\(3^{x+1}+3^{x+2}=2^2.3^4\)

\(3^{x+1}+3.3^{x+1}=4.81\)

\(4.3^{x+1}=4.81\)

\(3^{x+1}=81=3^4\)

\(x+1=4\Rightarrow x=3\)

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

Chọn C

Ta có: P(x) + Q(x) = x3+ x2+ 2x-1

⇒ Q(x) = (x3 + x2 + 2x-1) - P(x)

= 2x3 + 4x2 - 8x - 3.

\(720:\left(x\times2+x\times3\right)=2\times3\)

\(720:\left(x\times5\right)=6\)

\(x\times5=720:6\)

\(x\times5=120\)

\(x=120:5\)

\(x=24\)

3^x+25=26x2²+2x3⁰

3^x+25=26x4+2

3^x+25=106

3^x=106-25=81

3^x=3⁴

⇒ x=4

lê minh quang bị mắc lỗi ở chỗ 3^0

f'(x)=2*3x^2+3*2*(a+2)*x+6a^2

=6x^2+6x(a+2)+6a^2

Δ=(6a+12)^2-4*6*6a^2

=36a^2+144a+144-144a^2

=-108a^2+144a+144

f'(x)>0 với mọi x

=>-108a^2+144a+144<0

=>a<-2/3; a>2

f'(-1)=6

=>6*(-1)^2+6*(-1)*(a+2)+6a^2=6

=>6a^2+6-6a-12=6

=>6a^2-6a-12=0

=>a^2-a-2=0

=>a=2(loại) hoặc a=-1(nhận)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

\(1-\frac{1}{x+1}=\frac{99}{100}\)

=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

=> x+1 = 100

=> x = 100 - 1 

=> x = 99

mơ đi Nguyễn Đình Dũng

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{299}{600}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{299}{600}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{299}{600}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{299}{600}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{600}\)

\(\Rightarrow x+1=600\)

\(\Rightarrow x=600-1\)

\(\Rightarrow x=599\)

\(Vậy\) \(x=599\)