\(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\\ ĐKXĐ:x\ne-4;x\ne4\)

\(\Leftrightarrow\dfrac{5x^2-80}{x^2-16}=\dfrac{2x^2-8x-x+4}{x^2-16}+\dfrac{3x^2+12x-x-4}{x^2-16}\\ \Rightarrow5x^2-80=2x^2-9x+4+3x^2+11x-4\\ \Leftrightarrow5x^2-2x^2-3x^2+9x-11x=80-4+4\\ \Leftrightarrow-2x=80\\ \Leftrightarrow x=-40\left(nhận\right)\)

Vậy S = { -40 }

Sai chỗ nào vậy ? 

a. Đổi 500 ml = 0,05 lít

\(n_{H_2}=\dfrac{3.36}{22,4}=0,15\left(mol\right)\)

PTHH : Fe + H₂SO₄ -> FeSO₄ + H₂

         0,15   0,15

\(m_{Fe}=0,15.56=8,4\left(g\right)\)

b. \(C_{M_{H_2SO_4}}=\dfrac{0.15}{0,05}=3\left(M\right)\)

c. PTHH : 2H₂ + O₂ --t^o---> 2H₂O

            0,15  0,075 

\(V_{O_2}=0,075.22,4=1,68\left(l\right)\)

a. \(158+445+342+555\\ =\left(158+342\right)+\left(445+555\right)\\ =1500\)

b. \(53.\left(51+4\right)+53.\left(49+96\right)+53\\ =53.\left(51+4+49+96\right)\\ =10600\)

a. \(x^3-2x^2y+xy-9x\\ =x\left(x^2+2xy+y^2-9\right)\\ =x\left(x+y\right)^2-3^2\\ =x\left(x+y-3\right)\left(x+y+3\right)\)

a. Đổi 50 ml = 0,05 lít 

\(n_{H_2}=\dfrac{3.36}{22,4}=0,15\left(mol\right)\)

PTHH : Fe + H₂SO₄ ---t^o--> FeSO₄ + H₂

       0,15    0,15

b. \(m_{Fe}=0,15.56=8,4\left(g\right)\)

c. \(C_{M_{H_2SO_4}}=\dfrac{0.15}{0,05}=3\left(M\right)\)

Bài 8 :

\(A+B+C\\ =4x^2-5xy+3y^2+3x^2+2xy+y^2-x^2+3xy+2x^2\\ =8x^2+4y^2\)

\(B-C-A\\ =3x^2+2xy+y^2+x^2-3xy-2y^2-4x^2+5xy-3y^2\\ =4xy-4y^2\)