\(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)

Ta có: \(3x^2-6xy+3y^2-12z^2\)

    \(=3.\left(x^2-2xy+y^2-4z^2\right)\)

    \(=3.\left[\left(x-y\right)^2-4z^2\right]\)

    \(=3.\left(x-y-2z\right).\left(x-y+2z\right)\)

Theo bài ra ta có : 

\(x^3-4x^2+x=x\left(x^2-4x+1\right)\)

Gớm Tú ơi, làm gì mà Dis nhiều thế :)) Nghiếp khiếp vậy mày:))))

a) \(A=\left(2\sqrt{3}-5\sqrt{27}+4\sqrt{12}\right):\sqrt{3}\)

\(=2\sqrt{3}:\sqrt{3}-5\sqrt{27}:\sqrt{3}+4\sqrt{12}:\sqrt{3}\)

\(=2\sqrt{3:3}-5\sqrt{27:3}+4\sqrt{12:3}\)

\(=2\sqrt{1}-5\sqrt{9}+4\sqrt{4}=2.1-5.3+4.2=2-15+8=-5\)

\(B=\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\frac{\left(2+\sqrt{3}\right).\left(\sqrt{2-\sqrt{3}}\right)^2}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}\)

\(=\frac{\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=\frac{4-3}{\sqrt{4-3}}=\frac{1}{\sqrt{1}}=1\)

b) \(ĐKXĐ:x\ge\frac{7}{2}\)

Thay \(A=-5\), \(B=1\)vào biểu thức ta được:

\(1-3\sqrt{2x-7}=-5\)\(\Leftrightarrow3\sqrt{2x-7}=6\)

\(\Leftrightarrow\sqrt{2x-7}=2\)\(\Leftrightarrow2x-7=4\)

\(\Leftrightarrow2x=11\)\(\Leftrightarrow x=\frac{11}{2}\)( thỏa mãn ĐKXĐ )

Vậy \(x=\frac{11}{2}\)

(switch)\(\rightarrow\)switch

Bài 1 :

\(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\) ; \(n_{HCl}=\frac{3,65}{36,5}=0,1\left(mol\right)\)

PTHH : \(Mg+2HCl-->MgCl_2+H_2\)

Ta thấy : \(\frac{n_{HCl}}{2}< n_{Mg}\left(0,05< 0,1\right)\)=> Spu Mg còn dư

Theo pthh : \(n_{H_2}=n_{MgCl_2}=n_{Mg\left(pứ\right)}=\frac{1}{2}n_{HCl}=0,05\left(mol\right)\)

=> \(\hept{\begin{cases}m_{Mg\left(dư\right)}=\left(0,1-0,05\right)\cdot24=1,2\left(g\right)\\m_{MgCl_2}=95\cdot0,05=4,75\left(g\right)\\V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\end{cases}}\)

Bài 2 : 

\(n_{Al}=\frac{5,4}{27}=0,2\left(mol\right)\) ; \(n_{H_2SO_4}=\frac{14,7}{98}=0,15\left(mol\right)\)

PTHH : \(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\)

Ta thấy : \(\frac{n_{Al}}{2}>\frac{n_{H_2SO_4}}{3}\left(0,1>0,05\right)\) => Spu Al còn dư

Theo pthh : \(n_{Al\left(pứ\right)}=\frac{2}{3}n_{H_2SO_4}=0,1\left(mol\right)\)

                   \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2SO_4}=0,05\left(mol\right)\)

                   \(n_{H_2}=n_{H_2SO_4}=0,15\left(mol\right)\)

=> \(\hept{\begin{cases}m_{Al\left(dư\right)}=\left(0,2-0,1\right)\cdot27=2,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=342\cdot0,05=17,1\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{cases}}\)

Bài 3 :

\(n_{H_2}=\frac{4,704}{22,4}=0,21\left(mol\right)\)

PTHH : \(2M+6HCl-->2MCl_3+3H_2\)

Theo pthh : \(n_M=\frac{2}{3}n_{H_2}=0,14\left(mol\right)\)

=> \(\frac{3,78}{M_M}=0,14\)

=> \(M_M=27\) (g/mol)

=> Kim loại M là Nhôm (Al)

Bài 4 :

\(n_P=\frac{6,2}{31}=0,2\left(mol\right)\)

PTHH : \(2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\)   (1)

             \(4P+5O_2-t^o->2P_2O_5\)   (2)

Theo pthh (1); \(n_{O_2}=\frac{1}{2}n_{KMnO_4}=0,2\left(mol\right)\)

Xét pứ (2) , thấy : \(\frac{n_P}{4}>\frac{n_{O2}}{5}\left(0,05>0,04\right)\) => spu photpho còn dư

Theo pthh (2) : \(n_{P_2O_5}=\frac{2}{5}n_{O_2}=0,08\left(mol\right)\)

=> \(m_{P_2O_5}=0,08\cdot142=11,36\left(g\right)\)

bạn từng câu lên sẽ dễ nhìn hơn 

Ta có: \(2x^2+5x+3=0\)

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

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-1\end{cases}}\)

\(2x^2+5x+3=0\)

\(\Leftrightarrow2x^2+2x+3x+3=0\)

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{3}{2}orx=-1\)

Vậy nghiệm của phương trình là x = -3/2 ; -1