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7 tháng 3 2017
Ta có : A = 1.2 + 2.3 + 3.4 + ...... + 100.101
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 100.101.102
=> 3A = 100.101.102
=> A = 100.101.102/3
=> A = 343400
15 tháng 12 2021
1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)
2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)
3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)
Áp dụng t/c dtsbn:
\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
VT
0
CG
18 tháng 9 2019
S= (1+2-3-4)-(5+6-7-8)-...-(97+98-99-100)+101+102 S= (-4 -4 -... -4) +101+102 S=(-4).25+101+102 S=-100+101+102 S=103
JE
1
13 tháng 6 2018
Với mọi x,y ta có :
\(\left(\frac{3x+5}{9}\right)^{100}\ge0\)
\(\left(\frac{3y+0,4}{3}\right)^{102}\ge0\)
\(\Leftrightarrow\left(\frac{3x+5}{9}\right)^{100}+\left(\frac{3y+0,4}{3}\right)^{102}\ge0\)
Lại có : \(\left(\frac{3x+5}{9}\right)^{100}+\left(\frac{3y+0,4}{3}\right)^{102}=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(\frac{3x+5}{9}\right)^{100}=0\\\left(\frac{3y+0,4}{3}\right)^{102}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{3x+5}{9}=0\\\frac{3y+0,4}{3}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x+5=0\\3y+0,4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{0,4}{3}\end{cases}}\)
Vậy ..
S
16 tháng 7 2019
\(+,x< -2\Rightarrow\left\{{}\begin{matrix}x+2< 0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=-2-x\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow1-3x=5\Rightarrow x=-\frac{4}{3}\left(\text{loại}\right)\)
\(+,x\ge\frac{3}{2}\Rightarrow\left\{{}\begin{matrix}2x-3\ge0\\x+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|2x-3\right|=2x-3\\\left|x+2\right|=x+2\end{matrix}\right.\Rightarrow3x-1=5\Rightarrow x=2\left(\text{thoa man}\right)\)
\(+,-2\le x< \frac{3}{2}\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\2x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+2\right|=x+2\\\left|2x-3\right|=3-2x\end{matrix}\right.\Rightarrow5-x=0\Rightarrow x=0\left(\text{thoa man}\right)\)
\(2.\text{ Ta co:}\left\{{}\begin{matrix}\left|x-102\right|\ge102-x\\\left|2-x\right|\ge x-2\end{matrix}\right.\Rightarrow A\ge102-x+x-2=100.\Rightarrow A_{min}=100.\text{dâu "=" xay ra}\Leftrightarrow\left\{{}\begin{matrix}102-x\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow2\le x\le102\)
a) |5/3 - x| - |-5/6| = |-5/9|
=> |5/3 - x| - 5/6 = 5/9
=> |5/3 - x| = 5/9 + 5/6
=> |5/3 - x| = 25/18
=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)
a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)
\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)