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a) 262 + 52.24 + 242
= 262 + 2.26.24 + 242
= ( 26 + 24 )2
= 502 = 2500
b) 30032 - 32
= ( 3003 + 3 ) ( 3003 - 3 )
= 3006 . 3000 = 9018000
c) 872 + 732 - 272 -132
= ( 872 - 132 ) + ( 732 - 272 )
= [ ( 87 + 13 )( 87- 13 )] + [ ( 73 - 27 )( 73 + 27 ) ]
= ( 100 . 74 ) + ( 46 . 100 )
= 7400 + 4600 = 12000
d)792 - 79.58 + 292
= 792 - 2.79.29 + 292
= ( 79 - 29 )2
= 502 = 2500
a) 262 + 52 . 24 + 242 = 262 + 2 . 26 . 24 + 242
= ( 26 + 24 )2
= 502
= 2500
b) 30032 - 32 = ( 3003 - 3 ) . ( 3003 + 3 )
= 3000. 3006
= 9018000
c) 872 + 732 - 272 - 132 = ( 872 - 272 ) + ( 732 - 132 )
= ( 87 - 27 ) . ( 87 + 27 ) + ( 73 - 13 ) . ( 73+13)
= 60 . 114 + 60 . 86
= 60 . ( 114 + 86 )
= 60 . 200
= 12000
d) 792 - 79 . 58 + 292 = 792 - 2 . 79 . 29 + 292
= ( 79 - 29 )2
= 502
= 2500
Ribi Nkok Ngok lê thị hương giang Nguyễn Huy Tú Nguyễn Nam Vũ Elsa
B. \(\frac{x+4}{2015}+1+\frac{x+3}{2016}+1=\frac{x+2}{2017}+1+\frac{x+1}{2018}+1\)
<=> \(\frac{x+2019}{2015}+\frac{x+2019}{2016}=\frac{x+2019}{2017}+\frac{x+2019}{2018}\)
<=>(x+2019).(\(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}>0\)
Vì (\(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}>0\)
=> x+2019>0
=>x>-2019
a)\(\dfrac{3}{x^2+5x+4}+\dfrac{2}{x^2+10x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+3x-18}\left(đkxđ:x\ne-1;-4;-6;3\right)\)
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{\left(x+6\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{4}{3}+\dfrac{1}{x-3}-\dfrac{1}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{4}{3}+\dfrac{1}{x-3}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{-4}{\left(x+1\right)\left(x-3\right)}=\dfrac{4}{3}\)
\(\Leftrightarrow\left(x+1\right)\left(3-x\right)=3\)
\(\Leftrightarrow2x-x^2+3=3\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\left(tm\right)\)
b)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow x^2+2x+1-y^2-4y-4-7=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)
Mà x,yEN*=>x-y-1<x+y+3
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y-1=1\\x+y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=-7\\x+y+3=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy ...
a,Ta có A=\(\dfrac{\left(54-23\right)\left(54+23\right)}{\left(36,5-25,5\right)\left(36,5+25,5\right)}=\dfrac{31.77}{11.62}=\dfrac{7}{2}\)
b,Ta có B=\(\dfrac{\left(82-34\right)\left(82+34\right)}{\left(30,5-1,5\right)\left(30,5+1,5\right)}=\dfrac{48.116}{29.32}=\dfrac{6.8.4.29}{29.8.4}=6\)
c,Ta có C=\(\dfrac{\left(86-54\right)\left(86^2+86.54+54^2\right)}{32}+86.54=86^2+86.54+54^2+86.54=\left(86+54\right)^2=19600\)
mu là mũ nha giúp tui vứi
a) \(82^2-18^2=\left(82-18\right)\left(82+18\right)=64.100=6400\)
b) \(26^2+52.24+24^2=26^2+2.26.24+24^2=\left(26+24\right)^2=50^2=2500\)
c) \(\dfrac{93^3+78^3}{171}-93.78=\dfrac{\left(93+78\right)\left(93^2-93.78+73^2\right)}{171}-93.78\)
\(=\dfrac{171\left(93^2-93.78+73^2\right)}{171}-93.78\)
\(=93^2-93.78+73^2-93.78\)
\(=93^2-2.93.78+73^2=\left(93-73\right)^2=20^2=400\)