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1, a, \(\frac{3+4\sqrt{3}}{\sqrt{6}+\sqrt{2}-\sqrt{5}}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{\left(\sqrt{6}+\sqrt{2}\right)^2-5}\)
=\(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{8+4\sqrt{3}-5}\)
= \(\frac{\left(3+4\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}+\sqrt{5}\right)}{3+4\sqrt{3}}\)
=\(\sqrt{6}+\sqrt{2}+\sqrt{5}\)
b, M = \(\frac{\sqrt{3}\left(x-1\right)}{\sqrt{x^2}-x+1}\)(ĐKXĐ: \(x\ge0\))
= \(\frac{\sqrt{3}\left(x-1\right)}{x-x+1}\)
= \(\sqrt{3}\left(x-1\right)\)
Thay x = \(2+\sqrt{3}\)(TMĐK) vào M ta có:
M = \(\sqrt{3}\left(2+\sqrt{3}-1\right)=\sqrt{3}\left(1+\sqrt{3}\right)=3+\sqrt{3}\)
Vậy với x = \(2+\sqrt{3}\)thì M = \(3+\sqrt{3}\)
2, Mình chỉ giải câu a thôi nhé:
\(\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt{1+a}\)
\(\Leftrightarrow\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge\left(2\sqrt{1+a}\right)^2\)
\(\Leftrightarrow1+b+2\sqrt{\left(1+b\right)\left(1+c\right)}+1+c\ge4\left(1+a\right)\)
\(\Leftrightarrow2+b+c+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)\left(1\right)\)
Vì \(\left(\sqrt{1+b}-\sqrt{1+c}\right)^2\ge0\)
\(\Rightarrow2+b+c\ge2\sqrt{\left(1+b\right)\left(1+c\right)}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4+2\left(b+c\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4\left(1+a\right)+2\sqrt{\left(1+b\right)\left(1+c\right)}\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4\left(1+a\right)\)
\(\Leftrightarrow4+2\left(b+c\right)\ge4+4a\)
\(\Leftrightarrow2\left(b+c\right)\ge4a\)
\(\Leftrightarrow b+c\ge2a\)
4*. Thật ra cái này mình xài làm trội, làm giảm là được mà
Đặt A = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+....+\frac{1}{2\sqrt{n}}\)
\(\frac{1}{2}A=\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+....+\frac{1}{\sqrt{n}+\sqrt{n}}\)
Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2}}>\frac{1}{\sqrt{3}+\sqrt{2}}\)
\(\frac{1}{\sqrt{3}+\sqrt{3}}>\frac{1}{\sqrt{4}+\sqrt{3}}\)
+ .........................................................
\(\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
Cộng tất cả vào
\(\Rightarrow\frac{1}{\sqrt{2}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n}}>\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{n+1}+\sqrt{n}}\)\(\frac{1}{2}A>\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
\(\frac{1}{2}A>\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n+1}-\sqrt{n}\)
\(\frac{1}{2}A>\sqrt{n+1}-\sqrt{2}\)
\(A>2\sqrt{n+1}-2\sqrt{2}>2\sqrt{n+1}-3\)
\(A+1>2\sqrt{n+1}-3+1\)
\(A+1>2\sqrt{n+1}-2\)
\(A+1>2\left(\sqrt{n+1}-1\right)\)
Vậy ta có điều phải chứng minh.
#)Giải :
a) \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)
\(M=\frac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)
a.Ta co:\(x^2-x=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=1\left(n\right)\end{cases}}\)
\(\Rightarrow M=\frac{1-2}{1}=-1\)
b.De \(M\in Z\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}}\in Z\Rightarrow\sqrt{x}-2⋮\sqrt{x}\Rightarrow x=4\)
ĐKXĐ:...
\(A=\left(\frac{\sqrt{x}\left(x-1\right)-x-2}{x-1}\right):\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{x-1}\right)\)
\(A=\left(\frac{x\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)=\frac{x\left(\sqrt{x}-1\right)}{x-4}-\frac{1}{\sqrt{x}-2}\)
Câu B vt lại đề đi
\(C=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\left(x-1\right)^2}{2}\)
\(C=\frac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
\(C=\frac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=\sqrt{x}-x\)
\(A=5-\sqrt{x+\sqrt{x}+1}\)
ĐK: \(x\ge0\)
=> \(x+\sqrt{x}\ge0\)
=> \(x+\sqrt{x}+1\ge1\)
=> \(\sqrt{x+\sqrt{x}+1}\ge1\)
=> \(-\sqrt{x+\sqrt{x}+1}\le1\)
Do đó: \(A\le4\)
Dấu "=" xảy ra khi x=0
\(B=\frac{3x+6\sqrt{x}}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}+3}{1-\sqrt{x}}\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{3x+6\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\frac{3x+6\sqrt{x}-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3x+6\sqrt{x}-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+3}{\sqrt{x}+2}\ge\frac{3}{2}\)
Dấu "=" xảy ra khi x=0
a)A= \(5-\sqrt{x+\sqrt{x}+1}\). ĐKXĐ: \(x\ge0\)
Ta luôn có: \(x+\sqrt{x}\ge0\) với \(x\ge0\)
\(\Rightarrow x+\sqrt{x}+1\ge1\)
\(\Rightarrow\sqrt{x+\sqrt{x}+1}\ge1\)
\(\Rightarrow-\sqrt{x+\sqrt{x}+1}\le-1\)
\(\Rightarrow5-\sqrt{x+\sqrt{x}+1}\le4\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Vậy GTLN của A=4 khi x=0
b) B= \(\frac{3x+6\sqrt{x}}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}+2}{1-\sqrt{x}}\). ĐKXĐ: \(x\ge0; x\ne1\)
= \(\frac{3x+6\sqrt{x}-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
= \(\frac{3x+6\sqrt{x}-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) = \(\frac{x+2\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
= \(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) = \(\frac{\sqrt{x}+3}{\sqrt{x}+2}=\frac{\left(\sqrt{x+2}\right)+1}{\sqrt{x+2}}\)
= \(\frac{\sqrt{x}+2}{\sqrt{x}+2}+\frac{1}{\sqrt{x}+2}=1+\frac{1}{\sqrt{x}+2}\)
Ta luôn có: \(\sqrt{x}+2\ge2\) với \(x\ge0; x\ne1\)
\(\Rightarrow\frac{1}{\sqrt{x}+2}\le\frac{1}{2}\)
\(\Rightarrow1+\frac{1}{\sqrt{x}+2}\le\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Vậy GTLN của B=\(\frac{3}{2}\) khi x=0
\(B=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x+2}{\sqrt{x}}\)
Nếu không phiền, bạn có thể giải chi tiết cho mình được không ạ. Mình cảm ơn nhiều !
Ta có \(B=\frac{\sqrt{x}}{2}+\frac{2}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{2}+\frac{2}{\sqrt{x}-1}+\frac{1}{2}\)
Áp dụng bất đẳng thức Cosi được \(\frac{\sqrt{x}-1}{2}+\frac{2}{\sqrt{x}-1}\ge2\Rightarrow B\ge2+\frac{1}{2}=\frac{5}{2}\)
Dấu đẳng thức xảy ra <=> \(\sqrt{x}-1=2\Leftrightarrow x=9\)
Vậy Min B = \(\frac{5}{2}\Leftrightarrow x=9\)