Cho P=\(\left(\frac{2a+1}{a\sqrt{a}-1}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\times\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\left(ĐK:a\ge0,a\ne1\right)\)
a) Rút Gọn P
b) Xét Dấu Của Biểu Thức \(P\times\sqrt{1-a}\)
K
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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
\(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1+\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\frac{\sqrt{a}+1+a+\sqrt{a}}{\sqrt{a}+1}\right)\left(\frac{1-\sqrt{a}+a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\frac{a+2\sqrt{a}+1}{\sqrt{a}+1}.\frac{a-2\sqrt{a}+1}{1-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}.-\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}\)
\(=-\left(\sqrt{a}+1\right).\left(\sqrt{a}-1\right)\)
\(=1-a\)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
9 tháng 10 2020
a)
\(=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)
\(=\frac{4a\sqrt{a}}{a-1}.\frac{1}{2a\sqrt{a}}=\frac{2}{a-1}\)
b) \(\frac{2}{a-1}=a\Rightarrow a^2-a-2=0\)
Ta có: 1+1+(-2)=0, nên pt có 2 nghiệm a1=-1<0 (không thỏa mãn đk)=> loại
a2=2(thỏa mãn đk)=> chọn
Vậy a=2 thì P=a
31 tháng 7 2019
\(A=\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}+1\right).\)\(\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\)
\(=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\)\(\left(\frac{-\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-1\right)\)
\(=\left(\sqrt{a}+1\right)\left(-\sqrt{a}-1\right)\)
\(=-\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)=-\left(\sqrt{a}+1\right)^2\)
\(b,A=-a^2\Rightarrow-\left(\sqrt{a}+1\right)^2=a^2\)
\(\Leftrightarrow a=\sqrt{a}+1\Rightarrow a-\sqrt{a}-1=0\)
\(\Rightarrow4a-4\sqrt{a}-4=0\)
\(\Rightarrow4a-4\sqrt{a}+1-5=0\)
\(\Rightarrow\left(2\sqrt{a}-1\right)^2-\sqrt{5}^2=0\)
\(\Rightarrow\left(2\sqrt{a}-1+\sqrt{5}\right)\left(2\sqrt{a}-1-\sqrt{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2\sqrt{a}=1-\sqrt{5}\\2\sqrt{a}=1+\sqrt{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=\frac{1-\sqrt{5}}{2}\\\sqrt{a}=\frac{1+\sqrt{5}}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}a=\frac{\left(1-\sqrt{5}\right)^2}{4}\left(tm\right)\\a=\frac{\left(1+\sqrt{5}\right)^2}{4}\left(tm\right)\end{cases}}\)
Mong mọi người giúp đỡ mình , mình đang cần gấp , cảm ơn mọi người
Ta có HĐT : \(\hept{\begin{cases}a\sqrt{a}+b\sqrt{b}=\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\\a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\end{cases}\left(a,b\ge0\right)}\)
\(P=\left(\frac{2a+1}{a\sqrt{a}-1}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\times\left(\frac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\)
ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
\(=\left(\frac{2a+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{\sqrt{a}}{a+\sqrt{a}+1}\right)\times\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(\frac{2a+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\times\left(\frac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(\frac{2a+1-a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\times\left(1-\sqrt{a}+a-\sqrt{a}\right)\)
\(=\frac{a+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\times\left(a-2\sqrt{a}+1\right)\)
\(=\frac{1}{\sqrt{a}-1}\times\left(\sqrt{a}-1\right)^2\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\)
b) \(P\times\sqrt{1-a}\)
\(=\left(\sqrt{a}-1\right)\times\sqrt{1-a}\)
ĐKXĐ: \(0\le x< 1\)
Với \(0\le x< 1\)
Ta có :\(\hept{\begin{cases}\sqrt{a}\le\sqrt{1}=1\Rightarrow\sqrt{a}-1\le0\\\sqrt{1-a}\ge0\end{cases}}\)
\(\Rightarrow\left(\sqrt{a}-1\right)\left(\sqrt{1-a}\right)\le0\)