Câu 1:

\(3^{n+3}+2^{n+3}+3^{n+1}+2^{n+2}=3^{n+1}\left(3^2+1\right)+2^{n+2}\left(2+1\right)\)

\(=3^{n+1}.10+2^{n+2}.3\)

\(=2.3.\left(3^n+2^{n+1}\right)\)

\(=6.\left(3^n+2^{n+1}\right)\) chia hết chia hết cho 6

Câu 2:

\(\frac{x}{y}=\frac{9}{7}\Leftrightarrow\frac{x}{9}=\frac{y}{7}\)(1)

\(\frac{y}{z}=\frac{7}{3}\Leftrightarrow\frac{y}{7}=\frac{z}{3}\)(2)

Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:  \(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3\)

Hay: \(\frac{x}{9}=-3\Leftrightarrow x=-27\)

        \(\frac{y}{7}=-3\Leftrightarrow y=-21\)

         \(\frac{z}{3}=-3\Leftrightarrow z=-9\)

Vậy ...........................

biết mà không chi

???? cái j vậy

Ta có : B = 2x+1/x-3 = (2x-6)+7/x-3 = 2+ 7/x-3 

Để B nhận giá trị nguyên thì x-3 thuộc Ư(7) = (+-1;+-7)

suy ra : x-3=-1 => x=2                               x-3=1 => x=4

             x-3=-7 => x=-4                               x-3=7 => x=10

Vậy x =(-4;2;4;10) thì B nhận giá trị nguyên

a, Xét tam giác AOM và tam giác BOM

Ta có: OA = OB ( giả thiết)

         góc AOM = góc BOM ( Ot là tia phân giác góc xOy)

         OM cạnh chung

Do đó: tam giác AOM = tam giác BOM ( c-g-c)

làm tiếp đi bạn tôi k cho

a)\(F\left(0\right)=2016-3.0\left|2.0-5\right|\)

    \(F\left(0\right)=2016\)

     \(F\left(3\right)=2016-3.3\left|2.3-5\right|\)

      \(F\left(3\right)=2016-9\)

      \(F\left(3\right)=2007\)

bằng

\(\frac{x}{5\left(x+1\right)}\)-\(\frac{x}{10\left(x-1\right)}\)=\(x\left(\frac{1}{5\left(x+1\right)}-\frac{1}{10\left(x-1\right)}\right)\)=\(x\left(\frac{2\left(x-1\right)}{10\left(x+1\right)\left(x-1\right)}-\frac{x+1}{10\left(x-1\right)\left(x+1\right)}\right)\)

=\(x\left(\frac{2\left(x-1\right)-\left(x+1\right)}{10\left(x+1\right)\left(x-1\right)}\right)\)=\(x\left(\frac{2x-2-x+1}{10\left(x-1\right)\left(x+1\right)}\right)\)=\(x\left(\frac{x-1}{10\left(x-1\right)\left(x+1\right)}\right)\)=\(x\left(\frac{1}{10\left(x+1\right)}\right)\)=\(\frac{x}{10x+10}\)

\(\Rightarrow\frac{x}{5\left(x+1\right)}-\frac{x}{10\left(x-1\right)}\)

\(\Leftrightarrow\frac{2x\left(x-1\right)}{10\left(x+1\right)\left(x-1\right)}-\frac{x\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow\frac{2x\left(x-1\right)-x\left(x+1\right)}{10\left(x+1\right)\left(x-1\right)}\)

\(\Leftrightarrow\frac{2x^2-2x-x^2-x}{10\left(x+1\right)\left(x-1\right)}\)\(\Leftrightarrow\frac{x^2-3x}{10\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow\frac{x}{5x+5}-\frac{x}{10x-10}=\frac{x^2-3x}{10\left(x+1\right)\left(x-1\right)}\)

Ta có:\(\frac{x}{3}=\frac{y}{2};\frac{y}{4}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{6}=\frac{y}{4}=\frac{z}{10}=\frac{30}{20}=\frac{3}{2}\)

\(\Rightarrow x=\frac{3}{2}.6=9\)

\(\Rightarrow y=\frac{3}{2}.4=6\)

\(\Rightarrow z=\frac{3}{2}.10=15\)