Ta có: \(2x=9y=4z\Leftrightarrow\dfrac{2x}{36}=\dfrac{9y}{36}=\dfrac{4z}{36}\Leftrightarrow\dfrac{x}{18}=\dfrac{y}{4}=\dfrac{z}{9}\)

Đặt:

\(\dfrac{x}{18}=\dfrac{y}{4}=\dfrac{z}{9}=t\Leftrightarrow\left\{{}\begin{matrix}x=18t\\y=4t\\z=9t\end{matrix}\right.\)

Nên:

\(\dfrac{1}{18t}+\dfrac{1}{12t}+\dfrac{1}{9t}=\dfrac{1}{4}\)

Tính t rồi thay vào là ok

\(P=\dfrac{3n+2}{n-1}=\dfrac{3n-3+5}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{5}{n-1}=3+\dfrac{5}{n-1}\)

\(\Rightarrow5⋮n-1\Leftrightarrow n-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}n-1=1\\n-1=-1\\n-1=5\\n-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=2\\n=0\\n=6\\n=-4\end{matrix}\right.\left(tm\right)\)

Cho hình tam giác ABC có diện tích 20 cm2 . Giả sử rằng D là trung điểm của cạnh ABvà DECB là hình thang có đáy BC gấp 4 lần đáy DE . Hãy tính diện tích tam giác AEC

\(a=3+3^2+3^3+...+3^{2008}\)

\(3a=3^2+3^3+3^4+...+3^{2009}\)

\(3a-a=\left(3^2+3^3+3^4+...+3^{2009}\right)-\left(3+3^2+3^3+...+3^{2008}\right)\)

\(2a=3^{2009}-3\)

\(2a+3=3^{2009}=3^x\)

\(x=2009\)

\(\left(x-3\right)\left(x-4\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>4\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< 4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< 4\end{matrix}\right.\)

\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

\(\Leftrightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}=\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

\(\Rightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y}{b}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{y}{b}\end{matrix}\right.\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\left(đpcm\right)\)

\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

\(\Rightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)\(\Rightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)

\(\Rightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)

\(\Rightarrow100-x=0\Rightarrow x=100\)

\(A=1^2+3^2+6^2+9^2+12^2+...+39^2\)

\(A=1^2+3^2+3^2\left(2^2+3^2+4^2+....+13^2\right)\)

\(A=1+9+9.818\)

\(A=10+7362=7372\)

phân tích thành đa nhân tử?

\(x^4+4=\left(x^2\right)^2+2^2\)

\(=\left(x^2\right)^2+4x^2+2^2-4x^2\)

\(=\left[\left(x^2\right)^2+4x^2+2^2\right]-4x^2\)

\(=\left(x^2+2\right)^2-4x^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2\)

\(=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)