\(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2,với0\le a\ne1\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
31 tháng 7 2020
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
13 tháng 9 2018
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
1 tháng 10 2016
Điều kiện xác định \(x\ge0\)
\(A=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=x-\sqrt{x}-\left(x+\sqrt{x}\right)=-2\sqrt{x}\)
\(B=\frac{1}{3}-\sqrt{A+x+1}=\frac{1}{3}-\sqrt{x-2\sqrt{x}+1}=\frac{1}{3}-\sqrt{\left(\sqrt{x}-1\right)^2}=\frac{1}{3}-\left|\sqrt{x}-1\right|\)
\(=\frac{1}{3}-\left(1-\sqrt{x}\right)=\sqrt{x}-\frac{2}{3}\) (vì \(0\le x\le1\))
11 tháng 8 2020
a) Ta có: \(A=\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\sqrt{\left(\sqrt{5}-1\right)^2}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(2\sqrt{4+\left|\sqrt{5}-1\right|}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)(Vì \(\sqrt{5}>1\))
\(=\left(2\sqrt{4+\sqrt{5}-1}\right)\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\cdot\left(\sqrt{5}-1\right)\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{5+2\cdot\sqrt{5}\cdot1+1}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left|\sqrt{5}+1\right|\)
\(=2\cdot\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)\)
\(=2\cdot\left(5-1\right)\)
\(=2\cdot4=8\)
b) Ta có: \(B=\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}+\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\frac{2}{a+1}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}-1\right)^2+\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\right)\cdot\left(\frac{a+1-2}{a+1}\right)^2\)
\(=\frac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2a+2}{\left(a-1\right)}\cdot\frac{\left(a-1\right)^2}{\left(a+1\right)^2}\)
\(=\frac{2\left(a+1\right)\cdot\left(a-1\right)}{\left(a+1\right)^2}\)
\(=\frac{2a-2}{a+1}\)
KT
2 tháng 7 2018
ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}.\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\)
\(=\frac{1+2\sqrt{a}+a}{\sqrt{a}+1}.\frac{1-2\sqrt{a}+a}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.\frac{\left(1-\sqrt{a}\right)^2}{1-\sqrt{a}}\)
\(=\left(\sqrt{a}+1\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
10 tháng 3 2019
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}\right)\left(\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}.\frac{a-2\sqrt{a}+1}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.-\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(=-\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right)\)
\(=1-a\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 9 2019
\(P=\left(\frac{1}{a-1}+\frac{3\sqrt{a}+5}{a\left(\sqrt{a}-1\right)-\left(\sqrt{a}-1\right)}\right).\frac{\left(\sqrt{a}+1\right)^2}{4a}\)
\(=\left(\frac{\sqrt{a}-1}{\left(a-1\right)\left(\sqrt{a}-1\right)}+\frac{3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}\)
\(=\left(\frac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}\right).\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}\)
\(=\frac{4}{\left(\sqrt{a}-1\right)^2}.\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\)
Không rút gọn được nữa, chắc do bạn ghi sai đề
Ở đằng sau biểu thức là \(\left(\frac{\left(\sqrt{a}+1\right)^2}{4\sqrt{a}}-1\right)\) sẽ hợp lý hơn, khi đó sẽ rút gọn được