(y+1)+(y+2)+(y+3)+(y+4)+...+(y+37)=999
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( y+1)+(y+2)+...+(y+37) = 999
(y+y+y+...+y+y) + (1+2+3+4+..+37) = 999
y. 37 + (1+(2+37)+(3+36)+...) = 999
y.37 + (1+ 39 + 39 + 39+ ...) = 999
y.37 + (1 + 39 . 18) = 999
y . 37 + (1+702) =999
y.37+ 703 = 999
y.37 = 999-703
y.37= 296
y= 296 : 37
y = 8
Vật y = 8
2: Tọa độ giao điểm là:
\(\left\{{}\begin{matrix}2x-1=x+1\\y=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
hiệu 2 số là:
2-1=1
nên y sẽ là 1
nên ta có tổng là:
37x37x2=2738
đáp số:2738
37 x y + 1 + 2 + 3+ ...+37 = 37 x y + (1 + 37) : 2 x 37
= 37 x (y + 19)
\(12:y=\frac{3}{4}+\frac{4}{5}\)
\(12:y=\frac{15}{20}+\frac{16}{20}\)
\(12:y=\frac{31}{20}\)
\(y=12:\frac{31}{20}\)
\(y=12x\frac{20}{31}\)
\(y=\frac{240}{31}\)
\(\frac{4}{5}xY=1\)
\(y=1:\frac{4}{5}\)
\(y=1x\frac{5}{4}\)
\(y=\frac{5}{4}\)
\(\frac{37}{63}-y=\frac{1}{7}\)
\(y=\frac{37}{63}-\frac{1}{7}\)
\(y=\frac{37}{63}-\frac{9}{63}\)
\(y=\frac{28}{63}\)
\(y=\frac{4}{9}\)
\(y:\frac{3}{7}=\frac{3}{42}+\frac{5}{7}\)
\(y:\frac{3}{7}=\frac{3}{42}+\frac{30}{42}\)
\(y:\frac{3}{7}=\frac{33}{42}\)
\(y:\frac{3}{7}=\frac{11}{14}\)
\(y=\frac{11}{14}x\frac{3}{7}\)
\(y=\frac{33}{98}\)
\(12:y=\frac{3}{4}+\frac{4}{5}\) \(\frac{37}{63}-y=\frac{1}{7}\)
\(12:y=\frac{31}{20}\) \(y=\frac{37}{63}-\frac{1}{7}\)
\(y=12:\frac{31}{20}\) \(y=\frac{4}{9}\)
\(y=\frac{240}{31}\)
\(\frac{4}{5}\cdot y=1\) \(y:\frac{3}{7}=\frac{3}{42}+\frac{5}{7}\)
\(y=1:\frac{4}{5}\) \(y:\frac{3}{7}=\frac{11}{14}\)
\(y=\frac{5}{4}\) \(y=\frac{11}{14}.\frac{3}{7}\)
\(y=\frac{33}{98}\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Áp dụng t.c của dãy tỉ só bằng nhau,ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y}{3+4}=\dfrac{16}{7}\)
=>\(x=\dfrac{16}{7}.3=\dfrac{48}{7}\)
\(y=\dfrac{16}{7}.4=\dfrac{64}{7}\)
\(z=\dfrac{16}{7}.5=\dfrac{80}{7}\)
Vậy...
Các câu sau tương tự
ADTC dãy tỉ số bằng nhau ta cs
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x^2+y^2}{2^2+3^2}=\frac{52}{13}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=4\\\frac{y}{3}=4\\\frac{z}{4}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=16\end{matrix}\right.\)
\(\frac{x}{4}=\frac{y}{5}\Rightarrow\frac{x}{8}=\frac{y}{10}\) (1)
\(\frac{y}{2}=\frac{z}{3}\Rightarrow\frac{y}{10}=\frac{z}{15}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{x}{8}=\frac{y}{10}=\frac{z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được :
\(\frac{x}{8}=\frac{y}{10}=\frac{z}{15}=\frac{x-y-z}{8-10-15}=\frac{37}{-17}\)
\(\Rightarrow\frac{x}{8}=\frac{37}{-17}\Rightarrow x=-\frac{296}{17}\)
\(\Rightarrow\frac{y}{10}=-\frac{37}{17}\Rightarrow y=-\frac{370}{17}\)
\(\Rightarrow\frac{z}{15}=-\frac{37}{17}\Rightarrow z=-\frac{555}{17}\)
Bài 4:
\(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(\Rightarrow y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{2}{5}\cdot\dfrac{25}{16}\)
\(\Rightarrow y=\dfrac{5}{8}\)
________________
\(456+y:87=23987\)
\(\Rightarrow y:87=23987-456\)
\(\Rightarrow y:87=23531\)
\(\Rightarrow y=23531\cdot87\)
\(\Rightarrow y=2047197\)
a)\(\dfrac{4}{5}\times\dfrac{5}{8}:\dfrac{4}{5}\)
\(=\left(\dfrac{4}{5}:\dfrac{4}{5}\right)\times\dfrac{5}{8}\)
\(=1\times\dfrac{5}{8}=\dfrac{5}{8}\)
b)\(\dfrac{5}{6}+\left(\dfrac{1}{2}:\dfrac{3}{2}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\left(\dfrac{1}{3}+\dfrac{4}{5}\right)\)
\(=\dfrac{5}{6}+\dfrac{17}{15}\)
\(=\dfrac{59}{30}\)
Bài 2:
a) \(\dfrac{3}{4}+y:\dfrac{2}{5}=\dfrac{37}{16}\)
\(y:\dfrac{2}{5}=\dfrac{37}{16}-\dfrac{3}{4}\)
\(y:\dfrac{2}{5}=\dfrac{25}{16}\)
\(y=\dfrac{25}{16}\times\dfrac{2}{5}\)
\(y=\dfrac{5}{8}\)
b)\(456+y:87=23987\)
\(y:87=23987-456\)
\(y:87=23531\)
\(y=23531\times87\)
\(y=2047197\)
Tìm y giúp
\(\left(y+1\right)+\left(y+2\right)+\left(y+3\right)+\left(y+4\right)+...+\left(y+37\right)=999\)
\(y+y+y+y+...+y+1+2+3+4+...+37=999\) ( 37 số y )
Phần 1 + 2 + 3 + 4 + ... + 37 có : ( 37 - 1 ) / 1 + 1 = 37 ( số hạng )
\(\Rightarrow37y+\frac{\left(37+1\right)37}{2}=999\)
\(37y+703=999\)
\(37y=999-703\)
\(37y=296\)
\(y=\frac{296}{37}=8\)