2.a)\(-7+\left|x-4\right|=-3\\ \Leftrightarrow\left|x-4\right|=4\\ \Leftrightarrow\left\{{}\begin{matrix}x-4=4\left(x\ge4\right)\\-x+4=4\left(x< 4\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=8\\x=0\end{matrix}\right.\)

b)\(-\left|x+5\right|=0\Leftrightarrow x=-5\)

dau /../ là sao bạn

dễ mà tick mình đi mình trả lời

\(x^4+y^4\ge xy^3+x^3y\\ \Leftrightarrow x^4-xy^3+y^4-x^3y\ge0\\ \Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\\ \Leftrightarrow\left(x-y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\ge0\\ \Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\\ \)

Ma \(\left(x-y\right)^2\ge0\left(\forall x,y\right)\\ x^2+xy+y^2>0\left(\forall x,y\right)\)\(\Rightarrow x^4+y^4\ge xy^3+x^3y\left(\forall x,y\right)\left(dpcm\right)\)

3/7

tick đi đúng đó

\(n_{Al\left(pu\right)}=a\left(mol\right)\)

\(2Al+3Cu\left(NO_3\right)_3\rightarrow3Cu+2Al\left(NO_3\right)_3\)

b) Theo PTHH có

\(m\left(tang\right)=m_{Cu}-m_{Al\left(pu\right)}=64\cdot1,5a-27a=11,04\\ \Rightarrow a=0,16\left(mol\right)\\ m_{Al\left(pu\right)}=27\cdot0,16=4,32\left(g\right)\\ m_{Cu\left(spu\right)}=64\cdot1,5\cdot0,16=15,36\left(g\right)\)

c) \(C_{M\left(Cu\left(NO_3\right)_2\right)}=\dfrac{0,16\cdot1,5}{0,5}=0,48\left(M\right)\)

d)\(C_{M\left(Al\left(NO_3\right)_2\right)}=\dfrac{0,16}{0,5}=0,32\left(M\right)\)

(Điều kiện :Sau phản ứng coi Vdd thay đổi không đáng kể)

Theo DLBTKL ta có

\(m_{O_2}=6-3,6=2,4\left(g\right)\)

\(\Rightarrow n_{O_2}=\dfrac{2,4}{32}=0,075\left(mol\right)\\ 2R+O_2\underrightarrow{t^o}2RO\\ \Rightarrow n_R=0,075\cdot2=0,15\left(mol\right)\\ \Rightarrow M_R=\dfrac{3,6}{0,15}=24\)

a)Vậy M là Mg

b)\(Mg+H_2SO_4\left(d,n\right)\rightarrow MgSO_4+H_2S+H_2O\)

Ta có các quá trình

\(Mg^o-2e\rightarrow Mg^{+2}|||S^{+6}+4e\rightarrow S^{+2}\)

0,15 0,3 4a a

\(\Rightarrow0,3=4a\Rightarrow a=0,075\left(mol\right)\)

\(V_{H_2S}=0,075\cdot22,4=1,68\left(l\right)\)

...............chiu viet lung tung

\(n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\\ n_C=\dfrac{6}{12}=0,5\left(mol\right)\)

\(S+O_2\underrightarrow{t^o}SO_2\\ \\ C+O_2\underrightarrow{t^o}CO_2\)

Theo PTHH có \(V_{O_2\left(dktc\right)}=\left(0,3+0,5\right)\cdot22,4=17,92\left(l\right)\\ V_{kk}\approx5V_{O_2}=5\cdot17,92=89,6\left(l\right)\)

(Vì O2 chiếm khoảng 20% không khí)

a b doi nhau