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\(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
= \(\dfrac{a^3+a.c.b+b.d.c}{a.c.b+b.d.c+d^3}\)
= \(\dfrac{a^3}{d^3}=\dfrac{a}{d}\)
Đề có sai k bạn ??
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1. /x+1/=3
<=> x=2 hoặc x=-4
thay x=2 và x=-4 vào M, ta có:
M= 22-2.2+3= 3
M=(-4)2-2.(-4)+3=27
2.https://olm.vn/hoi-dap/question/127440.html
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Xét \(\left(a^3+b^3+c^3+d^3\right)-\left(a+b+c+d\right)\)
\(=\left(a^3-a\right)+\left(b^3-b\right)+\left(c^3-c\right)+\left(d^3-d\right)\)
Ta có \(a^3-a=a\left(a^2-1\right)=\left(a-1\right)a\left(a+1\right)⋮6\)(vì tích của 3 số nguyên/số tự nhiên liên tiếp)
Tương tự ta có \(\left(b^3-b\right)⋮6;\left(c^3-c\right)⋮6;\left(d^3-d\right)⋮6\)
\(\Rightarrow\left(a^3-a\right)+\left(b^3-b\right)+\left(c^3-c\right)+\left(d^3-d\right)⋮6\)
\(\Rightarrow\left(a^3+b^3+c^3+d^3\right)-\left(a+b+c+d\right)⋮6\)
Mà \(a+b+c+d⋮6\Rightarrow a^3+b^3+c^3+d^3⋮6\left(ĐPCM\right)\)
P/S: bt làm có bài này thôi :v
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Bài 1: a/ \(\left(2^9.16+2^9.34\right):2^{10}\)\(=2^9.\left(16+34\right):2^{10}\)\(=50.2^9:2^{10}\)\(=50.2^9:2^9:2=50.\dfrac{1}{2}=25\)
c/ \(A=3+3^2+3^3+...+3^{19}\)
\(3A=3^2+3^3+3^4+...+3^{20}\)
\(3A-A=\left(3^2+3^3+3^4+...+3^{20}\right)-\left(3+3^2+3^3+...+3^{19}\right)\)
\(2A=3^{20}-3\)
\(A=\left(3^{20}-3\right):2\)
b/\(\left(3^4.57-9^2.21\right):3^5=\left(3^4.57-3^4.21\right):3^5=3^4.\left(57-21\right):3^5=3^4:3^5.36=3^4:3^4:3.36=\dfrac{1}{3}.36=12\)
Bài 2:
a/ 44 + 66 + 72
Vì 44 \(⋮\) 4 : 72 \(⋮\) 4 và 66 \(⋮̸\) 4
nên tổng ( 44 + 66 + 72 ) \(⋮̸\) 4
b/ 58 - 20 +18
Vì 58 + 18 = 76 \(⋮\) 4 và 20 \(⋮\) 4
nên ( 58 - 20 + 18 ) \(⋮\) 4
c/ 2.5.4.11 - 56
Vì ( 2.5.4.11 ) \(⋮\) 4 và 56 \(⋮\) 4
nên \(\left(2.5.4.11-56\right)⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=2+2^2+2^3+2^4+.....+2^{98}+2^{99}\)
\(\Rightarrow2A=2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\)
\(\Rightarrow2A-A=\left(2^2+2^3+2^4+2^5.....+2^{99}+2^{100}\right)-\left(2+2^2+2^3+2^4+.....+2^{98}+2^{99}\right)\)
\(\Rightarrow A=2^{100}-2\)
b) \(B=2+2^4+2^7+......+2^{97}+2^{100}\)
\(\Rightarrow2^3B=2^4+2^7+......+2^{100}+2^{103}\)
\(\Rightarrow8.B-B=\left(2^4+2^7+......+2^{100}+2^{103}\right)-\left(2+2^4+2^7+......+2^{97}+2^{100}\right)\)
\(\Rightarrow7B=2^{103}-2\)
\(\Rightarrow B=\dfrac{2^{103}-2}{7}\)
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A = 20 + 21 + 22 + ... + 22006
2A = 2 + 22 + 23 +...+ 22006 + 22007
2A - A = ( 2 + 22 + 23 + ... + 22006 + 22007 ) - ( 20 + 21 + 22 +...+ 22006 )
A = 22007 - 1