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Ta có: \(\dfrac{A+C+E}{3}+\dfrac{A+B+D}{3}=40+28\)
\(\Rightarrow\dfrac{2A+B+C+D+E}{3}=68\)
\(\Rightarrow\dfrac{2A}{3}+\dfrac{B+C+D+E}{3}=68\)
Thay \(\dfrac{B+C+D+E}{3}=33\) được:
\(\dfrac{2A}{3}+33=68\)
\(\Rightarrow\dfrac{2}{3}A=68-33\)
\(\Rightarrow\dfrac{2}{3}A=35\)
\(\Rightarrow A=35:\dfrac{2}{3}\)
\(\Rightarrow A=\dfrac{105}{2}=52,5\)
Vậy \(A=52,5\)
\(\left(x-1\right)^2+\left(y+1\right)^2=0\)
Nhận xét:
\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
Do đó: Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x-1=0\\y+1=0\end{matrix}\right.\)
\(\Rightarrow x=1;y=-1\)
Vậy \(x=1;y=-1\)
a: \(2xy+x-y+xy^2+2xy\)
\(=x-y+xy^2+\left(2xy+2xy\right)\)
\(=x-y+xy^2+4xy\)
b: \(5xy^2+4y-4x\cdot2y^2\)
\(=4y+5xy^2-8xy^2\)
\(=4x-3xy^2\)
c: \(\sqrt{25}+\sqrt{36}+\sqrt{49}+...+\sqrt{100}\)
=5+6+7+8+9+10
=15+15+15
=45
d: Đặt \(A=1+4+9+16+...+9801+10000\)
Đặt \(B=1+8+27+...+729+1000\)
\(A=1+4+9+...+10000\)
\(=1^2+2^2+...+100^2\)
\(=\dfrac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}\)
\(=\dfrac{100\cdot101\cdot201}{6}\)
\(B=1+8+27+...+1000\)
\(=1^3+2^3+...+10^3=\left(1+2+...+10\right)^2\)
\(=55^2\)
=>\(A-B=\dfrac{100\cdot101\cdot201}{6}-55^2=335325\)
\(\dfrac{A+C+E}{3}\) + \(\dfrac{A+B+D}{3}\) = 40 + 28
\(\dfrac{2A}{3}\)+\(\dfrac{B+C+E+D}{3}\)= 68
\(\dfrac{2A}{3}\)+ 33 = 68
\(\dfrac{2A}{3}\)=35
2A = 35 X 3
2A = 105
A =\(\dfrac{105}{2}\)
Từ (1) \(\Rightarrow A+C+E=40\cdot3=120\)
Từ (2) \(\Rightarrow A+B+D=28\cdot3=84\)
Từ (3) \(\Rightarrow B+C+D+E=33\cdot3=99\)
Suy ra:
\(\left(A+C+E+A+B+D\right)-\left(B+C+D+E\right)=\left(120+84\right)-99\)
\(2A+\left(B+C+D+E\right)-\left(B+C+D+E\right)=105\)
\(2A=105\)
\(A=52,5\)
Vậy \(A=52,5\)
b: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{100^2}< \dfrac{1}{99}-\dfrac{1}{100}\)
Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=>\(A< 1-\dfrac{1}{100}\)
=>A<1
=>0<A<1
=>A không là số tự nhiên
a: \(A=1+4+9+...+10000\)
\(=1^2+2^2+...+100^2\)
\(=\dfrac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}\)
\(=\dfrac{100\cdot101\cdot201}{6}\)
\(B=1+8+27+...+1000\)
\(=1^3+2^3+...+10^3=\left(1+2+...+10\right)^2\)
\(=55^2\)
=>\(A-B=\dfrac{100\cdot101\cdot201}{6}-55^2=335325\)
a: \(1,25:\left(\dfrac{1}{2}-1\dfrac{1}{2}\right)-1,75\cdot\left(-20\%\right)\)
\(=\dfrac{5}{4}:\left(-1\right)-\dfrac{7}{4}\cdot\dfrac{-1}{5}\)
\(=-\dfrac{5}{4}+\dfrac{7}{20}=\dfrac{-25}{20}+\dfrac{7}{20}=-\dfrac{18}{20}=-\dfrac{9}{10}\)
b: \(\left(2,2+40\%\right):\left(\dfrac{1}{2}-1,25:20\%\right)\)
\(=\left(2,2+0,4\right):\left(0,5-1,25:0,2\right)\)
\(=2,6:\left(-5,75\right)=-\dfrac{52}{115}\)
c: \(\left[\dfrac{3}{4}-1,25:\left(-1\dfrac{1}{2}\right)\right]:\left(3,75-\dfrac{1}{2}:0,25\right)\)
\(=\left(\dfrac{3}{4}-\dfrac{5}{4}:\dfrac{-3}{2}\right):\left(\dfrac{15}{4}-\dfrac{1}{2}:\dfrac{1}{4}\right)\)
\(=\left(\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{2}{3}\right):\left(\dfrac{15}{4}-2\right)\)
\(=\left(\dfrac{3}{4}+\dfrac{5}{6}\right):\dfrac{7}{4}=\left(\dfrac{9}{12}+\dfrac{10}{12}\right):\dfrac{7}{4}\)
\(=\dfrac{19}{12}\cdot\dfrac{4}{7}=\dfrac{19}{21}\)
d: \(0,75\cdot\dfrac{-17}{13}-\dfrac{3}{4}\cdot\dfrac{-4}{13}-1,25\)
\(=0,75\cdot\dfrac{-17}{13}+\dfrac{3}{4}\cdot\dfrac{4}{13}-1,25\)
\(=0,75\cdot\left(-\dfrac{17}{13}+\dfrac{4}{13}\right)-1,25\)
=-0,75-1,25
=-2