132 :{ 41+ [28- (-67) -62)2 ] }
Câu 2 : Thực hiện phép tính:
a) ( -14 ) + ( - 16) + 24
b) /-25/ . 71 + 29 . 25
c) 132 : { 41 + [ 28 - ( 67 - 62 ) \(^2\) ] }
a: (-14)+(-16)+24
=-14-16+24
=-30+24
=-6
b: |-25|x71+29x25
\(=25\cdot71+25\cdot29\)
\(=25\left(71+29\right)=25\cdot100=2500\)
c: \(=132:\left(41+28-5^2\right)\)
\(=132:\left(69-25\right)=132:44=3\)
Thực hiện phép tính
a) (-14)+(-16)+24
b) giá trị tuyệt đối của -25*71+29*25
c) 132:{41+[28-(67-62)2]}
. Thực hiện các phép tính sau một cách hợp lí nhất:
a) 38 + 41 + 117 + 159 + 62 ;
b) 73 + 86 + 968 + 914 + 3032 ;
c) 341 . 67 + 341 . 16 + 659 . 83 ;
e) 563.23 + 23.36 + 23
f) 132 + 157 + 43 + 68
g) 49 . 68 + 100 . 34 + 68
Đáp án:
a, 38+41+117+159+6238+41+117+159+62
=(38+62)+(41+159)+117=(38+62)+(41+159)+117
=100 +200 +117=100 +200 +117
=417=417
b, 73+86+968+914+303273+86+968+914+3032
=73+(86+914)+(968+3032)=73+(86+914)+(968+3032)
=73+1000 +4000=73+1000 +4000
=5073=5073
c, 341.67+341.16+659.83341.67+341.16+659.83
=341.(67+16)+659.83=341.(67+16)+659.83
=341.83+659.83=341.83+659.83
=83.(341+659)=83.(341+659)
=83.1000=83000=83.1000=83000
d, 42.53+47.156−47.11442.53+47.156-47.114
=42.53+47.(156−114)=42.53+47.(156-114)
=42.53 +47.42=42.53 +47.42
=42.(53+47)=42.100=42.(53+47)=42.100
=4200=4200
e, 563.23+23.26+23563.23+23.26+23
=23.(563+26+1)=23.(563+26+1)
=23.600=23.600
=13800=13800
f, 132+157+43+68132+157+43+68
=(132+68)+(157+43)=(132+68)+(157+43)
=200 +200=400=200 +200=400
g, 49.68+100.34+6849.68+100.34+68
=49.68+50.68+68=49.68+50.68+68
=68.(49+50+1)=68.(49+50+1)
=68.100=6800=68.100=6800
h, 491.(267+53)+491.(151+67)491.(267+53)+491.(151+67)
=491.(267+53+151+67)=491.(267+53+151+67)
=491.538=491.538
=264158=264158
Vote mk nha.
(1/3+12/67+13/41)-(79/67-28/41)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)+\frac{1}{3}\)
\(=\left(-1\right)+1+\frac{1}{3}\)
\(=\frac{1}{3}\)
(1/3+12/67+13/41)-(79/67-28/41)
giúp mình với
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\\ =\left(\dfrac{103}{201}+\dfrac{13}{41}\right)-\dfrac{1363}{2747}\\ =\dfrac{6836}{8241}-\dfrac{1363}{2747}\\ =\dfrac{1}{3}\)
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\\ =\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\\ =\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\\ =\dfrac{1}{3}-\dfrac{67}{67}+\dfrac{41}{41}\\ =\dfrac{1}{3}-1+1\\ =\dfrac{1}{3}\)
tính bằng cách hợp lý
(1/3+12/67+13/41)-(79/67-28/41)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1\)
\(=\frac{1}{3}\)
(1/3+12/67+13/41)-(79/67-28/41). =1/3+12/613/41-79/67+28/41. =1/3+(12/67-79/67)+(13/41+28/41) =1/3+(-1)+1=1/3+0. =1/3
tính 4-(12/67+13/41)-(-12/67+28/41)
mng lm nhanh hộ mik nha.Mik cảm ơn ạ
2/15-2/65-4/39
32/13-(7/23+8/23)
(1/3+12/67+13/41) - (79/67 - 28/41)
38/45 - (8/45-17/51-3/11)
2/15-2/65-4/39=4/39-4/39=0
32/13-(7/23+8/23)=32/13-15/23=541/299
(1/3+12/67+13/41) - (79/67 - 28/41)=6836/8241-1363/2747=1/3
38/145 - (8/45-17/51-3/11)=38/145-(-212/495)=1982/2871
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(\left(\dfrac{15}{4}-5x\right)\cdot\left(9x^2-4\right)=0\)
\(\sqrt{x-2}+\dfrac{1}{3}=1\)
\(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}+\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\left(-1\right)+1=\dfrac{1}{3}+0=\dfrac{1}{3}\)
\(\left(\dfrac{15}{4}-5x\right).\left(9x^2-4\right)=0\)
\(\left[{}\begin{matrix}\dfrac{15}{4}-5x=0\\9x^2-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}5x=\dfrac{15}{4}\\9x^2=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)