Cho A = \(\dfrac{\sqrt{x}}{2\sqrt{x}+1}\); B = \(\dfrac{1}{2\sqrt{x}+1}\)(ĐKXĐ: x ≥ 0; x ≠ \(\dfrac{1}{4}\)). Tìm tất cả các giá trị của x để biểu thức: P = 5A + B nguyên.
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 \(A=\dfrac{x}{\sqrt{x}-1}\\ B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\div\left(\dfrac{2}{x}+\dfrac{x+2}{x\left(\sqrt{x}-1\right)}\right)\)
ĐKXĐ : x > 0 ; x ≠ 1
Tìm GTNN của \(\sqrt{A}\)
Bài 2 :
Cho \(A=\dfrac{\sqrt{x}-2}{3}\\ B=\dfrac{3x+4}{x-2\sqrt{x}}+\dfrac{2}{\sqrt{x}}-\dfrac{2\sqrt{x}}{\sqrt{x}-2}\)
Cho x ∈ N , tìm GTLN của \(\sqrt{B}\)
Cho biểu thức A:
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+1+\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2\sqrt{x}}\)
a) Rút gọn A.
b) cmr: \(A< \dfrac{2}{3}\)
a) ĐKXĐ: \(x>0,x\ne1\)
\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}=\dfrac{2\sqrt{x}}{x+\sqrt{x}+1}\)
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=')
1.cho biểu thức A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}\)với(x≥0;x≠4)
a)rút gọn A
b)tính A khi x=6+4\(\sqrt{2}\)
2.cho biểu thức P=\(\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)với x≥0;x≠1;x≠4
a)rút gọn P
b)tìm x để P=-4
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 \(\dfrac{B}{A}\)= \(\dfrac{1-\sqrt{x}}{2x^2}\)
1) Rút gọn và tính giá trị của biểu thức A=\(\sqrt{x}\).(\(\sqrt{x}\)+1)-(\(\sqrt{x}\)-1)2 - 2 , tại x=9
2) Cho biểu thức A=(\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) -\(\dfrac{2\sqrt{x}+7}{x-4}\) ) : (\(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1\) ) với x≥0 ; x≠4
a) Rút gọn biểu thức A
b) Tìm các giá trị của x để A≥ -2
1) \(A=\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)^2-2\)
\(A=\sqrt{x}\cdot\sqrt{x}+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-x+2\sqrt{x}-1-2\)
\(A=3\sqrt{x}-3\)
Thay \(x=9\) vào A ta có:
\(A=3\cdot\sqrt{9}-3=3\cdot3-3=9-3=6\)
Cho A=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-1}+4\sqrt{x}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x+1}}\right)\)
Rút gọn A
Sửa đề: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}+4\sqrt{x}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)+4\sqrt{x}\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2+4\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}\left[1+2\left(x+2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\right]}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{2\left[1+2\left(x\sqrt{x}-x+2x-2\sqrt{x}+\sqrt{x}-1\right)\right]}{x-1}\)
\(=\dfrac{2\left[1+2x\sqrt{x}+2x-2\sqrt{x}-2\right]}{x-1}=\dfrac{2\left(2x\sqrt{x}+2x-2\sqrt{x}-1\right)}{x-1}\)
cho biểu thức A=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
a,RGA
A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\) ;B=\(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
Cho P=\(\dfrac{A}{B}\) tìm x thỏa mãn: P.x≤\(10\sqrt{x}-29-\sqrt{x-25}\)
Ta có:
\(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) (ĐK: \(x\ne4;x\ge0\))
\(B=\dfrac{x}{\left(\sqrt{x}\right)^2-2^2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(\Rightarrow P=\dfrac{A}{B}=\dfrac{\dfrac{\sqrt{x}+2}{\sqrt{x}}}{\dfrac{\sqrt{x}}{\sqrt{x}-2}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\cdot\sqrt{x}}=\dfrac{x-4}{x}\) (ĐK: \(x\ne0\))
Theo đề ta có:
\(P\cdot x\le10\sqrt{x}-29-\sqrt{x}+25\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\dfrac{x-4}{x}\cdot x\le9\sqrt{x}-4\)
\(\Leftrightarrow x-4\le9\sqrt{x}-4\)
\(\Leftrightarrow x-9\sqrt{x}\le0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-9\right)\le0\)
Mà: \(\sqrt{x}\ge0\)
\(\Leftrightarrow\sqrt{x}-9\le0\)
\(\Leftrightarrow\sqrt{x}\le9\)
\(\Leftrightarrow x\le81\)
Kết hợp với đk:
\(0\le x\le81\)