Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
# Bài 1
* Ta cm BĐT sau \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\) (1) bằng cách biến đổi tương đương
* Với \(x,y>0\) áp dụng (1) ta có
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{\left(\sqrt{x}\right)^2}+\dfrac{1}{\left(\sqrt{y}\right)^2}\ge\dfrac{1}{2}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\)
Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\)
\(\Rightarrow\) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\le1\) \(\Leftrightarrow\) \(0< \dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\le1\) (I)
* Ta cm BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với \(a,b>0\) (2)
Áp dụng (2) với x , y > 0 ta có
\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\ge\dfrac{4}{\sqrt{x}+\sqrt{y}}\) (II)
* Từ (I) và (II) \(\Rightarrow\) \(\dfrac{4}{\sqrt{x}+\sqrt{y}}\le1\)
\(\Leftrightarrow\) \(\sqrt{x}+\sqrt{y}\ge4\)
Dấu "=" xra khi \(x=y=4\)
Vậy min \(\sqrt{x}+\sqrt{y}=4\) khi \(x=y=4\)
a: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
c: Để A=1/2 thì \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{1}{2}\)
=>\(-10\sqrt{x}+4=\sqrt{x}+3\)
=>x=1/121
d: \(A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)
\(=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}< =0\)
=>A<=2/3
a) \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{x-1}{x+\sqrt{x}+1}\right)=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
b) Khi x=5+2\(\sqrt{3}\Leftrightarrow P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}=\dfrac{4-2\sqrt{3}}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{16-12}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{2\left(2-\sqrt{3}\right)}{4}=\dfrac{2-\sqrt{3}}{2}\)
c) Ta có \(\left|A\right|\le1\Leftrightarrow\left|\dfrac{1}{x-1}\right|\le1\Leftrightarrow\dfrac{1}{\left|x-1\right|}\le1\Leftrightarrow\left|x-1\right|\ge1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-1\ge1\\1-x\le1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)
Kết hợp với ĐK
Vậy x\(\le0\) hoặc \(x\ge2\) thì \(\left|A\right|\le1\)
a) \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{x-1}{x+\sqrt{x}+1}\right)=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)
b) Khi x=5+2\(\sqrt{3}\Leftrightarrow P=\dfrac{1}{5+2\sqrt{3}-1}=\dfrac{1}{4+2\sqrt{3}}=\dfrac{4-2\sqrt{3}}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{16-12}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{2\left(2-\sqrt{3}\right)}{4}=\dfrac{2-\sqrt{3}}{2}\)
c) Ta có \(\left|A\right|\le1\Leftrightarrow\left|\dfrac{1}{x-1}\right|\le1\Leftrightarrow\dfrac{1}{\left|x-1\right|}\le1\Leftrightarrow\left|x-1\right|\ge1\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-1\ge1\\1-x\le1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)
Kết hợp với ĐK
Vậy x\(\le0\) hoặc \(x\ge2\) thì \(\left|A\right|\le1\)
3.a.Ta có :\(\sqrt{40^2-24^2}=\sqrt{\left(40-24\right)\left(40+24\right)}=\sqrt{16.64}=4.8=32\)
b.Ta có :\(\sqrt{52^2-48^2}=\sqrt{\left(52-48\right)\left(52+48\right)}=\sqrt{4.100}=2.10=20\)
4.a)Ta có :
\(\sqrt{4x}=8\Leftrightarrow4x=8^2\Leftrightarrow4x=64\Leftrightarrow x=16\left(tm\right)\)
Vậy x=16
b)Ta có :
\(\sqrt{0,7x}=6\Leftrightarrow0,7x=36\Leftrightarrow x=\dfrac{36}{0.7}\left(tm\right)\)
Vậy x=\(\dfrac{36}{0.7}\)
c)Ta có:
\(9-4\sqrt{x}=1\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
Vậy x=4
d)Ta có :
\(\sqrt{5x}< 6\Leftrightarrow5x< 36\Leftrightarrow x< \dfrac{36}{5}\)
vậy 0≤x<\(\dfrac{36}{5}\)
Bài 3
a) \(\sqrt{40^2-24^2}\)
\(=\sqrt{\left(40+24\right)\left(40-24\right)}\)
=\(\sqrt{64.16}=\sqrt{64}.\sqrt{16}\)
\(=8.4=24\)
b)\(\sqrt{52^2-48^2}\)
\(=\sqrt{\left(52+48\right)\left(52-48\right)}\)
\(=\sqrt{100.4}=\sqrt{100}.\sqrt{4}\)
=10.2=20
Bài 4
a)\(\sqrt{4x}=8\)
\(\Leftrightarrow2\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)(TM)
b)\(\sqrt{0,7x}=6\)
\(\Leftrightarrow\sqrt{\left(0,7x\right)^2}=6^2\)
\(\Leftrightarrow\left|0,7x\right|=36\)
\(\Leftrightarrow\left[{}\begin{matrix}0,7x=36\\0,7x=-36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{360}{7}\left(TM\right)\\x=-\dfrac{360}{7}\left(KTM\right)\end{matrix}\right.\)
c)\(9-4\sqrt{x}=1\)
\(\Leftrightarrow4\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(TM)
d)\(\sqrt{5x}< 6\)
= x-2 -2\(\sqrt{ }\)x-2 +1+4 đk: x\(\ge\)2
= (\(\sqrt{ }\)x-2 -1)2 +4
Vì (\(\sqrt{ }\)x-2 -1)2\(\ge\)0=>(\(\sqrt{ }\)x-2 -1)2+4\(\ge\)4
Vậy Min A= 4 ,dấu "=" xảy ra <=> (\(\sqrt{ }\)x-2 -1)2 =0 đk :x\(\ge\)2
<=> x=3 TM