PTTNT Viết công thức tổng quát để tìm ( PHẢI VIẾT ) Bài 1 : x^2 – 11x + 28 Bài 2 : 6x^2 + x – 12 Bài 3 : x^3 – 19x + 30 11/08/2021 Bởi Daisy PTTNT Viết công thức tổng quát để tìm ( PHẢI VIẾT ) Bài 1 : x^2 – 11x + 28 Bài 2 : 6x^2 + x – 12 Bài 3 : x^3 – 19x + 30
Bài 1: $x^{2}$ $-11x+28^{}$ $=(x^{2}$ $-7x)-(4x+28)^{}$ $=x(x-7)-4(x-7)^{}$ $=(x-7)(x-4)^{}$ Bài 2: $6x^{2}$ $+x-12^{}$ $=(6x^{2}$ $-8x)+(9x-12)^{}$ $=2x(3x-4)+3(3x-4)^{}$ $=(3x-4)(2x+3x)^{}$ Bài 3: $x^{3}$ $-19x+30^{}$ $=x^{3}$ $+2x^{2}$ $-8x^{2}$ $-16x+15x+30^{}$ $=x^{2}$ $(x+2)-2x(x+2)+15(x+2)^{}$ $=(x+2)(x^{2}$ $-8x+15)^{}$ $(x+2)^{}$ $=[(x^{2}$ $-5x)-(3x-15)]^{}$ $=(x+2)^{}$$[x(x-5)-3(x-5)]^{}$ $=(x+2)(x-5)(x-3)^{}$ @thuyylinhh20042007 Bình luận
`Bài 1 : x^2 – 11x + 28 ` ` ⇔ ( x^2 – 7x ) – ( 4x + 28 )` ` ⇔ x ( x – 7 ) – 4 ( x – 7 ) ` ` ⇔ ( x – 4 ) ( x – 7 ) ` `x^2 – ( a + b )x + a*b` `Bài 2 : 6(x^2) + x – 12 ` ` ⇔ ( 6(x^2) + 9x ) – ( 8x + 12 ) ` ` ⇔ 3x ( 2x + 3 ) – 4 ( 2x + 3 )` ` ⇔ ( 2x + 3 ) ( 3x – 4 )` ` a(x^2) + b^2 + c` Tách b = m + n | Đ/K ` ac = mn` ` Bài 3 : x^3 – 19x + 30 ` ` ⇔ ( x^3 – 4x ) – ( 15x + 30 ) ` ` ⇔ x ( x – 2 ) ( x + 2 ) – 15 ( x – 2 ) ` ` ⇔ ( x – 2 ) [ x ( x + 2 ) – 15 ]` ` ⇔ ( x – 2 ) [ x^2 + 5x – 3x – 15 ] ` ⇔ ( x – 2 ) ( x + 3 ) ( x + 5 ) ` ` x^3 – ( a^2 + b )x – ab ` ` color(blue){ # KAITO # } ` Bình luận
Bài 1:
$x^{2}$ $-11x+28^{}$
$=(x^{2}$ $-7x)-(4x+28)^{}$
$=x(x-7)-4(x-7)^{}$
$=(x-7)(x-4)^{}$
Bài 2:
$6x^{2}$ $+x-12^{}$
$=(6x^{2}$ $-8x)+(9x-12)^{}$
$=2x(3x-4)+3(3x-4)^{}$
$=(3x-4)(2x+3x)^{}$
Bài 3:
$x^{3}$ $-19x+30^{}$
$=x^{3}$ $+2x^{2}$ $-8x^{2}$ $-16x+15x+30^{}$
$=x^{2}$ $(x+2)-2x(x+2)+15(x+2)^{}$
$=(x+2)(x^{2}$ $-8x+15)^{}$ $(x+2)^{}$
$=[(x^{2}$ $-5x)-(3x-15)]^{}$
$=(x+2)^{}$$[x(x-5)-3(x-5)]^{}$
$=(x+2)(x-5)(x-3)^{}$
@thuyylinhh20042007
`Bài 1 : x^2 – 11x + 28 `
` ⇔ ( x^2 – 7x ) – ( 4x + 28 )`
` ⇔ x ( x – 7 ) – 4 ( x – 7 ) `
` ⇔ ( x – 4 ) ( x – 7 ) `
`x^2 – ( a + b )x + a*b`
`Bài 2 : 6(x^2) + x – 12 `
` ⇔ ( 6(x^2) + 9x ) – ( 8x + 12 ) `
` ⇔ 3x ( 2x + 3 ) – 4 ( 2x + 3 )`
` ⇔ ( 2x + 3 ) ( 3x – 4 )`
` a(x^2) + b^2 + c`
Tách b = m + n | Đ/K ` ac = mn`
` Bài 3 : x^3 – 19x + 30 `
` ⇔ ( x^3 – 4x ) – ( 15x + 30 ) `
` ⇔ x ( x – 2 ) ( x + 2 ) – 15 ( x – 2 ) `
` ⇔ ( x – 2 ) [ x ( x + 2 ) – 15 ]`
` ⇔ ( x – 2 ) [ x^2 + 5x – 3x – 15 ]
` ⇔ ( x – 2 ) ( x + 3 ) ( x + 5 ) `
` x^3 – ( a^2 + b )x – ab `
` color(blue){ # KAITO # } `