giúp mik vs ạ :( cảm ơn ạ:<
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B(34):{0;34;68;102;136;...}
Mà x<102
=> x E {0;34;68;102}
x € B(34)= {0;34;68;102;...}
Mà x<102 nên : x € {0;34;68}
beautifully
excitedly
interesting
collected
arrangements
peacefully
excited
different
a,\(x^2-7x+6=x^2-x-6x+6\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x-6\right)\left(x-1\right)\)
a) x2-7x+6=(x2-x)-(6x-6)=x(x-1)-6(x-1)=(x-6)(x-1)
b) x2-6x+3=(x2-6x+9)-6=(x-3)2-\(\sqrt{6^2}\)=(x-3-\(\sqrt{6}\))(x-3+\(\sqrt{6}\))
c) x2-4x+3=(x2-x)-(3x-3)=x(x-1)-3(x-1)=(x-3)(x-1)
d) 3x2-5x+2=(3x2-3x)-(2x-2)=3x(x-1)-2(x-1)=(3x-2)(x-1)
e) 7x2-x-6=(7x2-7x)+(6x-6)=7x(x-1)+6(x-1)=(7x+6)(x-1)
f) 3x2-5x-8=(3x2+3x)-(8x+8)=3x(x+1)-8(x+1)=(3x-8)(x+1)
g) x2-6x+5=(x2-x)-(5x-5)=x(x-1)-5(x-1)=(x-5)(x-1)
h) x2-2x-3=(x2-2x+1)-4=(x-1)2-22=(x-1-2)(x-1+2)=(x-3)(x+1)
i) x2-x-12=(x2+3x)-(4x+12)=x(x+3)-4(x+3)=(x-4)(x+3)
Lời giải:
ĐKĐB $\Rightarrow \frac{2}{c}=\frac{a+b}{ab}\Rightarrow c(a+b)=2ab$
Khi đó:
$\frac{a}{b}-\frac{a-c}{c-b}=\frac{a(c-b)-b(a-c)}{b(c-b)}=\frac{ac-ab-ab+bc}{b(c-b)}=\frac{c(a+b)-2ab}{b(c-b)}=\frac{2ab-2ab}{b(c-b)}=0$
$\Rightarrow \frac{a}{b}=\frac{a-c}{c-b}$ (đpcm)