Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
a: \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6}{\sqrt{x}-1}-\dfrac{2\sqrt{3}}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-6-2\sqrt{3}}{\sqrt{x}-1}\)
b: \(=\dfrac{3-\sqrt{x}-1+\sqrt{x}+5\sqrt{x}}{\sqrt{x}-2}=\dfrac{5\sqrt{x}+2}{\sqrt{x}-2}\)
c: \(=\dfrac{2-6\sqrt{x}-1+\sqrt{x}-3+\sqrt{x}}{\sqrt{x}-4}\)
\(=\dfrac{-4\sqrt{x}-4}{x-4}\)
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) Để \(P< \dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Rightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\sqrt{x}+2}< 0\Rightarrow\dfrac{\sqrt{x}-3}{2\sqrt{x}+2}< 0\)
mà \(2\sqrt{x}+2>0\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
\(\Rightarrow0\le x< 9\left(x\ne1\right)\)
\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{\sqrt{3}-x}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Ta có: \(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
`a)P=(x^2+sqrtx)/(x-sqrtx+1)-(2x+sqrtx)/sqrtx`
`P=(sqrtx(sqrtx+1)(x-sqrtx+1))/(x-sqrtx+1)-(sqrtx(2sqrtx+1))/sqrtx`
`P=x+sqrtx-2sqrtx-1`
`P=x-sqrtx-1`
a: Ta có: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=x+\sqrt{x}-2\sqrt{x}-1\)
\(=x-\sqrt{x}-1\)
Câu 2:
2) Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
Câu 2 :
Gọi : vận tốc của người đi chậm là : x (km/h) ( x > 0 )
Vận tốc của người đi nhanh : x + 4 (km/h)
Vi : người đi chậm đến muộn hơn : 45 phút \(=\dfrac{3}{4}\left(h\right)\)
Khi đó :
\(\dfrac{36}{x}-\dfrac{36}{x+4}=\dfrac{3}{4}\)
\(\Leftrightarrow\left[36\cdot\left(x+4\right)-36x\right]\cdot4=3x\cdot\left(x+4\right)\)
\(\Leftrightarrow3x^2+12x-144=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\left(n\right)\\x=16\left(l\right)\end{matrix}\right.\)
Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+3}+\dfrac{2-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3\sqrt{x}+2\sqrt{x}-2+2-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)