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\(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|=0\)
Vì \(\left|x-\frac{2}{3}\right|\ge0\)và \(\left|y+\frac{5}{9}\right|\ge0\)nên \(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|\ge0\)
(Dấu "="\(\Leftrightarrow\)\(\left|x-\frac{2}{3}\right|=0\)và \(\left|y+\frac{5}{9}\right|=0\))
\(\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{-5}{9}\end{cases}}\)
vì \(\left|x-\frac{2}{3}\right|>0\)hoặc =0 ;\(\left|y+\frac{5}{9}\right|>0\)hoặc =o
mà\(\left|x-\frac{2}{3}\right|+\left|y+\frac{5}{9}\right|=0\)
nên |x-2/3| =0 và |y+5/9|=0
\(\Rightarrow\hept{\begin{cases}x-\frac{2}{3}=0\\y+\frac{5}{9}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{-5}{9}\end{cases}}}\)
a.\(A=\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\)
Ta có: \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\)
Dấu = xảy ra khi :
\(\frac{x}{5}+\frac{23}{2}=0\Leftrightarrow\frac{x}{5}=-\frac{23}{2}\Leftrightarrow x=-\frac{115}{2}\)
\(y-\frac{14}{3}=0\Leftrightarrow y=\frac{14}{3}\)
Vậy ..............
Ta có:
a) \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall y\)
=> \(\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\forall x;y\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{x}{5}+\frac{23}{2}=0\\y-\frac{14}{3}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
Vậy Min của A = 2019 tại \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
câu b tượng tự
\(a)\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x(x+3)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3})\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{x+3}\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
\(b)x-(\frac{50x}{100}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-(\frac{100x}{200}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\)
\(\Rightarrow3x=\frac{45}{4}\cdot8\)
\(\Rightarrow3x=90\Rightarrow x=30\)
\(c)1+2+3+4+...+x=820\)
Ta có : \(1+2+3+4+...+x=\frac{(1+x)\cdot x}{2}\)
Do đó : \(\frac{(1+x)\cdot x}{2}=820\)
\(\Rightarrow(1+x)\cdot x=820\cdot2\)
\(\Rightarrow(1+x)\cdot x=1640\)
\(\Rightarrow(1+x)\cdot x=40\cdot41\)
Vì x và x + 1 là hai số tự nhiên liên tiếp nên => x = 40
Chúc bạn học tốt :3
a)
⇒ \(\frac{11x-1}{4}=\frac{10}{4}\)
⇒ 11x - 1 = 10
11x = 10 + 1 = 11
x = 11 : 11 = 1
b)
\(\left[{}\begin{matrix}3x-6=0\\\frac{x}{9}-\frac{1}{3}=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3x=0+6\\\frac{x}{9}=0+\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}3x=6\\\frac{x}{9}=\frac{1}{3}\end{matrix}\right.\)⇒ \(\left[{}\begin{matrix}x=6:3\\\frac{x}{9}=\frac{3}{9}\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy x = 2 hoặc x = 3
c)
\(M=c\left(\frac{5}{7}+\frac{7}{14}-\frac{17}{14}\right)\)
\(M=c\left(\frac{10}{14}+\frac{7}{14}-\frac{17}{14}\right)\)
\(M=\left(\frac{2018}{2019}-\frac{2019}{2020}\right).0\)
M = 0
d)
\(N=\frac{-7}{13}+2-\frac{19}{13}+\frac{2020}{2018}.\frac{2018}{202}\)
\(N=\left(\frac{-7}{13}-\frac{19}{13}\right)+2+10\)
N = \(-2+2+10\)
N = 10
Tim x biet
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)
=> \(1-\frac{2}{x+1}=\frac{2019}{2020}\)
=> \(\frac{2}{x+1}=\frac{1}{2020}=\frac{2}{4040}\)
=> x + 1 = 4040 => x = 4039
\(\hept{\begin{cases}\left(x+\frac{2019}{2020}\right)^{100}\ge0\\\left(y-\frac{9}{11}\right)^{200}\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x+\frac{2019}{2020}=0\\y-\frac{9}{11}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-2019}{2020}\\y=\frac{9}{11}\end{cases}}\)
Ta có : \(\left[x+\frac{2019}{2020}\right]^{100}\ge0\forall x\)
\(\left[y-\frac{9}{11}\right]^{200}\ge0\forall y\)
\(\Leftrightarrow\left[x+\frac{2019}{2020}\right]^{100}+\left[y-\frac{9}{11}\right]^{200}\ge0\forall x,y\)
Dấu " = " xảy ra khi : \(\hept{\begin{cases}x+\frac{2019}{2020}=0\\y-\frac{9}{11}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{2019}{2020}\\y=\frac{9}{11}\end{cases}}\)