`#3107.101107`

a)

\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)

Vậy, `x = 4`

b)

\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)

Vậy, `x = 3`

c)

\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)

Vậy, `x = 4.`

a) \(27< 3^x< 243\)

\(\Rightarrow3^3< 3^x< 3^5\)

\(\Rightarrow3< x< 5\)

c) \(3^x+3^{x+2}=810\)

\(\Rightarrow3^x\left(1+3^2\right)=810\)

\(\Rightarrow3^x.10=810\)

\(\Rightarrow3^x=810:10\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

Ta có  \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

           \(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

                 \(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)

                 \(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

                 \(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

                 \(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)

                  \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

                   \(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

                     \(\frac{1}{x+1}=\frac{1}{18}\left(1\right)\)

                   Có \(\left(1\right)\Leftrightarrow\left(x+1\right).1=1.18\) 

                                       \(\Rightarrow x+1=18\)

                                       \(\Rightarrow x=18-1\)

                                       \(\Rightarrow x=17\)

\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+...+\frac{1}{Xx\left(X+1\right)}=\frac{1}{9}\)

\(\frac{7-6}{6x7}+\frac{8-7}{7x8}+\frac{9-8}{8x9}+...+\frac{\left(X+1\right)-X}{Xx\left(X+1\right)}=\frac{1}{9}\)

\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{1}{9}\)

\(\frac{1}{6}-\frac{1}{X+1}=\frac{1}{9}\)

\(\frac{1}{X+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)

X+1=18

X=17

a) x = 6

b) x = 2

c) x = 7

d) x = -1

a: \(8x\left(2023+x\right)-8x\left(x+2024\right)=56\)

=>\(8x\left(x+2023-x-2024\right)=56\)

=>-8x=56

=>\(x=\frac{56}{-8}=-7\) (nhận)

b: \(5-2025x=9-2026\left(x-1\right)\)

=>-2025x+5=9-2026x+2026

=>-2025x+5=-2026x+2035

=>x=2035-5=2030(nhận)

c: \(\left(-12\right)^2\cdot x=56+10\cdot13x\)

=>144x=56+130x

=>14x=56

=>x=4(nhận)

d: -(x-32+11)=(21-33-x+7)

=>-(x-21)=(-x-5)

=>-x+21=-x-5

=>21=-5(vô lý)

=>x∈∅

e: \(-2\left(x+6\right)+6\left(x+10\right)=8\)

=>-2x-12+6x+60=8

=>4x+52=8

=>4x=8-52=-44

=>\(x=-\frac{44}{4}=-11\) (nhận)

Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)

<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)

<=> \(\frac{28}{15}x=-\frac{56}{125}\)

<=> \(x=-\frac{2}{15}\)

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)

<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

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