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chứng minh = phản chứng . giả sử trong 25 số tự nhiên ko có 2 số nào bằng nhau . ko mất tính tổng quát , giả sử\(a_11,a_22,..,a_{25}25\)

thế thì

\(\frac{1}{\sqrt{a_1}}+\frac{1}{\sqrt{a_2}}+...+\frac{1}{\sqrt{a_{25}}}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+..+\frac{1}{\sqrt{25}}\)

ta lại có \(\frac{1}{\sqrt{25}}+..+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{1}}=\frac{1}{\sqrt{25+\sqrt{25}}}+\frac{1}{\sqrt{2+\sqrt{2}}}+1\)

\(< \frac{2}{\sqrt{24+\sqrt{24}}}+.+\frac{2}{\sqrt{2+\sqrt{2}}}+1\)

\(=2\left(\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\right)+1=2\left(\sqrt{25}-\sqrt{1}\right)+1=9\left(2\right)\)

từ (1) zà 2 suy ra \(\frac{1}{\sqrt{a_1}}+\frac{1}{\sqrt{a_2}}+..+\frac{1}{\sqrt{a_{25}}}< 9\)trái zới giả thiết , suy ra ko tồn tại 2 số nào = nhau trong 25 số

\(A=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+12\)

\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+12\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+12\)

Đặt \(x^2+3x=a\) nên \(A=a\left(a+2\right)+12=a^2+2a+1+11=\left(a+1\right)^2+11\ge11\)

Dấu "=" xảy ra \(\Leftrightarrow a+1=0\Leftrightarrow x^2+3x=0\Leftrightarrow x\left(x+3\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

Vậy \(A_{min}=11\) tại \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

cảm mơn b nhiều nha

Ta có:

\(a< b+c\)

\(\Leftrightarrow2a< a+b+c=2\)

\(\Leftrightarrow a< 1\)

Tương tự ta cũng có:

\(\hept{\begin{cases}b< 1\\c< 1\end{cases}}\)

\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)

\(\Leftrightarrow-abc+ab+bc+ca-a-b-c+1>0\)

\(\Leftrightarrow abc< \left(ab+bc+ca\right)-1\)

\(\Leftrightarrow2abc< 2\left(ab+bc+ca\right)-2\)

\(\Leftrightarrow a^2+b^2+c^2+2abc< a^2+b^2+c^2+2\left(ab+bc+ca\right)-2\)

\(\Leftrightarrow a^2+b^2+c^2+2abc< \left(a+b+c\right)^2+2=4-2=2\)

\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)

\(\Leftrightarrow-x^2+\frac{2\sqrt{x}}{3}+1=-x+\sqrt{x+1}\)

\(\Leftrightarrow-x^2+\frac{2\sqrt{x}}{3}+1\)

\(\Leftrightarrow-x+\sqrt{x+1}\)

\(\Leftrightarrow\frac{1}{3}\left(-3x^2+2\sqrt{x+3}\right)=x+\sqrt{x+1}\)

\(\Leftrightarrow3\sqrt{x}\left(x-1\right)+1=0\)

\(\Rightarrow\)Phương trình có nghiệm bằng 0

lại thg xàm loiz này

\(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)

\(pt\Leftrightarrow\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}-x+\sqrt{1-x}+x-1\)

\(\Leftrightarrow\frac{2}{3}\sqrt{-x\left(x-1\right)}=\frac{x-x^2}{\sqrt{x}+x}+\frac{1-x-\left(x-1\right)^2}{\sqrt{1-x}+x-1}\)

\(\Leftrightarrow\frac{2}{3}\sqrt{-x\left(x-1\right)}-\frac{-x\left(x-1\right)}{\sqrt{x}+x}-\frac{-x\left(x-1\right)}{\sqrt{1-x}+x-1}=0\)

\(\Leftrightarrow-x\left(x-1\right)\left(\frac{\frac{4}{9}}{\frac{2}{3}\sqrt{-x\left(x-1\right)}}-\frac{1}{\sqrt{x}+x}-\frac{1}{\sqrt{1-x}+x-1}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-x=0\\x-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(a^4+a^2-2\)

\(=a^4-a^3+a^3-a^2+2a^2-2a+2a-2\)

\(=a^3\left(a-1\right)+a^2\left(a-1\right)+2a\left(a-1\right)+2\left(a-1\right)\)

\(=\left(a-1\right)\left(a^3+a^2+2a+2\right)\)

\(=\left(a-1\right)\left[a^2\left(a+1\right)+2\left(a+1\right)\right]\)

\(=\left(a-1\right)\left(a+1\right)\left(a^2+2\right)\)

=a^2-1+a^4-1

=a2-1+(a2-1)(a2+1)

=(a2-1)(a2+2)