Toán đại số lớp 8

H

heoxinh9kc

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M

maimailabaoxa01

a. $x^2-13x+36$
=$x^2-2.x.\frac{13}{2}+\frac{169}{4}-\frac{169}{4}+36$
=$(x-\frac{13}{2})^2-\frac{25}{4}$
=$(x-\frac{13}{2}+\frac{5}{2})(x-\frac{13}{2}-\frac{5}{2})$
=$(x-4)(x-9)$
b. $3x^2-16x+5$
=$3(x^2-2.x.\frac{8}{3}+\frac{64}{9})+5-\frac{64}{3}$
=$3(x-\frac{8}{3})^2-\frac{49}{3}$
=$3[(x-\frac{8}{3})^2-\frac{49}{9})$
=$3(x-\frac{8}{3}+\frac{7}{3})(x-\frac{8}{3}-\frac{7}{3})$
=$3(x-\frac{1}{3})(x-5)$
c. $xy(x-y)+yz(y-z)+zx(z-x)$
=$xy(x-y)+yz(y-x+x-z)+zx(z-x)$
=$xy(x-y)+yz(y-x)+yz(x-z)+zx(z-x)$
=$xy(x-y)-yz(x-y)-yz(z-x)+zx(z-x)$
=$(xy-yz)(x-y)-(yz-zx)(z-x)$
=$y(x-z)(x-y)-z(y-x)(z-x)$
=$y(x-z)(x-y)-z(x-y)(x-z)$
=$(y-z)(x-z)(x-y)$
 
P

phamhuy20011801

a. $x^2-13x+36$
=$x^2-2.x.\frac{13}{2}+\frac{169}{4}-\frac{169}{4}+36$
=$(x-\frac{13}{2})^2-\frac{25}{4}$
=$(x-\frac{13}{2}+\frac{5}{2})(x-\frac{13}{2}-\frac{5}{2})$
=$(x-4)(x-9)$
b. $3x^2-16x+5$
=$3(x^2-2.x.\frac{8}{3}+\frac{64}{9})+5-\frac{64}{3}$
=$3(x-\frac{8}{3})^2-\frac{49}{3}$
=$3[(x-\frac{8}{3})^2-\frac{49}{9})$
=$3(x-\frac{8}{3}+\frac{7}{3})(x-\frac{8}{3}-\frac{7}{3})$
=$3(x-\frac{1}{3})(x-5)$

Tách ra thôi, đây là phương pháp quá quen thuộc
$a, x^2-13x+36=x^2-9x-4x+36=(x-9)(x-4)\\
b, 3x^2-16x+5=3x^2-15x-x+5=(3x-1)(x-5)$
 
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