2. Gọi Kim loại có hoá trị 2 là A => CTHH oxit là AO

\(m_O=11,2-8=3,2g\)

\(n_O=\dfrac{3,2}{16}=0,2mol\)

\(PT:2A+O_2-t^o>2AO\)

\(0,2mol\) \(0,4mol\)

Ta có: \(n_{AO}=\dfrac{m_{AO}}{M_{AO}}\Leftrightarrow0,2=\dfrac{11,2}{A+16}\Leftrightarrow0,2A+3,2=11,2\)

\(\Leftrightarrow0,2A=8\)

\(\Leftrightarrow A=40\)

\(\Rightarrow A\) là \(Ca\Rightarrow CTHH\) của \(Oxit\) là \(CaO\)

\(\)

\(2Fe+6H_2SO_4\) đặc \(->Fe_2\left(SO_4\right)_3+6H_2O+3SO_4\uparrow\)

Còn lại là Fe₂O₃

2. \(a+b+c=0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3a^{2c}+3ac^2+3b^2c+3bc^2+6abc\)

\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)+\left(3b^2c+3bc^2+3abc\right)-3abc\)

\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3ac\left(a+c+b\right)+3bc\left(b+c+a\right)-3abc\)

Ta có: \(a+b+c=0\)

\(a^3+b^3+c^3+3ab.0+3ac.0+3bc.0=3abc\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

\(\sin^4a.\left(3-2\sin^2a\right)+\cos^4a\left(3-2\cos^2a\right)\)

\(=3\sin^4a-2\sin^6a+3\cos^4a-2\cos^6a\)

\(=3\left(\sin^4a+\cos^4a\right)-2\left(\sin^6a+\cos^6a\right)\)

\(=3\left(\left(\sin^2a\right)^2+\left(\cos^2a\right)^2\right)-2\left(\left(\sin^2a\right)^3+\left(\cos^2a\right)^3\right)\)

\(=3.1-2\left(sin^2a+\cos^2a\right)\left(\sin^4-sin^2.\cos^2+\cos^4\right)\)

\(=3-2.1\left(\left(\sin^2a\right)^2+\left(\cos^2a\right)^2\right).\left(-\sin^2.\cos^2\right)\)

\(=3-2\left(-\sin^2.\cos^2\right)\)

\(k_1=k_2=k_3\)

Cắt lò xo làm 3 phần bằng nhau => \(l_1=l_2=l_3\)

\(\Rightarrow k_1.l_1=k_2.l_2=k_3.l_3=k.l\)

kl=k₁.l.300=>k₁=k₃=k₂=300N/m

Aps dụng bất đẳng thức cô si cho 2 số 1-x và 1-x ta có:

\(\dfrac{1-x+1-z}{2}\ge\sqrt{\left(1-x\right)\left(1-z\right)}\)

\(\Leftrightarrow\left(1-z\right)\left(1-x\right)\le\left(\dfrac{1-z+1-x}{2}\right)^2\)

\(\Leftrightarrow4\left(1-z\right)\left(1-x\right)\le\left(1+y\right)^2\)

\(\Leftrightarrow4\left(1-x\right)\left(1-y\right)\left(1-z\right)\le\left(1+y\right)^2\left(1-y\right)\)

Ta có: \(1-y^2\le1\)

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

Do đó: \(4\left(1-x\right)\left(1-y\right)\left(1-z\right)\le x+2y+z\)

Áp dụng BĐT cô-si cho 2 số 1-x và 1-z ta được:

\(\dfrac{1-x+1-z}{2}\ge\sqrt{\left(1-x\right)\left(1-z\right)}\)

\(\Leftrightarrow\text{ ( 1 − x ) ( 1 − z )\le(\dfrac{\text{1 − x + 1 −}z}{2})^2 }\)

\(\Leftrightarrow\text{4 ( 1 − x ) ( 1 − z ) ≤ ( 1 + y ) ^2}\)

\(\Leftrightarrow\text{ 4 ( 1 − x ) ( 1 − z ) ( 1 − y ) ≤ ( 1 + y ) ^2 ( 1 − y )}\)

mặt khác\(\text{ 1 − y ^2 ≤ 1}\)

\(\text{( 1 + y ) ^2 ( 1 − y ) = ( 1 + y ) ( 1 − y ^2) = ( x + 2y + z ) ( 1 − y^2 ) (1+y)^2(1−y)=(1+y)(1−y^2)=(x+2y+z)(1−y^2)}\)Do đó: 4(1−x)(1−y)(1−z)≤x+2y+z