cho A= 3+3\(^2\)+\(3^3+3_{ }^4+...+3^{2016}\)
a) Tính a
b) tim x, biet 2A+3=3x
NQ
Những câu hỏi liên quan
Cho A= 3+3^2+3^3+...+3^2016 a,Tính A b, Tìm x để 2A+3=3^x
a)
Ta có 3A=\(3^2+3^3+3^4+...+3^{2017}\)
3A-A=\(\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)
2A=\(3^{2017}-3\)
A=\(\frac{3^{2017}-3}{2}\)
b)
A=\(\frac{3^{2017}-3}{2}\)
2A=\(3^{2017}-3\)
2A+3=\(3^{2017}-3+3=3^{2017}\)
=>x=2017
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Cho A = 3 +3^2 + 3^3 + .....+ 3^2008. tim x biet 2a+3=3^x
\(A=3+3^2+...+3^{2008}\)
\(3A=3.\left(3+3^2+...+3^{2008}\right)\)
\(3A-A=\left(3^2+3^3+...+3^{2009}\right)-\left(3+3^2+...+3^{2008}\right)\)
\(2A=3^{2009}-3\)
\(2A+3=3^{2009}-3+3\)
\(2A+3=3^{2009}\)
Vì \(2A+3=3^x\)hay \(3^{2009}=3^x\)
\(\Rightarrow x=2009\)
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Thank you to kick me ooooooooooooooooooo
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tim x biet : a,-12/21×x + 3/7=-4/3, b, 2x +-3^2=5/4 , c, 75% x- 2/5=3/10, d, (3x-2)×(x+7)
Tinh gia tri bieu thuc
1/(2016^-2016+1)+1/(2016^-2015+1)+.....+1/(2016^-1+1)+1/(2016^0+1)+1/(2016^1+1)+....+1/(2016^2016+1)
2. Tim bo 3 so nguyen duong a,b,c biet
ac=b(a-b+c) và 1/a+1/b+1/c=1
3. Tim cac cap so thuc x,y co tich la 1 sao cho
(x+y-2)/4=1/(x-1)+1/(y-1)
cho A=3^1+3^2+3^3+....+3^2015 tim n biet 2A+3=3^n
Bạn ơi, A + 3 + ... hay là A = 3 + 32+... hả bạn?
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tim a b biet x ^4-3x +2=(x-1)(x^3+ax^2+bx-2)
a > tim cap so nguyen x , y biet (x-1) x ( y+2 ) = 5
b> cho A = 31 + 32+33 +...+32006 . tìm x để 2A + 3 = 3x
Cho a,b,c>0 t/m \(a^2+b^2+c^2=3\). Tìm max
P\(P=\dfrac{a}{a^2+2b+3}+\dfrac{b}{b^2+2c+3}+\dfrac{c}{c^2+2a+3_{ }}\le\dfrac{1}{2}\)
Ta có: \(P\le\dfrac{a}{2a+2b+2}+\dfrac{b}{2b+2c+2}+\dfrac{c}{2c+2a+2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{a}{a+b+1}+\dfrac{b}{b+c+1}+\dfrac{c}{c+a+1}\le1\)
\(\Rightarrow\dfrac{a}{a+b+1}-1+\dfrac{b}{b+c+1}-1+\dfrac{c}{c+a+1}-1\le-2\)
\(\Leftrightarrow\dfrac{b+1}{a+b+1}+\dfrac{c+1}{b+c+1}+\dfrac{a+1}{c+a+1}\ge2\)
Thật vậy, ta có:
\(VT=\dfrac{\left(a+1\right)^2}{\left(a+1\right)\left(a+c+1\right)}+\dfrac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}+\dfrac{\left(c+1\right)^2}{\left(c+1\right)\left(b+c+1\right)}\)
\(VT\ge\dfrac{\left(a+b+c+3\right)^2}{ab+bc+ca+3\left(a+b+c\right)+6}=\dfrac{2\left(ab+bc+ca\right)+6\left(a+b+c\right)+12}{ab+bc+ca+3\left(a+b+c\right)+6}=2\)
Dấu "=" xảy ra khi \(a=b=c=1\)
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cho 2 đa thức
A(x) = 1/3(x^3-6x^4+3x^2-1) + 2(x^2-x^5+x)
B(x) = x^6-4x^5+2x^2+x^3+2/3
a, tính a(x)+b(x), 2a(x)-b(x), 3a(x)-6b(x)
b, tính a(4), a(-1), b(2), a(-1)-2b(1)