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a,=(2a + b - 3c).(2a + b - 3c)
=4a\(^2\)+2ab-6ac+2ab+b\(^2\)-3bc-6ac-3cb+9c\(^2\)
=4a\(^2\)+b\(^2\)+9c\(^2\)+4ab
=2\(^2\).a\(^2\)+4ab+b\(^2\)+9c\(^2\)
=(2a+b)\(^2\)+9c\(^2\)( đáng lẽ chỗ này nó phải là -9c\(^2\) nhưng t ko ra đc )
b,=(a + 2b + 3c - 4d)(a + 2b + 3c - 4d)
=a\(^2\)+2ab+3ac-4ad+2ab+4b\(^2\)+6bc-8bd+3ac+6bc+9c\(^2\)-12cd-4ad-8bd-12cd+16d\(^2\)
=a\(^2\)+4b\(^2\)+9c\(^2\)+16d\(^2\)+4ab+6ac-8ad+12bc-16bd-24cd
=(a\(^2\)+4ab+4b\(^2\))+(9c\(^2\)-24cd+16d\(^2\))+6ac-8ad+12bc-16bd
=(a+2b)\(^2\)+(3c-4d)\(^2\)+2(3ac-4ad+6bc-8bd)
=(a+2b)\(^2\)+(3c-4d)\(^2\)+2[a(3c-4d)+2b(3c-4d)]
=(a+2b)\(^2\)+(3c-4d)\(^2\)+2(a+2b)(3c-4d)
khiếp bài dài nghoằng ra ý :(
a: \(\left(2a+b-3c\right)^2\)
\(=4a^2+b^2+9c^2+4ab-12ac-6bc\)
mik ko bít
I don't now
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Áp dụng tc dstbn:
\(\widehat{A}=2\widehat{B}=2\widehat{C}=4\widehat{D}\Rightarrow\dfrac{\widehat{A}}{4}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{2}=\dfrac{\widehat{D}}{1}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{4+2+2+1}=\dfrac{360^0}{9}=40^0\\ \Rightarrow\widehat{A}=40^0\cdot4=160^0\)
c) 93.32+14.16=93.2.16+14.16=16.(186+14)=16.200=3200
e)-8.40+2.108+24=24(9+1)-320=240-320=-80
b)86.15+150.1,4=86.15+15.14=15(86+14)=15.100=1500
d)98,6.199-990.9,86=98,6.199-99.98,6=98,6(199-99)=98,6.100=9860
f)993.98+21.331-50.99,3=993.(98+7-5)=993.100=99300
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