LT
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HN
0
13 tháng 10 2018
2) Mình nghĩ nên nhỏ hơn 3 thì dễ tính hơn... @@
Ta có :
\(\dfrac{x}{x+y+z}< \dfrac{x}{x+y}< \dfrac{x}{x}\\ \dfrac{y}{x+y+z}< \dfrac{y}{y+z}< \dfrac{y}{y}\\ \dfrac{z}{x+y+z}< \dfrac{z}{z+x}< \dfrac{z}{z}\)
\(\Rightarrow\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< \dfrac{x}{x}+\dfrac{y}{y}+\dfrac{z}{z}\\ \Rightarrow\dfrac{x+y+z}{x+y+z}< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< 1+1+1\\ \Rightarrow1< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< 3\)
25 tháng 2 2020
Bài 1 :
Ta có : \(15x^4y^n.\left(-2x^5y^9\right)=30x^9y^{17}\)
=> \(15x^4.\left(-y\right)^n.\left(-2\right).\left(-x\right)^5.\left(-y\right)^9=30\left(-x\right)^9.\left(-y\right)^{17}\)
=> \(30\left(-x\right)^9.\left(-y\right)^{n+9}=30.\left(-x\right)^9\left(-y\right)^{17}\)
=> \(\left(x\right)^9.\left(-y\right)^{n+9}=\left(-x\right)^9\left(-y\right)^{17}\)
=> \(x^9y^{n+9}=x^9y^{17}\)
- TH1 : \(x,y=0\)
=> \(0^{n+9}=0^{17}\) ( Luôn đúng \(\forall n\) )
=> \(n\in R\)
- TH2 : \(x,y\ne0\)
=> \(y^{n+9}=y^{17}\)
=> \(n+9=17\)
=> \(n=8\)
NN
0
23 tháng 6 2018
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
Ta có:
`|x-1|+|x+4|+|x-9|+|x-3|`
`=|-(x-1)|+|x+4|+|-(x-9)|+|x-3|`
`=|1-x|+|x+4|+|9-x|+|x-3|`
`=(|1-x|+|x-3|)+(|x+4|+|9-x|)`
Áp dụng BĐT: `|a|+|b|>=|a+b|` ta được:
`|1-x|+|x-3|>=|1-x+x-3|=|-2|=2`
`|x+4|+|9-x|>=|x+4+9-x|=|13|=13`
Do đó: `|1-x|+|x-3|+|x+4|+|9-x|>=2+13=15`
Hay: `|x-1|+|x+4|+|x-9|+|x-3|>=15` (đpcm)
∣x−1∣+∣x+4∣+∣x−9∣+∣x−3∣
\(= \mid - \left(\right. x - 1 \left.\right) \mid + \mid x + 4 \mid + \mid - \left(\right. x - 9 \left.\right) \mid + \mid x - 3 \mid\)
\(= \mid 1 - x \mid + \mid x + 4 \mid + \mid 9 - x \mid + \mid x - 3 \mid\)
\(= \left(\right. \mid 1 - x \mid + \mid x - 3 \mid \left.\right) + \left(\right. \mid x + 4 \mid + \mid 9 - x \mid \left.\right)\)
Áp dụng BĐT: \(\mid a \mid + \mid b \mid > = \mid a + b \mid\) ta được:
\(\mid 1 - x \mid + \mid x - 3 \mid > = \mid 1 - x + x - 3 \mid = \mid - 2 \mid = 2\)
\(\mid x + 4 \mid + \mid 9 - x \mid > = \mid x + 4 + 9 - x \mid = \mid 13 \mid = 13\)
Do đó: \(\mid 1 - x \mid + \mid x - 3 \mid + \mid x + 4 \mid + \mid 9 - x \mid > = 2 + 13 = 15\)
Hay: \(\mid x - 1 \mid + \mid x + 4 \mid + \mid x - 9 \mid + \mid x - 3 \mid > = 15\) (đpcm)