Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(P=\dfrac{\sqrt{a}\left[\left(\sqrt{a}\right)^3+1\right]}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(P=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(P=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(P=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(P=a-\sqrt{a}\)
b) Với a > 1 thì \(a>\sqrt{a}\) , do đó \(P=a-\sqrt{a}>0\), suy ra \(\left|P\right|=P\)
c) \(A=a-\sqrt{a}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Vậy A nhỏ nhất bằng \(-\dfrac{1}{4}\) khi cà chỉ khi \(\sqrt{a}=\dfrac{1}{2}\) hay \(a=\dfrac{1}{4}\)
a: \(P=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}\)
b: a>1 nên P>0
\(\Leftrightarrow P=\left|P\right|\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
\(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{a-1}\)
\(=\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-a-\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b: \(P=\sqrt{a}+7\)
=>\(2\left(a+2\sqrt{a}+1\right)=a+7\sqrt{a}\)
=>\(2a+4\sqrt{a}+2-a-7\sqrt{a}=0\)
=>\(a-3\sqrt{a}+2=0\)
=>\(\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>\(\left[{}\begin{matrix}a=1\left(loại\right)\\a=4\left(nhận\right)\end{matrix}\right.\)
c: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}=\dfrac{2a-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}\right)}{\sqrt{a}}=\dfrac{2\left[\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}{\sqrt{a}}>0\)
=>P>6
a) \(P=\dfrac{1-2\sqrt{a}+a}{1-\sqrt{a}}\cdot\dfrac{1+2\sqrt{a}+a}{1+\sqrt{a}}\) \(=\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)\) \(=1-a\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)
Để P>0 \(\Leftrightarrow1-a>0\) \(\Leftrightarrow a< 1\)
Vậy \(0\le a< 1\)
`a)P=((1-asqrta)/(1-sqrta)+sqrta).((1+asqrta)/(1+sqrta)-sqrta)`
`=(((1-sqrta)(a+sqrta+1))/(1-sqrta)+sqrta).(((1+sqrta)(a-sqrta+1))/(1+sqrta)-sqrta)`
`=(a+sqrta+1+sqrta)(a-sqrta+1-sqrta)`
`=(a+2sqrta+1)(a-2sqrta+1)`
`=(sqrta+1)^2(sqrta-1)^2`
`=(a-1)^2`
`b)a<7-4sqrt3`
`<=>(a-1)^2<(2-sqrt3)^2`
`<=>sqrt3-2<a-1<2-sqrt3`
`<=>sqrt3-1<a<3-sqrt3`
\(=\dfrac{a+1-1}{\sqrt{a+1}}\cdot\dfrac{a^2+3\sqrt{a+1}-2a+2a-a^2}{a}\)
\(=\dfrac{3\sqrt{a+1}}{\sqrt{a+1}}=3\)
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
a)
ĐK: \(a>0\)
\(P=\dfrac{\sqrt{a}\left(\sqrt{a}^3+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\\ =\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\\ =a+\sqrt{a}-2\sqrt{a}-1+1\\ =a-\sqrt{a}\)
b)
\(a>1\Rightarrow\sqrt{a}-1>0\Rightarrow\sqrt{a}\left(\sqrt{a}-1\right)>0\)
\(\Rightarrow\left|P\right|=P\)
Cho tớ hỏi sao lại khử hết mẫu vậy cậu?