a, \(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^{10}+6^7\)

\(\Leftrightarrow\frac{1}{6}\cdot6^x+6^x\cdot6^2=6^{10}+6^7\)

\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(6^3+1\right)\)

\(\Leftrightarrow6^{x-1}=6^7\Leftrightarrow x-1=7\)

\(\Leftrightarrow x=8\)

b, \(\left(\frac{1}{2}-\frac{1}{6}\right)\cdot3^{x+4}-4\cdot3^x=3^{16}-4\cdot3^{13}\)

\(\Leftrightarrow\frac{1}{3}\cdot3^{x+4}-4\cdot3^x=3^{13}\left(3^3-4\right)\)

\(\Leftrightarrow3^x\cdot3^3-4\cdot3^x=3^{13}\left(3^3-4\right)\)

\(\Leftrightarrow3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)

\(\Leftrightarrow3^x=3^{13}\Leftrightarrow x=13\)

a. x=8

b. x=13

còn cách tính thì mình quên rồi vì minh học cái này lâu lắm rồi ko nhớ đc.

có chỗ x^7 hả

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{2}{3}\)

a)\(0,2:1\frac{1}{5}=\frac{2}{3}:\left(6.x+7\right)\)

\(\frac{2}{3}:\left(6.x+7\right)=0,2:1\frac{1}{5}\)

\(\frac{2}{3}:\left(6.x+7\right)=0,2:\frac{6}{5}\)

\(\frac{2}{3}:\left(6.x+7\right)=\frac{1}{6}\)

\(6.x+7=\frac{2}{3}:\frac{1}{6}\)

\(6.x+7=4\)

      \(6.x=4-7\)

       \(6.x=-3\)

           \(x=-3:6\)

            \(x=-0,5\)

  Vậy x=-0,5 hay \(\frac{-1}{2}\)

d)\(\frac{x}{y}=\frac{2}{3};x.y=96\)

Từ \(\frac{x}{y}=\frac{2}{3}\)suy ra \(\frac{x}{3}=\frac{y}{2}\)

 Đặt k=\(\frac{x}{3}=\frac{y}{2}\)

\(\Rightarrow x=3.k;y=2.k\)

Vì \(x.y=96\)nên \(2k.3k=96\)

                                            \(\Rightarrow6.k^2=96\)

                                              \(\Rightarrow k^2=96:6\)

                                               \(\Rightarrow k^2=16\)

                                                 \(\Rightarrow k=4\)hoặc\(k=-4\)

+)Với \(k=4\)thì \(x=2\);\(y=3\)

+)Với \(k=-4\)thì \(x=-2\);\(y=-3\)

               Vậy \(x=2;y=3\)hoặc \(x=-2;y=-3\)

e) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)và \(x.y.z=810\)

    Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)

\(\Rightarrow x=2k;y=3k;z=5k\)

Vì \(x.y.z=810\)nên \(2k.3k.5k=810\)

                                \(\Rightarrow30.k^3=810\)

                                 \(\Rightarrow k^3=810:30\)

                                  \(\Rightarrow k^3=27\)

                                   \(\Rightarrow k=3\)

Với \(k=3\)thì \(x=6\); \(y=9\); \(z=15\)

            Vậy \(x=6\); \(y=9\); \(z=15\)

Mk chỉ làm đc vậy thui bn à! Xin lỗi thật nhiều nha

bài ở sách mô đây mi