Ta có: \(\left\{{}\begin{matrix}x\left(x+2y+3z\right)=-5\\y\left(x+2y+3z\right)=27\\z\left(x+2y+3z\right)=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-5}=x+2y+3z\\\dfrac{y}{27}=x+2y+3z\\\dfrac{z}{5}=x+2y+3z\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{-5}=\dfrac{y}{27}=\dfrac{z}{5}\Rightarrow\left\{{}\begin{matrix}y=\dfrac{-27}{5}x\\z=-x\end{matrix}\right.\)

Ta có: \(x\left(x+2y+3z\right)=-5\Rightarrow x\left(x+2.\dfrac{-27}{5}x-3x\right)=-5\)

\(\Rightarrow\dfrac{-64}{5}x^2=-5\Rightarrow x^2=\dfrac{25}{64}\Rightarrow x=\dfrac{5}{8}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{27}{5}x=-\dfrac{27}{8}\\z=-x=-\dfrac{5}{8}\end{matrix}\right.\)

\(10.1000.100000=10.10^3.10^5=10^9\)

\(\sqrt{5}+2\sqrt{6}-\left(\sqrt{5}-2\sqrt{6}\right)=4\sqrt{6}\)

\(B=\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)

\(\Rightarrow2B=\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2B=\dfrac{1}{25}-\dfrac{1}{75}=\dfrac{2}{75}\Rightarrow B=\dfrac{1}{75}\)

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{10}-1\right)=\dfrac{-1}{2}.\dfrac{-2}{3}...\dfrac{-9}{10}=\dfrac{-1}{10}>\dfrac{-1}{9}\)

\(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)=\dfrac{-3}{2^2}.\dfrac{-8}{3^2}...\dfrac{-99}{10^2}=-\left(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}...\dfrac{9.11}{10^2}\right)=-\left(\dfrac{1.2...9}{2.3...10}.\dfrac{3.4...11}{2.3...10}\right)=-\left(\dfrac{1}{10}.\dfrac{11}{2}\right)=-\dfrac{11}{20}< -\dfrac{11}{21}\)

a) \(x^2-2xy-4z^2+y^2=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)=\left(6+4-2.45\right)\left(6+4+2.45\right)=-8000\)b) \(3\left(x-3\right)\left(x+7\right)+\left(x-4\right)^2+48=3\left(x^2+4x-21\right)+\left(x^2-8x+16\right)+48=4x^2+4x+1=\left(2x+1\right)^2=\left(2.0,5+1\right)^2=4\)

b) \(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)

\(\Leftrightarrow x-1=0\)   hoặc   \(x+1=0\)

\(\Leftrightarrow x=1\)         hoặc    \(x=-1\)

c) \(x^2-6x+8=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

a) \(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

(do \(x^2+1\ge1>0\))

Gọi thương số đó khi chia cho 15 là a

=> Số đó có dạng 15a + 9

1) \(15a+9=3\left(5a+3\right)⋮3\)

2) Ta có: \(15a⋮5\)

               \(9⋮̸5\)

\(\Rightarrow15a+9⋮̸5\)

\(C+O_2\rightarrow CO_2\)