c) Ta có: \(\left(3x-5\right)\left(2x+1\right)=12\)

\(\Leftrightarrow6x^2+3x-10x-5-12=0\)

\(\Leftrightarrow6x^2-7x-17=0\)

\(\Delta=\left(-7\right)^2-4\cdot6\cdot\left(-17\right)=49+408=457\)

Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7+\sqrt{457}}{12}\\x_2=\dfrac{7-\sqrt{457}}{12}\end{matrix}\right.\)

1.A

2.C

3.D

4.D

5.B

6.B

Bạn tham khảo nhé:3

uses crt;

var st:array[1..100]of string;

a,b,c:array[1..100]of real;

i,n:integer;

max:real;

begin

clrscr;

readln(n);

for i:=1 to n do readln(st[i],a[i],b[i],c[i]);

max=(a[1]+b[1]+c[1])/3;

for i:=1 to n do 

if (max<(a[i]+b[i]+c[i])/3) then max:=(a[i]+b[i]+c[i])/3;

writeln(max:4:2);

readln;

end.

a) 3x.(x-5) - x.(3x+2) = 4

3x^2 - 15 -3x^2 -2x = 4

-15 - 2x = 4

2x = -19

x = -19/2

b) x^2 -  x = 0

x.(x-1) = 0

=> x = 0

x - 1 = 0 => x = 1

KL:...

Bài 1 : 

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ S + O_2 \xrightarrow{t^o} SO_2\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\)

Bài 2 : 

\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ a) n_{Fe_3O_4} = \dfrac{2,32}{232} = 0,01(mol)\\ n_{Fe} = 3n_{Fe_3O_4} = 0,03(mol)\Rightarrow m_{Fe} = 0,03.56 = 1,68(gam)\\ n_{O_2} = 2n_{Fe_3O_4} = 0,02(mol) \Rightarrow m_{O_2} = 0,02.32 = 0,64(gam)\\ b) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,04(mol)\Rightarrow m_{KMnO_4} = 0,04.158= 6,32(gam)\)