>0. so sánh A với 3
\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
`b,` Tớ tính mãi ko ra, xl cậu nha=')
Cho A= \(\dfrac{\sqrt{x}}{\sqrt{x}-1}\) và B= \(\dfrac{2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{x-1}\)
a) rút gọn B
b) Tìm x để \(\dfrac{B}{A}\)= \(\dfrac{1-\sqrt{x}}{2x^2}\)
Rút gọn :
B=\(\dfrac{2a\sqrt{1+x^2}}{\sqrt{1+x^2}-x}\) với x=\(\dfrac{1}{2}\left(\sqrt{\dfrac{1-a}{a}}-\sqrt{\dfrac{a}{1-a}}\right)\) và 0<a<1
Với \(x>0\) cho 2 biểu thức \(A=\dfrac{2+\sqrt{x}}{\sqrt{x}}\) và \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
1) Tính giá trị của biểu thức A khi \(x=64\)
2) Rút gọn biểu thức B
3) Tìm x để \(\dfrac{A}{B}>\dfrac{3}{2}\)
1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
2: \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3: A/B>3/2
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)
=>\(-\sqrt{x}+2>0\)
=>-căn x>-2
=>căn x<2
=>0<x<4
1) Thay x=64 vào A ta có:
\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)
2) \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
3) Ta có:
\(\dfrac{A}{B}>\dfrac{3}{2}\) khi
\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)
Mà: \(2\sqrt{x}\ge0\forall x\)
\(\Leftrightarrow2-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Kết hợp với đk:
\(0< x< 4\)
Cho biểu thức A= \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\) với x>0 và x\(\ne\)1. Rút gọn biểu thức A
Sửa đề: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
Bài 11:
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) CMR: 0 < A < 2
\(a,A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(x\ge0;x\ne1\right)\\ A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
\(b,x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\\ \Rightarrow\dfrac{2}{x+\sqrt{x}+1}>0\left(1\right)\)
\(\sqrt{x}+\dfrac{1}{2}\ge\dfrac{1}{2}\\ \Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2\ge\dfrac{1}{4}\\ \Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge1\\ \Leftrightarrow\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{2}{1}=2\\ \Rightarrow A< 2\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrow0< A< 2\)
A=\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}\)-\(\dfrac{1}{\sqrt{3}-\sqrt{2}}\)+\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
B=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}\)+\(\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)-\(\dfrac{3x+9}{x-9}\)với x≥0;x≠9
a. Rút gọn biểu thức A và B
b. Tìm x để một phần ba giá trị của A bằng giá trị của biểu thức B
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)
=2
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
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
Bài 1: Cho biểu thức A= \(\dfrac{x-4}{\sqrt{x}+2}\) B= \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) - \(\dfrac{\sqrt{x}-1}{2-\sqrt{x}}\)-\(\dfrac{9-x}{4-x}\) (x ≥ 0, x ≠ 4 )
a) Tính A khi x = \(\dfrac{1}{4}\)
b) Rút gọn B
c) Tìm các giá trị x nguyên sao cho A.B ≤ 2
(mink đag cần gấp)
a) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ
nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(A=\dfrac{x-4}{\sqrt{x}+2}\), ta được:
\(A=\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\left(\dfrac{1}{4}-\dfrac{16}{4}\right):\left(\dfrac{1}{2}+2\right)=\dfrac{-15}{4}:\dfrac{5}{2}\)
\(\Leftrightarrow A=\dfrac{-15}{4}\cdot\dfrac{2}{5}=\dfrac{-30}{20}=\dfrac{-3}{2}\)
Vậy: Khi \(x=\dfrac{1}{4}\) thì \(A=\dfrac{-3}{2}\)
b) Ta có: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-1}{2-\sqrt{x}}-\dfrac{9-x}{4-x}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{9-x}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2+x+2\sqrt{x}-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x-4+9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
Thay x = \(\dfrac{1}{4}\)vào bt A ta có: A= \(\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\dfrac{-15}{4}:\dfrac{5}{2}=\dfrac{-3}{2}\)
Vậy x = \(\dfrac{1}{4}\)vào bt A nhận giá trị là -3/2
b)
Rút gọn:
1) \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\) với x >1
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\) với x ≠ 4, x ≠16, x >0
a: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-\left(2\sqrt{x}+1\right)+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)
b: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
\(=\dfrac{x-4+5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)