Có 2 nghiệm 

Đặt B=0

=>x^2-9=0

=>x^2=9

=>x=3 hoặc x=-3

`B=x^2-9=0`

`-> x^2=0+9`

`-> x^2=9`

`-> x^2=(+-3)^2`

`-> x=+-3`

Vậy, đa thức `B` có `2` nghiệm là `x={3 ; -3}`.

Hoa (write) _____write____ to her pen pal in the USA every month.
When being a child, she received a good (educate) _____education______ .

Cảm ơn bạn nhiều nha

hữu danh vô thực

Hữu danh vô thực

c: E nằm giữa C và D

=>CE+DE=CD

=>ED=5cm

d: I là trung điểm của ED

=>IE=ID=ED/2=2,5cm

e: CE=2cm

IE=2,5cm

=>CE<IE

Ckao. Rất vui đc làm quen

hi bn

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.

Bài 2:

a: \(\left(2x-5\right)^2-4x\left(x-3\right)=0\)

=>\(4x^2-20x+25-4x^2+12x=0\)

=>-8x+25=0

=>-8x=-25

=>\(x=\frac{25}{8}\)

b: \(2\left(x+5\right)-x^2-5x=0\)

=>2(x+5)-x(x+5)=0

=>(x+5)(2-x)=0

=>\(\left[\begin{array}{l}x+5=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=2\end{array}\right.\)

c: \(6x^2-7x+2=0\)

=>\(6x^2-3x-4x+2=0\)

=>3x(2x-1)-2(2x-1)=0

=>(2x-1)(3x-2)=0

=>\(\left[\begin{array}{l}2x-1=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12\\ x=\frac23\end{array}\right.\)

Bài 1:

a: x(x-5)+(x+3)(x-3)

\(=x^2-5x+x^2-9\)

\(=2x^2-5x-9\)

b: \(\frac{x}{x-1}+\frac{2x-4}{x^2-1}-\frac{5}{x+1}\)

\(=\frac{x\left(x+1\right)+2x-4-5\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+x+2x-4-5x+5}{\left(x-1\right)\left(x+1\right)}=\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x-1}{x+1}\)

c: \(\left(20x^2+7x-6\right):\left(5x-2\right)\)

\(=\left(20x^2-8x+15x-6\right):\left(5x-2\right)\)

\(=\frac{4x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=4x+3\)