cho biểu thức A=\(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (với a>0)
a.rút gọn biểu thức A
b.tính giá trị nhỏ nhất của A
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
6 tháng 2 2022
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 \(a=\dfrac{1}{4}\)
6 tháng 2 2022
a: \(P=\sqrt{a}\left(\sqrt{a}+1\right)-2\sqrt{a}-1+1=a-\sqrt{a}\)
b: a>1 nên P>0
\(\Leftrightarrow P=\left|P\right|\)
12 tháng 4 2021
a) \(A=\dfrac{3\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{3\sqrt{a}-3-\sqrt{a}-1-\sqrt{a}+3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{1}{\sqrt{a}+1}\)
b) Với \(a=3-2\sqrt{2}\)(tmđk)
\(A=\dfrac{1}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\dfrac{\sqrt{a}-1}{a-1}\)
\(=\dfrac{\sqrt{3-2\sqrt{2}}-1}{3-2\sqrt{2}-1}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}-1}{2-2\sqrt{2}}=\dfrac{\left|\sqrt{2}-1\right|-1}{2-2\sqrt{2}}=\dfrac{\sqrt{2}-1-1}{2-2\sqrt{2}}=\dfrac{\sqrt{2}-2}{2-2\sqrt{2}}=\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{2\left(1-\sqrt{2}\right)}=\dfrac{\sqrt{2}}{2}\)
14 tháng 2 2021
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x-1}\right)^2}\)
\(=\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{x-1}\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
b) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{2\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}\)
\(\Leftrightarrow\sqrt{x}+1-6\sqrt{x}=0\)
\(\Leftrightarrow-5\sqrt{x}+1=0\)
\(\Leftrightarrow-5\sqrt{x}=-1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{25}\)(nhận)
Vậy: Để \(A=\dfrac{1}{3}\) thì \(x=\dfrac{1}{25}\)
6 tháng 11 2021
trình bày rõ ràng ra bạn còn câu b nữa
5 tháng 7 2021
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
AT
5 tháng 7 2021
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
12 tháng 8 2021
a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Thay x=3+2\(\sqrt{2}\)
A=\(\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}\)=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2-2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
A=\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0
\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)
9 tháng 12 2023
a: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{4}{\sqrt{x}+1}\)
b: Để A=-B thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{-4}{\sqrt{x}+1}\)
=>\(\left(\sqrt{x}+1\right)^2=-4\left(\sqrt{x}-1\right)\)
=>\(x+2\sqrt{x}+1+4\sqrt{x}-4=0\)
=>\(x+6\sqrt{x}-3=0\)
=>\(x+6\sqrt{x}+9-12=0\)
=>\(\left(\sqrt{x}+3\right)^2=12\)
=>\(\left[{}\begin{matrix}\sqrt{x}+3=2\sqrt{3}\\\sqrt{x}+3=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\sqrt{3}-3\left(vôlý\right)\\\sqrt{x}=2\sqrt{3}-3\end{matrix}\right.\)
=>\(\sqrt{x}=2\sqrt{3}-3\)
=>\(x=\left(2\sqrt{3}-3\right)^2=21-12\sqrt{3}\)
Lời giải:
a.
\(A=\frac{\sqrt{a}(a\sqrt{a}+1)}{a-\sqrt{a}+1}-\frac{\sqrt{a}(2\sqrt{a}+1)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}(\sqrt{a}+1)(a-\sqrt{a}+1)}{a-\sqrt{a}+1}-(2\sqrt{a}+1)+1\)
\(=\sqrt{a}(\sqrt{a}+1)-(2\sqrt{a}+1)+1=a-\sqrt{a}\)
b.
$A=a-\sqrt{a}=(\sqrt{a}-0,5)^2-0,25\geq -0,25$ với mọi $a>0$
Vậy $A_{\min}=-0,25$ khi $\sqrt{a}-0,5=0$
$\Leftrightarrow a=0,25$
cho mình hỏi làm sao để tách\(a\sqrt{a}+1=\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)\)