кαвαиє ѕнιяσ sao bt hay z :)

Trích mẫu thử, dùng dung dịch NaOH \(\rightarrow\) nhận biết được Al (tan, sủi bọt khí)

Dùng dung dịch HCl \(\rightarrow\) nhận biết được Fe (tan, sủi bọt khí)

Dùng dung dịch \(AgNO_3\rightarrow\) Cu tác dụng tạo Ag còn  Ag ko phản ứng

\(Al+NaOH+H_2O\to NaAlO_2+\dfrac{3}{2}H_2\\ Fe+2HCl\to FeCl_2+H_2\\ Cu+2AgNO_3\to Cu(NO_3)_2+2Ag\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{99}+2^{100}\right)\\ A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{99}\right)=3\left(2+2^3+...+2^{99}\right)⋮3\)

Bài 2 đa thức bậc 2 chia đa thức bậc 2 dư đa thức bậc 1 ??

Bài 1:

Đặt \(a=\sqrt[7]{\dfrac{3}{5}};b=\sqrt[7]{\dfrac{5}{3}}\Rightarrow\left\{{}\begin{matrix}a+b=x\\ab=1\end{matrix}\right.\)

Ta có \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\)

\(\Rightarrow a^3+b^3=x\left(x^2-3\right)=x^3-3x\)

Ta có \(a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2\)

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

\(\Rightarrow\left(a^3+b^3\right)\left(a^4+b^4\right)=\left(x^3-3x\right)\left(x^4-4x^2+2\right)\\ =x^7-3x^5-4x^5+12x^3+2x^3-6x\\ =x^7-7x^5+14x^3-6x\)

Lại có \(\left(a^4+b^4\right)\left(a^3+b^3\right)=a^7+b^7+\left(ab\right)^3\left(a+b\right)=\dfrac{3}{5}+\dfrac{5}{3}+x=\dfrac{34}{15}+x\)

\(\Rightarrow x^7-7x^5+14x^3-6x=\dfrac{34}{15}+x\\ \Rightarrow15x^7-105x^5+210x^3-105x-34=0\left(1\right)\)

Vậy (1) nhận \(x=\sqrt[7]{\dfrac{3}{5}}+\sqrt[7]{\dfrac{5}{3}}\) làm nghiệm

\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\\ M=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(a^2+b^2\right)+6a^2b^2\\ M=1-3ab+3ab\left(a^2+b^2+2ab\right)=1-3ab+3ab\left(a+b\right)^2\\ M=1-3ab+3ab=1\)

\(=\dfrac{30\cdot a^{2010}+30\cdot b^{2010}-4\cdot a^{2009}-4\cdot b^{2009}}{a^{2008}+b^{2008}}\\ =\dfrac{a^{2008}\left(30a^2-4b\right)+b^{2008}\left(30b^2-4b\right)}{a^{2008}+b^{2008}}\)

\(k,\Rightarrow\left[{}\begin{matrix}4,2-2x=0,2\\2x-4,2=0,2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=4,4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,2\end{matrix}\right.\\ m,\Rightarrow\dfrac{6}{x}=\dfrac{8}{3}\Rightarrow x=\dfrac{6\cdot3}{8}=\dfrac{9}{4}\\ n,\Rightarrow\dfrac{5}{3}:\dfrac{x}{4}=20\\ \Rightarrow\dfrac{x}{4}=\dfrac{5}{3}:20=\dfrac{1}{12}\Rightarrow x=\dfrac{4}{12}=\dfrac{1}{3}\\ p,\Rightarrow\dfrac{8}{3}:x=\dfrac{16}{9}:\dfrac{8}{3}=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{8}{3}:\dfrac{2}{3}=4\)

nhân ra xong đặt nhân tử chung -,-