Ta có : 3x + 3x + 2 = 810
=> 3x(1 + 32) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
ta có \(3^3+3^x+2=810\)
=>\(3^x\left(1+3^2\right)=810\)
=>\(3^x.10=810\)
=>\(3^x=81\)
=>\(3^x=3^4\)
=>x=4
Vậy x=4
Ta có : 3x + 3x + 2 = 810
=> 3x(1 + 32) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
ta có \(3^3+3^x+2=810\)
=>\(3^x\left(1+3^2\right)=810\)
=>\(3^x.10=810\)
=>\(3^x=81\)
=>\(3^x=3^4\)
=>x=4
Vậy x=4
Bài 1: Thực hiện phép tính:
a, \(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
b, \(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
c, \(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}\)
d, \(\frac{4^{2002}.9^{1001}}{16^{1001}.3^{2003}}\)
e, \(\sqrt{25-16}-\left|-3,7+0,7\right|\)
Bài 2: Tìm x
a, \(\frac{1}{3}x+\frac{4}{5}=3\frac{4}{5}\)
b, \(\left|x+\frac{3}{4}\right|-2,25=1\frac{3}{4}\)
c, \(\left(-x+\frac{2}{5}\right)^4=\frac{1}{16}\)
d, \(\left(\frac{2}{5}\right)^{3x}:\left(\frac{4}{3}\right)^{21}=\left(\frac{6}{20}\right)^{21}\)
e, \(\frac{-x}{\frac{3}{5}}=\frac{\frac{27}{5}}{-x}\)
g, \(x:1\frac{1}{2}=-2,5:2\frac{1}{5}\)
Chứng minh : \(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{64}{125}\)
Bài 1: Tìm n
1/4*2/6*3/8*4/10*5/12*.....* 30/62*31/64=2^n
Bài 2: Tính
a) \(\frac{\left(-5\right)^{60}\cdot30^5}{15^5\cdot5^{61}}\)
b)\(2^3+3\cdot\left(\frac{1}{3}\right)^0-2^{-2}\cdot4+\left[\left(-2\right)^3:\frac{1}{2}\right]\cdot8\)
(GỢI Ý CÂU B:\(^{2^{-2}=\frac{1}{2^2}}\)
Bài 3: Tìm x
a)\(^{2^x=16}\)
b)\(^{\left(\frac{x}{13}\right)^2=\frac{49}{169}}\)
c)\(\left(\frac{-1}{5}\right)^x=\left(\frac{-1}{125}\right)^3\cdot x^4=\frac{-16}{625}\)
d)\(^{6^{4-x}=216}\)
Bài 4:Tìm n
a)\(3^n\cdot3^{-2}=3^5\)
b)Tìm x:
1)\(\frac{2^{4-x}}{16^5}=32^6\)
2)\(^{9\cdot5^x=6\cdot5^6+3\cdot5^6}\)
3)\(\frac{2^3}{2^x}=4^5\)
Các bạn giúp mik!!!Mik sắp kiểm tra 45p r
a) \(\frac{\left(-1\right)^3}{15}+\left(-\frac{2}{3}\right):2\frac{2}{3}-\left|-\frac{5}{6}\right|\)
b) \(1\frac{5}{13}-0,\left(3\right)-\left(1\frac{4}{9}+\frac{18}{13}-\frac{1}{3}\right)\)
c) \(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{5}-125\frac{1}{3}\)
d) \(\frac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
TÌM X
a,\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=6\)
b,\(\left(x^2-4\right).\left(2x+\frac{4}{3}\right)=0\)
\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=-\frac{64}{125}\)
Chứng minh rằng:
a)\(12^8.9^{12}=18^{16}\)
b)\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
c)\(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
Làm nhanh giúp mình nhá. Thanks. ^_^
a) C/m: \(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow a=b=c\)
b) C/m: \(T=x\left(x-a\right)\left(x+a\right)\left(x+2a\right)+a^4\ge0\) \(\forall x,a\in R\)
c) Tìm x sao cho: \(\frac{x+5}{2015}+\frac{x+4}{2016}+\frac{x+3}{2017}+\frac{x+2}{2018}=\frac{x+2015}{5}+\frac{x+2016}{4}+\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\frac{\left(5^4-5^3\right)}{125^4}-\frac{64}{125}\)