3^x.9+3^x+1=2².3⁹

3^x.9+3^x.3=4.19683

3^x.(9+3)=78732

3^x.12=78732

3^x=78732:12

3^x=6561

3^x=3⁸

=>x=8

Vậy x=8

a) \(\Leftrightarrow2.\left(\frac{2.3^x}{3}+3^x.3^2\right)=2.3^6\left(2+3^3\right)\)

\(\Leftrightarrow2.\left(\frac{2.3^x+3.3^x.3^2}{3}\right)=2.3^6.29\)

\(\Leftrightarrow2.\left[\frac{3^x.\left(2+3.3^2\right)}{3}\right]=2.3^6.19\)

\(\Leftrightarrow2.3^{x-1}.29=2.3^6.29\Leftrightarrow3^{x-1}.29=\frac{2.3^6.29}{2}=3^6.29\Leftrightarrow3^{x-1}=\frac{3^6.29}{29}=3^6\)

\(\Leftrightarrow3^{x-1}=3^6\Leftrightarrow x-1=6\Leftrightarrow x=6+1=7\)

vậy x=7 . Chọn mình nha

mấy bài sao tương tự nếu ko biết thì nhắn tin mình chỉ típ nha

bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)

\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)

b)

ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

Ta có: P=AB

\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)

\(=\dfrac{3x}{x+1}\)

Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)

\(\Leftrightarrow9\left(x+1\right)=6x\)

\(\Leftrightarrow9x-6x=-9\)

\(\Leftrightarrow3x=-9\)

hay x=-3(loại)

Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)

kết quả là: 6

                x=1;2

                x=7 y=0

                n=1;3;5;7

                x=1  y=3

1)\(\frac{11\cdot3^{29}-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)

2)\(|2x-3|+2^3\cdot3=25\Rightarrow x=1;2\)

3)  \(x183y=61831\Rightarrow x=6;y=1\)

4)\(B=\frac{n-1}{n-4}\Rightarrow n=1;3;5;7\)

5)\(\left(2x+1\right)\cdot\left(y^2-5\right)=12\Rightarrow x=1;y=3\)

mình là người đúng nhất ở bài 3 vì 61831 mới chia 2,5,9 dư 1

k cho mình nhé

a: \(x-\frac25+\frac{7}{12}=\frac12\cdot\frac34\cdot\frac{-9}{16}\)

=>\(x-\frac{24}{60}+\frac{35}{60}=\frac38\cdot\frac{-9}{16}=\frac{-27}{128}\)

=>\(x=-\frac{27}{128}-\frac{11}{60}=\frac{-27\cdot15-11\cdot32}{1920}=-\frac{757}{1920}\)

b: \(\frac49x-\frac12=-\frac59\)

=>\(\frac49x=-\frac59+\frac12=\frac{-10+9}{18}=\frac{-1}{18}\)

=>\(x=-\frac{1}{18}:\frac49=-\frac{1}{18}\cdot\frac94=\frac{-1}{2\cdot4}=-\frac18\)

c: \(-\frac23x+\frac58=\frac{7}{12}\)

=>\(-\frac23x=\frac{7}{12}-\frac58=\frac{14}{24}-\frac{15}{24}=-\frac{1}{24}\)

=>\(x=\frac{1}{24}:\frac23=\frac{1}{24}\cdot\frac32=\frac{1}{8\cdot2}=\frac{1}{16}\)

d: \(\frac25x:\frac73\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac25\cdot\frac37\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac{3}{4\cdot5\cdot7}=\frac{29}{15}\)

=>\(x=\frac{29}{15}:\frac{3}{140}=\frac{29}{15}\cdot\frac{140}{3}=\frac{29\cdot28}{3\cdot3}=\frac{812}{9}\)

1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)

2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)

\(2^x=2.2^8=2^9;x=9\)

4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)

\(\Leftrightarrow3^5.x=3^5.5;x=5\)