Tìm x;y biết
(x-11+y)2+(x+4-y)2=0
Giúp em với ạ, em cảm ơn nhìu:33
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Diện tích đáy thùng:
2,1 x 1,6= 3,36(m2)
Diện tích xung quanh thùng:
2 x 1,3 x (2,1+1,6)= 9,62(m2)
Diện tích gỗ tối thiểu cần dùng đóng thùng:
9,62 + 3,36 = 12,98(m2)
Đ.số: 12,98m2
\(\dfrac{x}{2}=-\dfrac{5}{4}\\ \Rightarrow4x=-5\cdot2\\ \Rightarrow4x=-10\\ \Rightarrow x=-\dfrac{10}{4}\\ \Rightarrow x=-\dfrac{5}{2}\)
\(\dfrac{x}{2}=\dfrac{-5}{4}\)
\(\Rightarrow\dfrac{2x}{4}=\dfrac{-5}{4}\)
\(\Rightarrow x=-\dfrac{5}{2}=-2,5\)
\(a,2\left|3x-1\right|+1=5\\ \Rightarrow2\left|3x-1\right|=5-1\\ \Rightarrow2\left|3x-1\right|=4\\ \Rightarrow\left|3x-1\right|=4:2\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(b,\left|\dfrac{x}{2}-1\right|=3\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-1=3\\\dfrac{x}{2}-1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{x}{2}=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)
\(c,\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3\\ \Rightarrow\left[{}\begin{matrix}-x+\dfrac{2}{5}=3\\-x+\dfrac{2}{5}=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-x=\dfrac{13}{5}\\-x=-\dfrac{17}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{13}{5}\\x=\dfrac{17}{5}\end{matrix}\right.\)
\(d,\left|x-\dfrac{1}{3}\right|=2\dfrac{3}{5}\\ \Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{13}{5}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{13}{5}\\x-\dfrac{1}{3}=-\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}+\dfrac{1}{3}\\x=-\dfrac{13}{5}+\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{44}{15}\\x=-\dfrac{34}{15}\end{matrix}\right.\)
\(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=\dfrac{7}{5}+\dfrac{7}{10}x\\ \Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}+\dfrac{13}{5}=\dfrac{7}{5}+\dfrac{7x}{10}\\ \Rightarrow\dfrac{5x}{10}-\dfrac{6x}{10}+\dfrac{26}{10}=\dfrac{14}{10}+\dfrac{7x}{10}\\ \Rightarrow\dfrac{5x}{10}-\dfrac{6x}{10}-\dfrac{7x}{10}=\dfrac{14}{10}-\dfrac{26}{10}\\ \Rightarrow\dfrac{-8x}{10}=-\dfrac{12}{10}\\ \Rightarrow\dfrac{-4x}{5}=-\dfrac{6}{5}\\ \Rightarrow-20x=-30\\ \Rightarrow x=\dfrac{3}{2}\)
a. \(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) + \(\dfrac{1}{6.8}\) + ...... + \(\dfrac{1}{20.22}\)
= 1/2 ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ..... + 1/20 - 1/22)
=1/2 ( 1/2 - 1/22)
= 1/2 . 5/11
= 5/22
b. 1+ 2/3 + 2/6 + 2/10 +...+ 2/45
=>1/2.(1+2/3+2/6+....+2/45)=1/2+2/6+2/12+...+2/90
=1/2+2/2.3+2/3.4+...+2/9.10
=2.(1/4+3-2/2.3+4-3/3.4+...+10-9/9.10)
=2. ( 1/4+1/2-1/3+1/3-1/4+.....+1/9-1/10)
= 2.( 1/4-1/10)=2.3/20=3/10
=> vì 1/2.*=3/10
=> *=3/10:1/2=3/5
tick mình nhé
B = 1 + \(\dfrac{2}{3}\) + \(\dfrac{2}{6}\) +\(\dfrac{2}{10}\) + \(\dfrac{2}{15}\)+...+ \(\dfrac{2}{45}\)
B = 1 + 2.(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{45}\))
B = 1 + \(\dfrac{4}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) +\(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{45}\))
B = 1 + 4.( \(\dfrac{1}{6}\) +\(\dfrac{1}{12}\)+ \(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+...+ \(\dfrac{1}{90}\))
B = 1 + 4.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{9.10}\))
B = 1 + 4 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\)+...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\))
B = 1 + 4.( \(\dfrac{1}{2}\) - \(\dfrac{1}{10}\))
B = 1 + 4. \(\dfrac{2}{5}\)
B = \(\dfrac{13}{5}\)
C = 1/(9.10) - 1/(8.9) - 1/(7.8) - ... - 1/(2.3) - 1/(1.2)
= 1/9 - 1/10 - 1/8 + 1/9 - 1/7 + 1/8 - ... - 1/2 + 1/3 - 1 + 1/2
= 1/9 - 1/10 + 1/9 - 1
= 2/9 - 11/10
= -79/90
\(\left(3-\dfrac{1}{2}\right)^3-\dfrac{5^4}{25}+\left[\left(\dfrac{1}{2}\right)^2\right]^3\)
\(=\left(\dfrac{5}{2}\right)^3-\dfrac{5^4}{5^2}+\left[\left(\dfrac{1}{2}\right)^2\right]^3\)
\(=\left(\dfrac{5}{2}\right)^3+\left[\left(\dfrac{1}{2}\right)^2\right]^3-5\)
\(=\left(\dfrac{5}{2}+\dfrac{1}{2}\right)\left[\left(\dfrac{5}{2}\right)^2-\dfrac{5}{2}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-5\)
\(=3\left[\dfrac{25}{4}+\dfrac{1}{4}-\dfrac{5}{4}\right]-5\)
\(=3.\dfrac{21}{4}-5\)
\(=\dfrac{63}{4}-5=\dfrac{43}{4}\)
Đính chính \(\dfrac{5^4}{5^2}=25\)
\(...=\dfrac{63}{4}-25=-\dfrac{37}{4}\)
Cô làm rồi mà em
TA CÓ 0=02
⇒X-11+Y+X+4-Y=0
⇒(X+X)+(-11+4)+(Y-Y)=0
⇒2X+(-7)+0=0
⇒2X=0-(-7)
⇒2X=7
⇒X=7:2
⇒X=3,5
VẬY X =3,5