1.Tính:(-147).13+(-13).(-47)
2.Tìm X:-3.(x-2)3=81
MAi mk đi thi rùi,giúp nha
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47. (23 + 50)- 23. ( 47+50 )
= 47. 23+ 47. 50- 23. 47 - 23. 50
=47. 50 - 23. 50
= 50 .( 47 -23)
=50. 24 = 1200
để A có GTLN thì 2(x-1)2 + 3 phải bé nhất
mà 2(x-1)2 luôn > hoặc = 0
=> A có GTLN thì 2(x-1)2 + 3 = 3
=> x=1
GTLN of A là 1/3 khi và chỉ khi x = 1
để B có GTLN thì 17-x > 0 và bé nhất
=> 17-x = 1
=> x = 16
GTLN của B = 1 khi và chỉ khi x=16
Ta có:\(x.\frac{1}{5}+x.\frac{4}{5}=2\)
=>\(x.\frac{1}{5}+\frac{4}{5}=2\)
=>\(x.\frac{5}{5}=2\)
=>\(x.1=2\)
=>\(x=2:1\)
=>\(x=2\)
Cho \(x=\frac{1}{4}\Rightarrow2.f\left(\frac{1}{\frac{1}{4}}\right)=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow2f\left(4\right)=\frac{1}{16}\Rightarrow f\left(4\right)=\frac{1}{32}\)
a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(=\frac{1}{5}+\frac{2}{7}\)
\(=\frac{7}{35}+\frac{10}{35}\)
\(=\frac{17}{35}\)
Vậy \(A=\frac{17}{35}\)
b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)
\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)
\(=5.\frac{50}{671}\)
\(=\frac{250}{671}\)
Vậy \(B=\frac{250}{671}\)
2010 x 2010 - 2009 x 2011
= ( 2011 - 2010 ) x 2010 - 2009 x 2011
= 2011 x 2010 - 1 x 2010 - 2009 x 2011
= 2011 x ( 2010 - 2009 ) - 1 x 2010
= 2011 x 1 - 2010
= 2011 - 2010
= 1
( 456 x 11 + 912 ) x 37 : ( 13 x 74 )
= ( 456 x 11 + 456 x 2 ) x 37 : ( 13 x 74 )
= [ 456 x ( 11 + 2 ) ] x 37 : ( 13 x 74 )
= [ 456 x 13 ] x 37 : ( 13 x 74 )
= 13 x 456 x 37 : ( 13 x 37 x 2 )
= 13 x 37 x 2 x 228 : ( 13 x 37 x 2 )
= ( 13 x 37 x 2 ) x 228 : ( 13 x 37 x 2 )
= ( 13 x 37 x 2 ) : ( 13 x 37 x 2 ) x 228
= 1 x 228
= 228
(44 x 52 x 60) : (11 x13 x 15 )
= [ ( 4 x 11 ) x ( 4 x 13 ) x ( 4 x 15 ) ] : (11 x13 x 15 )
= [ 4 x 11 x 4 x 13 x 4 x 15 ] : (11 x13 x 15 )
= [ ( 4 x 4 x 4 ) x ( 11 x 13 x 15 ) ] : (11 x13 x 15 )
= 64 x (11 x13 x 15 ) : (11 x13 x 15 )
= 64 x 1
= 64
a) 2010 x 2010 - 2009 x 2011
= 2010 x 2010 - [ 2009 x ( 2010 +1) ]
=2010 x 2010 - 2009 x 2010 - 2009
=2010( 2010 - 2009 ) -2009
= 2010 -2019
= 1
c) ( 44 x 52 x60 ) : (11x13x15)
\(\frac{44\cdot52\cdot60}{11\cdot13\cdot15}=\frac{4\cdot11\cdot4\cdot13\cdot4\cdot15}{11\cdot13\cdot15}=64\)
B) (456 x 11 +912) x 37 : ( 13 x 74 )
= \(\frac{\left(456\cdot11+912\right)\cdot37}{13\cdot74}=\frac{\left[456\left(11+2\right)\right]\cdot37}{13\cdot37}=\frac{456\cdot13}{13\cdot2}=\frac{456}{2}=228\)
Nhớ k mik nha bn
Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
Giải:
a) \(\left(-147\right).13+\left(-13\right).\left(-47\right)\)
\(=\left(-147\right).13+13.47\)
\(=13\left(-147+47\right)\)
\(=13\left(-100\right)=-1300\)
Vậy ...