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: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
dễ mà bạn quy đồng biến đỗi là ra chứ làm đánh mấy bài này ra tốn tg lắm
Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)
\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
\(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{9x-1}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}=\dfrac{3x+3\sqrt{x}-1}{9x-1}.\dfrac{3\sqrt{x}+1}{3}=\dfrac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
\(1,P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(dkxd:x\ge0,x\ne9\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)
\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{-\sqrt{x}+5}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{5-\sqrt{x}}\)
\(=-\dfrac{x}{5-\sqrt{x}}\)
\(2,x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)
\(=2+\sqrt{3}+2-\sqrt{3}=4\)
\(x=4\Rightarrow P=-\dfrac{4}{5-\sqrt{4}}=\dfrac{-4}{5-2}=-\dfrac{4}{3}\)
a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
\(=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}-\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\right)\cdot\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\right)\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\)
b) Ta có: \(x=3-2\sqrt{2}\)
\(=2-2\cdot\sqrt{2}\cdot1+1\)
\(=\left(\sqrt{2}-1\right)^2\)
Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được:
\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\dfrac{\sqrt{2}-\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{2}-\sqrt{2}+1}{\sqrt{2}-1}\)
\(=\dfrac{1}{\sqrt{2}-1}\)
\(=\sqrt{2}+1\)
Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}}{2x-1}-1\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{2x}-1+2x-2x+1}{2x-1}=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}\)
\(B=\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=1+\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{2x-1}\)
\(=1+\dfrac{-2\sqrt{x}-1-2x}{2x-1}\)
\(=\dfrac{2x-1-2\sqrt{x}-1-2x}{2x-1}=\dfrac{-2-2\sqrt{x}}{2x-1}\)
\(P=A:B=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}:\dfrac{-2\sqrt{x}-2}{2x-1}\)
\(=\dfrac{2\sqrt{x}\left(x+\sqrt{2}\right)}{2x-1}\cdot\dfrac{2x-1}{-2\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}\left(x+\sqrt{2}\right)}{\sqrt{x}+1}\)
b: Thay \(\sqrt{x}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\) vào P, ta được:
\(P=\left[-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\left(\dfrac{3+2\sqrt{2}}{2}+\sqrt{2}\right)\right]:\left[\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}+1\right]\)
\(=\left[\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\dfrac{3+4\sqrt{2}}{2}\right]:\left[\dfrac{2+\sqrt{2}+2}{2}\right]\)
\(=\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{4}\cdot\dfrac{2}{4+\sqrt{2}}\)
\(=\dfrac{-\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{2\cdot\left(2\sqrt{2}+1\right)}=\dfrac{-\left(4\sqrt{2}+3\right)}{3\cdot\left(3+\sqrt{2}\right)}\)
a
ĐK: \(1< x\ne10\)
Đặt \(t=\sqrt{x-1}\Rightarrow x=t^2+1;0< t\ne3\)
Khi đó:
\(P=\left(\dfrac{t}{3+t}+\dfrac{t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t^2-3t}-\dfrac{1}{t}\right)\\ =\left(\dfrac{t\left(3-t\right)+t^2+9}{\left(3-t\right)\left(3+t\right)}\right):\left(\dfrac{3t+1}{t\left(t-3\right)}-\dfrac{1}{t}\right)\\ =\dfrac{3t+9}{\left(3-t\right)\left(3+t\right)}:\dfrac{3t+1-t+3}{t\left(t-3\right)}=\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}:\dfrac{2t+4}{t\left(t-3\right)}\\ =\dfrac{3\left(t+3\right)}{\left(3-t\right)\left(3+t\right)}.\dfrac{t\left(t-3\right)}{2t+4}=\dfrac{-3t}{2t+4}=\dfrac{-3\sqrt{x-1}}{2\sqrt{x-1}+4}\)
b
Ta có:
\(x=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{5}+1\right)\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{5}\left|1-\sqrt{2}\right|\)
\(=\sqrt{2}+1-\left(\sqrt{5}+1\right)\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\)
\(=\sqrt{2}+1-\sqrt{5}\left|1-\sqrt{2}\right|-\left|1-\sqrt{2}\right|+\sqrt{5}\left|1-\sqrt{2}\right|\\ =\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)
Vậy \(P=\dfrac{-3\sqrt{2-1}}{2\sqrt{2-1}+4}=-\dfrac{1}{2}\)
a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
\(=\dfrac{2x-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{x\left(x-1\right)}{x+\sqrt{x}+1}\)