\(\left\{\begin{matrix}2x=y^2-4y+5\left(1\right)\\2y=x^2-4x+5\left(2\right)\end{matrix}\right.\)

Lấy (1) - (2) ta được:

\(2x-2y=y^2-4y+5-x^2+4x-5\)

\(\Leftrightarrow y^2-x^2-2y+2x=0\)

\(\Leftrightarrow\left(y-x\right)\left(y+x-2\right)=0\)

Tới đây thì đơn giản rồi chỉ cần thế ngược lại là ra nhé

\(x^2-2xy+2y^2-2x+6y+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(-2x+2y\right)+1+\left(y^2+4y+4\right)=0\)

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

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

\(\Leftrightarrow\left\{\begin{matrix}x-y-1=0\\y+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)

\(\Rightarrow A=\frac{3x^2y-1}{4xy}=\frac{3.\left(-1\right)^2.\left(-2\right)-1}{4.\left(-1\right).\left(-2\right)}=-\frac{7}{8}\)

Đặt \(1\left\{\begin{matrix}\frac{a}{x}=m\\\frac{b}{y}=n\\\frac{c}{z}=p\end{matrix}\right.\)

Thì bài toán thành:

Cho \(\left\{\begin{matrix}\frac{1}{m}+\frac{1}{n}+\frac{1}{p}=0\left(1\right)\\m+n+p=2\left(2\right)\end{matrix}\right.\)

Tính A = m² + n² + p²

Từ (1) ta suy ra được: mn + np + pm = 0

Từ (2) ta bình phương 2 vế được:

\(m^2+n^2+p^2+2\left(mn+np+pm\right)=4\)

\(\Rightarrow A=4\)

Có cần giải câu này nữa không b?

\(M=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^2-xy+y^2=\left(1-y\right)^2-\left(1-y\right)y+y^2\)

\(=3y^2-3y+1=\left(3y^2-3y+\frac{3}{4}\right)+\frac{1}{4}\)

\(=3\left(y-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

Vậy GTNN là \(\frac{1}{4}\) dạt được khi \(x=y=\frac{1}{2}\)

 \(a^2=b.c\Rightarrow\frac{b}{a}=\frac{a}{c}=k\Rightarrow b=a.k;a=c.k\)

a, ta có:\(\frac{a+b}{a-b}=\frac{c.k+a.k}{c.k-a.k}=\frac{k.\left(c+a\right)}{k.\left(c-a\right)}=\frac{c+a}{c-a}\) 

b, ta có : \(\frac{2015c+a}{2015a+b}=\frac{2015c+c.k}{2015a+a.k}=\frac{c.\left(2015+k\right)}{a.\left(2015+k\right)}=\frac{c}{a}\)

\(Fe\left(0,2\right)+2HCl\rightarrow FeCl_2+H_2\left(0,2\right)\)(1)

\(Fe_2O_3\left(0,2\right)+3H_2\rightarrow2Fe\left(0,4\right)+3H_2O\) (2)

\(n_{H_2}=\frac{4,48}{22,4}=0,2\)

\(\Rightarrow n_{Fe}=0,2\)

\(\Rightarrow m_{Fe}=0,2.56=11,2\)

\(\Rightarrow m_{Fe\left(2\right)}=33,6-11,2=22,4\)

\(\Rightarrow n_{Fe\left(2\right)}=\frac{22,4}{56}=0,4\)

\(\Rightarrow m_{Fe_2O_3}=0,2.160=32\)

\(\Rightarrow\%Fe=\frac{11,2}{11,2+32}.100\%=25,93\%\)

\(\Rightarrow\%Fe_2O_3=100\%-25,93\%=74,07\%\)

a/ \(4Al\left(\frac{1}{15}\right)+3O_2\left(0,05\right)\rightarrow2Al_2O_3\left(\frac{1}{30}\right)\)

b/ \(n_{Al}=\frac{1,8}{27}=\frac{1}{15}\)

\(\Rightarrow m_{Al_2O_3}=102.\frac{1}{30}=3,4\)

c/ \(V_{O_2}=0,05.22,4=1,12\)

\(\Rightarrow V_{kk}=1,12.5=5,6\)

d/ \(2KMnO_4\left(0,1\right)\rightarrow K_2MnO_4+MnO_2+O_2\left(0,05\right)\)

\(\Rightarrow m_{KMnO_4}=0,1.158=15,8\)

\(4Al\left(0,4\right)+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2\left(0,3\right)\)

Chất rắn không tan là Ag.

\(n_{H_2}=\frac{6,72}{22,4}=0,3\)

\(\Rightarrow m_{Al}=0,4.27=10,8\)

\(\Rightarrow\%Al=\frac{10,8}{10,8+4,6}.100\%=70,13\%\)

\(\Rightarrow\%Ag=100\%-70,13\%=29,87\%\)

a/ \(9^{2n+1}+1=\left(9+1\right)\left(9^{2n}-9^{2n-1}+...\right)=10\left(9^{2n}-9^{2n-1}+...\right)\)

Chia hết cho 10

b/ \(3^{4n+1}+2=3^{4n+1}-3+5=3\left(3^{4n}-1\right)+5\)

\(=3\left(81^n-1\right)+5=3.80\left(81^{n-1}+...\right)+5\)

Cái này chia hết cho 5