219023:467=469

KQ là 4657 và 2735

k nha !!!

\(T=\frac{3^{108}-1}{26}\)

1 + 1 x1 + 1 x 1 + 2 x 3 - 4 x 0 = 1 + 1 + 1 + 6 - 0 =9

Ta có:

\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{\left(a+b+c\right)+\left(a-b+c\right)}{\left(a+b-c\right)+\left(a-b-c\right)}=\frac{a+b+c+a-b+c}{a+b-c+a-b-c}=\frac{2a+2c}{2a-2c}=\frac{2\left(a+c\right)}{2\left(a-c\right)}=\frac{a+c}{a-c}\left(1\right)\)\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{\left(a+b+c\right)-\left(a-b+c\right)}{\left(a+b-c\right)-\left(a-b-c\right)}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\Rightarrow\frac{a+c}{a-c}=1\)

\(\Leftrightarrow a+c=a-c\Leftrightarrow a+c-a+c=0\Leftrightarrow2c=0\Leftrightarrow c=0\)(đpcm)

b) \(b.x^2+8x-33=x^2-3x+11x-33x=x\left(x-3\right)+11\left(x-3\right)=\left(x-3\right)\left(x+11\right)\)

\(c.x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)

a,bx6-a,b=1a,b

a,bx(6-1)=10+a,b

a,bx5=10+a,b

a,bx5-a,b=10

a,bx4=10

a,b=10:4

a,b=2,5

Chúc bạn học tốt!!!