[Toán 8] Rút gọn

H

huuthuyenrop2

2X^3 -7X^2-12X+45 / 3X^3 -19X^2+33X-9

[laTEX]\frac{(2x+5)(x-3)^2}{3x-1)(x-3)^2} = \frac{2x+5}{3x-1}[/laTEX]
 
S

stronger1952000

mọi ngưòi ơi!
Giúp em Bài 3.3/SBT Toán 8 (tập 1) vớiiiiiiiiiiiiiiiiiiiiiiiiiiiii
 
E

eunhyuk_0330

Bài 3.3 sbt toán 8 t1:
a) $P= (5x-1) + 2(1-5x)(4+5x) +(5x+4)^2
= [(5x-1) + (1-5x)(4+5x)] + [(1-5x)(4+5x) + (5x+4)^2]
= [(1-5x)(4+5x)-(1-5x)] + [(1-5x)(5x+4) + (5x+4)^2]
= (1-5x)(4+5x-1) + (5x+4)(1-5x+5x+4)
= (1-5x)(5x+3) + (5x+4).5
= 5x + 3 - 25x^2 -15x + 25x + 20
= -25x^2 +15x + 23$
b) $ Q= (x-y)^3 + (y+x)^3 + (y-x)^3 -3xy(x+y)
= (x-y)^3 + (x+y)^3 - (x-y)^3 - 3xy(x+y)
= (x+y)^3 - 3xy(x+y)
= x^3+y^3+ 3xy(x+y) - 3xy(x+y)
= x^3+y^3$
 
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C

cbgrjfgt

Bài 3.3 sbt toán 8 t1:
a) $P= (5x-1) + 2(1-5x)(4+5x) +(5x+4)^2
= [(5x-1) + (1-5x)(4+5x)] + [(1-5x)(4+5x) + (5x+4)^2]
= [(1-5x)(4+5x)-(1-5x)] + [(1-5x)(5x+4) + (5x+4)^2]
= (1-5x)(4+5x-1) + (5x+4)(1-5x+5x+4)
= (1-5x)(5x+3) + (5x+4).5
= 5x + 3 - 25x^2 -15x + 25x + 20
= -25x^2 +15x + 23$
b) $ Q= (x-y)^3 + (y+x)^3 + (y-x)^3 -3xy(x+y)
= (x-y)^3 + (x+y)^3 - (x-y)^3 - 3xy(x+y)
= (x+y)^3 - 3xy(x+y)
= x^3+y^3+ 3xy(x+y) - 3xy(x+y)
= x^3+y^3$
 
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