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5/
Đặt \(\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=a\ge0\\\sqrt{\frac{6}{x}-2x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=\frac{3}{x}\)
Pt trở thành:
\(a-1=\frac{a^2+b^2}{2}-b\)
\(\Leftrightarrow a^2+b^2-2a-2b+2=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x-\frac{3}{x}}=1\\\sqrt{\frac{6}{x}-2x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-x-3=0\\2x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow x=\frac{3}{2}\)
4/
ĐKXĐ: \(x\ge\frac{1}{5}\)
\(\Leftrightarrow\frac{4x-3}{\sqrt{5x-1}+\sqrt{x+2}}=\frac{4x-3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\Rightarrow x=\frac{3}{4}\\\sqrt{5x-1}+\sqrt{x+2}=5\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{5x-1}-3+\sqrt{x+2}-2=0\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\sqrt{5x-1}+3}+\frac{x-2}{\sqrt{x+2}+2}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{5}{\sqrt{5x-1}+3}+\frac{1}{\sqrt{x+2}+2}\right)=0\)
\(\Leftrightarrow x=2\)
Mấy bài này dài vật vã ghê =)))))))))))))
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.
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
Tìm miền xác định phải không
a)
\(1-\sqrt{2x-x^2}\)
a xác định \(\Leftrightarrow2x-x^2\ge0\)
\(0\le x\le2\)
b)
\(\sqrt{-4x^2+4x-1}\)
b xác định
\(\Leftrightarrow-4x^2+4x-1\ge0\)
\(-\left(4x^2-4x+1\right)\ge0\)
\(4x^2-4x+1\le0\)
\(\left(2x-1\right)^2\le0\)
2x - 1 = 0
x = 1/2
c)
\(\frac{x}{\sqrt{5x^2-3}}\)
c xác định
\(\Leftrightarrow5x^2-3>0\)
\(5x^2>3\)
\(x^2>\frac{3}{5}\)
\(\orbr{\begin{cases}x< -\frac{\sqrt{15}}{5}\\x>\frac{\sqrt{15}}{5}\end{cases}}\)
d)
d xác định
\(\Leftrightarrow\sqrt{x-\sqrt{2x-1}}>0\)
\(x-\sqrt{2x-1}>0\)
\(x>\sqrt{2x-1}\)
\(\hept{\begin{cases}2x-1\ge0\\x^2>2x-1\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\x^2-2x+1>0\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\\left(x-1\right)^2>0\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\x-1\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ge\frac{1}{2}\\x\ne1\end{cases}}\)
e)
e xác định
\(\Leftrightarrow\frac{-2x^2}{3x+2}\ge0\)
\(3x+2< 0\) ( vì \(-2x^2\le0\forall x\) )
\(x< -\frac{2}{3}\)
f)
f xác định
\(\Leftrightarrow x^2+x-2>0\)
\(\orbr{\begin{cases}x< -2\\x>1\end{cases}}\)
http://123link.pro/U9cCj2bj
\(\frac{2x-\sqrt{x}}{\sqrt{x}-1}+\frac{x}{\sqrt{x}-1}\)
\(=\frac{2x+x-\sqrt{x}}{\sqrt{x}-1}\)
\(=\frac{3x-\sqrt{x}}{\sqrt{x-1}}\)
P/s đề thiếu hay đủ vậy em : ))