1. Rút gọn biểu thức B = ((x + 2sqrt(x))/(sqrt(x)) + sqrt(x) - 2) / (sqrt(x)) với x > 0.
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22 tháng 7 2023
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right)\cdot\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3+2\sqrt{x}}{x-9}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2}{\sqrt{x}-3}=\dfrac{6}{\sqrt{x}-3}\)
3 tháng 8 2023
a: \(B=\dfrac{x-2\sqrt{x}}{\sqrt{x}-2}-\dfrac{2x+12\sqrt{x}+18}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}-\dfrac{2\left(x+6\sqrt{x}+9\right)}{\sqrt{x}+3}\)
=căn x-2(căn x+3)
=-căn x-6
b: B+8>0
=>-căn x-6+8>0
=>-căn x+2>0
=>-căn x>-2
=>căn x<2
=>0<=x<4
15 tháng 12 2023
a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=-\dfrac{2}{\sqrt{x}+1}\)
c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)
d: |B|=A
=>\(\left|-\dfrac{2}{\sqrt{x}+1}\right|=4\)
=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)
=>\(\sqrt{x}+1=\dfrac{1}{2}\) hoặc \(\sqrt{x}+1=-\dfrac{1}{2}\)
=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)
AH
Akai Haruma
Giáo viên
19 tháng 8 2023
Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn.
16 tháng 10 2023
1: Khi x=36 thì \(A=\dfrac{6}{2\cdot6-4}=\dfrac{6}{12-4}=\dfrac{6}{8}=\dfrac{3}{4}\)
2:
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >4\end{matrix}\right.\)
\(C=B:A\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{3\sqrt{x}-x}{x-4}\right):\dfrac{\sqrt{x}}{2\sqrt{x}-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+3\sqrt{x}-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}-x}{\sqrt{x}+2}\cdot\dfrac{2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+2}\)
3: \(C\cdot\sqrt{x}< \dfrac{4}{3}\)
=>\(\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{3}< 0\)
=>\(\dfrac{2\sqrt{x}\cdot3-4\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)
=>\(6\sqrt{x}-4\sqrt{x}-8< 0\)
=>\(2\sqrt{x}-8< 0\)
=>\(\sqrt{x}< 4\)
=>\(0< =x< 16\)
Kết hợp ĐKXĐ của C, ta được: \(\left\{{}\begin{matrix}0< x< 16\\x< >4\end{matrix}\right.\)
27 tháng 8 2023
a: Khi x=16 thì \(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{4}=\dfrac{3}{2}\)
b: P=A:B
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{6}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
c: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt{x}}>0\)
=>P>1
4 tháng 1
1: \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\cdot\left(\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
2: Thay x=9 vào A, ta được:
\(A=\dfrac{3}{3+1}=\dfrac{3}{4}\)
7 tháng 12 2021
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
TP
8 tháng 12 2021
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế
31 tháng 8 2023
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
H9
HT.Phong (9A5)
CTVHS
31 tháng 8 2023
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:
\(0< x< 4\)
\(B=\left(\dfrac{x+2\sqrt{x}}{\sqrt{x}}+\sqrt{x}-2\right):\sqrt{x}\left(x>0\right)\\ =\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}+\sqrt{x}-2\right].\dfrac{1}{\sqrt{x}}\\ =\left(\sqrt{x}+2+\sqrt{x}-2\right).\dfrac{1}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)