3^n+2 - 2^n+2 + 3^n - 2^n = (3ⁿ⁺²+3ⁿ)+(-2ⁿ⁺²-2ⁿ)

=3ⁿ.(3²+1)-2ⁿ.(2²+1)

=3ⁿ.10-2ⁿ.5

=3ⁿ.10-2^n-1.2.5

=3ⁿ.10-2^n-1.10

=10.(3ⁿ-2^n-1)

Vậy 3^n+2 - 2^n+2 + 3^n - 2^n chia hết cho 10

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}=\frac{2x^2+2y^2-3z^2}{2\cdot9+2\cdot16-3\cdot25}=\frac{-100}{-25}=4\)

\(\Rightarrow x^2=36;y^2=64;z^2=100\)

\(\Rightarrow\) x = + 6; y = + 8; z = + 10

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+4}{2001}=\frac{x+4}{2002}+\frac{x+4}{2003}\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\Rightarrow x+2004=0\)

=>x=-2004

vậy x=-2004

tham khảo bài của mình tại  http://olm.vn/hoi-dap/question/133172.html

\(=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)+\left(-\frac{3}{4}-\frac{2}{9}-\frac{1}{36}\right)+\frac{1}{64}\)

= 1 + -1 + 1/64 

= 0 +1/64 

= 1/64

=> 5x - 1 = 0 hoặc 2x - 1 / 3 = 0 

=> x = 1/5 hoạc x = 1/6

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