Cho A = \(\frac{a-1}{\sqrt{a}}\)
Tính A theo a = \(4+2\sqrt{3}\)
K
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HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023
\(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}=\dfrac{a^2\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{4}{5}}}{a^{\dfrac{1}{4}}}=\dfrac{a^{\dfrac{47}{15}}}{a^{\dfrac{1}{4}}}=a^{\dfrac{173}{60}}\)
\(\Rightarrow log_a\left(\dfrac{a^2\cdot\sqrt[3]{a}\cdot\sqrt[5]{a^4}}{\sqrt[4]{a}}\right)=log_a\left(a^{\dfrac{173}{60}}\right)=\dfrac{173}{60}\)
\(a^{2log_a\left(\dfrac{\sqrt{105}}{30}\right)}=a^{log_a\left(\dfrac{7}{60}\right)}=\dfrac{7}{60}\)
Vậy \(B=\dfrac{173}{60}+\dfrac{7}{60}=\dfrac{180}{60}=3\)
31 tháng 10 2016
a/ Điều kiện \(\hept{\begin{cases}a\ge0\\a\ne\frac{1}{9}\end{cases}}\) \(\Rightarrow0\le a\ne\frac{1}{9}\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(a-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(1-3\sqrt{a}\right)+\left(\sqrt{a}-2\right)\left(1+3\sqrt{a}\right)+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}:\left(\frac{3a\sqrt{a}-2\sqrt{a}+6-a}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}-6a+\sqrt{a}+3a-2-6\sqrt{a}+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}.\left(\frac{3\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\right)\)
\(=\frac{3a-2\sqrt{a}-1}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
\(=\frac{\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)
Hình như đề sai rồi bạn :(
1 tháng 11 2016
a/ Điều kiện xác định : \(\hept{\begin{cases}a\ge0\\a\ne9\end{cases}\Leftrightarrow}0\le a\ne9\)
b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(1-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)
\(=\frac{2\sqrt{a}\left(3\sqrt{a}-1\right)+\left(2-\sqrt{a}\right)\left(3\sqrt{a}+1\right)-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}:\frac{\sqrt{a}+5}{3\sqrt{a}-1}\)
\(=\frac{6a-2\sqrt{a}+6\sqrt{a}+2-3a-\sqrt{a}-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}.\frac{3\sqrt{a}-1}{\sqrt{a}+5}\)
\(=\frac{3a-2\sqrt{a}-1}{3\sqrt{a}+1}.\frac{1}{\sqrt{a}+5}\)
\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(3\sqrt{a}+1\right)\left(\sqrt{a}+5\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+5}\)
c/ \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\) thay vào M được
\(\frac{\sqrt{5}-2-1}{\sqrt{5}-2+5}=\frac{\sqrt{5}-3}{\sqrt{5}+3}=\frac{-7+3\sqrt{5}}{2}\)
d/ \(M=\frac{\sqrt{a}-1}{\sqrt{a}+5}=\frac{\sqrt{a}+5-6}{\sqrt{a}+5}=1-\frac{6}{\sqrt{a}+5}\)
Với mọi \(0\le a\ne9\) thì ta luôn có \(\sqrt{a}+5\ge5\Leftrightarrow\frac{6}{\sqrt{a}+5}\le\frac{6}{5}\Leftrightarrow-\frac{6}{\sqrt{a}+5}\ge-\frac{6}{5}\Leftrightarrow1-\frac{6}{\sqrt{a}+5}\ge1-\frac{6}{5}\)
\(\Rightarrow M\ge-\frac{1}{5}\)
Đẳng thức xảy ra khi a = 0
Vậy giá trị nhỏ nhất của M bằng \(-\frac{1}{5}\) khi a = 0
3 tháng 8 2017
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
AH
13 tháng 8 2018
mình biết nội quy rồi nên đưng đăng nội quy
ai chơi bang bang 2 kết bạn với mình
mình có nick có 54k vàng đang góp mua pika
ai kết bạn mình cho
KT
13 tháng 8 2018
\(A=\frac{2\sqrt{3}-4}{\sqrt{3}-1}+\frac{2\sqrt{2}-1}{\sqrt{2}-1}-\frac{1+\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
\(=\frac{2\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\frac{\left(2\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{\left(1+\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)+\left(2\sqrt{2}-1\right)\left(\sqrt{2}+1\right)-\left(1+\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=3-\sqrt{3}-2+4+\sqrt{2}-1-\left(\sqrt{3}-\sqrt{2}+3\sqrt{2}-2\sqrt{3}\right)\)
\(=4-\sqrt{3}+\sqrt{2}+\sqrt{3}-2\sqrt{2}=4-\sqrt{2}\)
KT
13 tháng 8 2018
\(A=\frac{2\sqrt{3}-4}{\sqrt{3}-1}+\frac{2\sqrt{2}-1}{\sqrt{2}-1}-\frac{1+\sqrt{6}}{\sqrt{2}+3}\)
\(=\frac{2\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\frac{\left(2\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{\left(1+\sqrt{6}\right)\left(3-\sqrt{2}\right)}{\left(\sqrt{2}+3\right)\left(3-\sqrt{2}\right)}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)+\left(2\sqrt{2}-1\right)\left(\sqrt{2}+1\right)-\frac{3-\sqrt{2}+3\sqrt{6}-2\sqrt{3}}{7}\)
\(=3-\sqrt{3}-2+4+\sqrt{2}-1-\frac{3-\sqrt{2}+3\sqrt{6}-2\sqrt{3}}{7}\)
\(=4-\sqrt{3}+\sqrt{2}-\frac{3-\sqrt{2}+3\sqrt{6}-2\sqrt{3}}{7}\)
\(=\frac{28-7\sqrt{3}+7\sqrt{2}-3+\sqrt{2}-3\sqrt{6}+2\sqrt{3}}{7}\)
\(=\frac{25-5\sqrt{3}+8\sqrt{2}-3\sqrt{6}}{7}\)
p/s: mk lm đc đến đây thôi,
ĐKXĐ : a > 0
Ta có :
\(a=4+2\sqrt{3}=3+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)( Thỏa mãn ĐKXĐ )
Vậy tại \(a=\left(\sqrt{3}+1\right)^2\)thì giá trị biểu thức A là :
\(A=\frac{\left(\sqrt{3}+1\right)^2-1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\frac{4+2\sqrt{3}-1}{\sqrt{3}+1}=\frac{3+2\sqrt{3}}{\sqrt{3}+1}\)
\(A=\frac{a-1}{\sqrt{a}}\)( a > 0 )
Với \(a=4+2\sqrt{3}\)( tm )
\(A=\frac{4+2\sqrt{3}-1}{\sqrt{4+2\sqrt{3}}}\)
\(=\frac{3+2\sqrt{3}}{\sqrt{3+2\sqrt{3}+1}}\)
\(=\frac{3+2\sqrt{3}}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\frac{3+2\sqrt{3}}{\left|\sqrt{3}+1\right|}\)
\(=\frac{3+2\sqrt{3}}{\sqrt{3}+1}\)
\(=\frac{\left(3+2\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{3\sqrt{3}-3+6-2\sqrt{3}}{3-1}\)
\(=\frac{3+\sqrt{3}}{2}\)