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3 tháng 11 2016

Đặt \(a=x,b=\frac{1}{x}\) thì ta có ab = 1

\(a-b=x-\frac{1}{x}=\frac{x^2-1}{x}=\frac{\left(x-1\right)\left(x+1\right)}{x}\). Vì \(x>1\) nên ta có \(a-b>0\)

\(3\left(a^2-b^2\right)< 2\left(a^3-b^3\right)\)

\(\Leftrightarrow3\left(a-b\right)\left(a+b\right)< 2\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(\Leftrightarrow\left(a^2+ab+b^2\right)>\frac{3}{2}\left(a+b\right)\) (chia cả hai vế cho \(a-b>0\))

\(\Leftrightarrow\left(a^2-\frac{3}{2}a+\frac{9}{16}\right)+\left(b^2-\frac{3}{2}b+\frac{9}{16}\right)+\frac{7}{8}>0\)(vì ab = 1)

\(\Leftrightarrow\left(a-\frac{3}{4}\right)^2+\left(b-\frac{3}{4}\right)^2+\frac{7}{8}>0\) (luôn đúng)

Vậy có đpcm.

3 tháng 11 2016

koooooooiuyfdfguhgfswaxrwgszdsxrfdtfg

NV
3 tháng 5 2019

Đặt \(x=\left[x\right]+\left\{x\right\}\)

\(\Rightarrow\left[3x\right]=\left[3\left[x\right]+3\left\{x\right\}\right]=3\left[x\right]+\left[3\left\{x\right\}\right]\)

\(\left[x+\frac{2}{3}\right]=\left[\left[x\right]+\left\{x\right\}+\frac{2}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]\)

\(\left[x+\frac{1}{3}\right]=\left[x\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

\(\Rightarrow\left[x+\frac{2}{3}\right]+\left[x+\frac{1}{3}\right]+\left[x\right]=3\left[x\right]+\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

Ta cần chứng minh \(\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

- Nếu \(\frac{2}{3}\le\left\{x\right\}< 1\Rightarrow\left\{{}\begin{matrix}2\le\left[3\left\{x\right\}\right]< 3\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\1\le\left[\left\{x\right\}+\frac{1}{3}\right]< 2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=2\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=1\end{matrix}\right.\)

\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

- Nếu \(\frac{1}{3}\le\left\{x\right\}< \frac{2}{3}\Rightarrow\left\{{}\begin{matrix}1\le\left[3\left\{x\right\}\right]< 2\\1\le\left[\left\{x\right\}+\frac{2}{3}\right]< 2\\0\le\left[\left\{x\right\}+\frac{1}{3}\right]< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=1\\\left[\left\{x\right\}+\frac{2}{3}\right]=1\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\)

\(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

- Nếu \(0< \left\{x\right\}< \frac{1}{3}\) tương tự trên ta có:

\(\left\{{}\begin{matrix}\left[3\left\{x\right\}\right]=0\\\left[\left\{x\right\}+\frac{2}{3}\right]=0\\\left[\left\{x\right\}+\frac{1}{3}\right]=0\end{matrix}\right.\) \(\Rightarrow\left[3\left\{x\right\}\right]=\left[\left\{x\right\}+\frac{2}{3}\right]+\left[\left\{x\right\}+\frac{1}{3}\right]\)

NV
21 tháng 10 2019

\(x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0\)

\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right)+\left(x+y+2\right)=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)+1\right)=0\)

\(\Leftrightarrow x+y+2=0\)

(phần trong ngoặc \(\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\frac{\left(y+1\right)^2}{4}+\frac{3\left(y+1\right)^2}{4}+1\)

\(=\left(x+1-\frac{y+1}{4}\right)^2+\frac{3\left(y+1\right)^2}{4}+1\) luôn dương)

\(\Rightarrow x+y=-2\)

\(xy>0\Rightarrow\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-x>0\\-y>0\end{matrix}\right.\)

Ta có: \(\frac{1}{-x}+\frac{1}{-y}\ge\frac{4}{-\left(x+y\right)}=2\) \(\Leftrightarrow\frac{1}{x}+\frac{1}{y}\le-2\) (đpcm)

Dấu "=" xảy ra khi và chỉ khi \(x=y=-1\)

NV
21 tháng 10 2019

2/ \(x;y;z\ne0\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{xz+yz+z^2}=0\)

\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(\frac{xy+yz+xz+z^2}{xyz\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\) dù trường hợp nào thì thay vào ta đều có \(B=0\)

3/ \(\Leftrightarrow mx-2x+my-y-1=0\)

\(\Leftrightarrow m\left(x+y\right)-\left(2x+y+1\right)=0\)

Gọi \(A\left(x_0;y_0\right)\) là điểm cố định mà d đi qua

\(\Leftrightarrow\left\{{}\begin{matrix}x_0+y_0=0\\2x_0+y_0+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_0=-1\\y_0=1\end{matrix}\right.\)

Vậy d luôn đi qua \(A\left(-1;1\right)\) với mọi m

18 tháng 11 2015

toán lớp mấy đó???

 

1 tháng 5 2020

bài nào cx hỏi z :))

BĐT cần chứng minh tương đương với :

\(x+\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}+1>4\)

Áp dụng BĐT Cô-si,

Ta có : \(x+\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}+1=\frac{x^2-1}{x}+\frac{x+1}{2x}+\frac{x+1}{2x}+\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}\)

\(\ge4\)

Dấu "=" xảy ra khi \(\frac{x^2-1}{x}=\frac{x+1}{2x}=\frac{4x^3}{\left(x-1\right)\left(x+1\right)^3}\)

giải đc cái trên là vô nghiệm nên dấu "=" không xảy ra

bạn thanh tùng làm giống mình đó