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cab = 3 x ab + 8
c x 100 + ab = 3 x ab + 8
c x 100 = ab x (3 - 1) + 8
c x 100 = ab x 2 + 8
Ta tìm được c chỉ có thể bằng 1 hoặc 2
Vì nếu c = 3 thì được
300 = ab x 2 + 8
300 - 8 = ab x 2
292 = ab x 2
ab = 292 : 2
ab = 146 (không được vì có 3 chữ số)
ab = (200 - 8) : 2 = 96
ab = (100 - 8) : 2 = 46
=> a=4,9 b=6,6 c=1,2
Ta có:\(\widehat{cab}=3\times\widehat{ab}+8\)
\(100\times c+\widehat{ab}=3\times\widehat{ab}+8\)
\(100\times c=3\times\widehat{ab}+8-\widehat{ab}\)
\(\Rightarrow100\times c=2\times\widehat{ab}+8\)
Vì \(2\times\widehat{ab}+8\le206\)
nên \(100\times c\le206\)
\(\Rightarrow c\le2\)
\(\Rightarrow c=0;1;2\)
Với c=0 ta có:
\(100\times0=2\times\widehat{ab+8}\)
\(\Rightarrow0=2\times\widehat{ab}+8\left(L\right)\)
Với c=1 ta có:
\(100\times1=2\times\widehat{ab}+8\)
\(100=2\times\widehat{ab}+8\)
\(100-8=2\times\widehat{ab}\)
\(92=2\times\widehat{ab}\)
\(\widehat{ab}=92:2\)
\(\Rightarrow\widehat{ab}=46\)
\(\Rightarrow\widehat{cab}=146\)
Với c=2 ta có:
\(100\times2=2\times\widehat{ab}+8\)
\(200=2\times\widehat{ab}+8\)
\(200-8=2\times\widehat{ab}\)
\(192=2\times\widehat{ab}\)
\(192:2=\widehat{ab}\)
\(\Rightarrow\widehat{ab}=96\)
\(\Rightarrow\widehat{cab}=296\)
vậy \(\widehat{cab}=146\)
hoặc 296
Nhớ tích cho mk nha
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right):2}=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times\left(x+1\right)}\times\frac{1}{2}=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4016}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
Vậy x = 2012
Ta có: x-(\(\frac{31}{5}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}\)\(+\frac{31}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)-\(\frac{31}{2}\)x(\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\))=\(\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{11}-\frac{1}{13}\right)\)=\(\frac{9}{13}\)
x-\(\frac{31}{5}\)\(-\frac{31}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{31}{2}.\frac{10}{39}\)\(=\frac{9}{13}\)
x-\(\frac{31}{5}-\frac{155}{39}=\frac{9}{13}\)
x-\(\frac{434}{195}\)=\(\frac{9}{13}\)
x =\(\frac{9}{13}+\frac{434}{195}=\frac{569}{195}\)
nhé
Giải:
( x - 15 ) x 7 - 270 : 45 = 169
( x - 15 ) x 7 - 6 =169
( x - 15 ) x 7 = 169 + 6
( x - 15 ) x 7 = 175
x - 15 = 175 : 7
x - 15 = 25
x = 25 +15
x = 40
Vậy x = 40
(x - 15 )x 7 - 270 : 45 =169
(x-5)x7 -6 =169
(x-5)x7 = 169 +6
(X-5)x7 = 175
X-5 = 175:7
X-5 = 25
X =25+5
X=30
Áp dụng công thức tính dãy số ta có
\(\left(x+1\right).\left[\left(x-1\right):1+1\right]:2=666\)
\(\Rightarrow\left(x+1\right)x=1332\)
\(\Rightarrow x^2+x-1332=0\)
\(\Rightarrow x^2+37x-36x-1332=0\)
\(\Rightarrow x\left(x+37\right)-36\left(x+37\right)=0\)
\(\Rightarrow\left(x+37\right)\left(x-36\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+37=0\\x-36=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-37\\x=36\end{cases}}}\)
\(\Rightarrow x=36\)
<=> (x+1).[(x-1):1+1]:2=666
<=> (x+1).x:2=666
<=>(x+1).x=666.2
<=>( x+1).x=1332
<=>(x+1).x=37.36
<=>x=36
Vậy x=36