Câu hỏi của Đinh Thị Minh Ánh

Question

a,Rút gọn

$\left(x-2\right)^2+\left(x+1\right)^2+2.\left(x-2\right).\left(-1-x\right)$

b,Thực hiện phép tính sau

$\left(x^3-3x^2+3x-6\right):x-3$

c,Tìm x biết

$x^3+3x^2+3x=100+4$

kudo shinichi · Accepted Answer

a) ta có: (x - 2)2 + (x + 1)2 + 2(x - 2)(-1 - x)

= (x - 2)2 - 2(x - 2)(x + 1) + (x + 1)2

= (x - 2 - x - 1)2 = 9

b) (x3 - 3x2 + 3x - 6) : x - 3

= x2 - 3x + 3 - 6/x - 3

= x2 - 3x - 6/xCâu trả lời: a) ta có: (x - 2)2 + (x + 1)2 + 2(x - 2)(-1 - x)

= (x - 2)2 - 2(x - 2)(x + 1) + (x + 1)2

= (x - 2 - x - 1)2 = 9

b) (x3 - 3x2 + 3x - 6) : x - 3

= x2 - 3x + 3 - 6/x - 3

= x2 - 3x - 6/x

Vũ Minh Tuấn · Answer

c)

$x^3+3x^2+3x=100+4$

$\Rightarrow x^3+3x^2+3x=104$

$\Rightarrow x^3+3x^2+3x-104=0$

$\Rightarrow x^3+3x^2+3x+1-105=0$

$\Rightarrow\left(x^3+3x^2+3x+1\right)-105=0$

$\Rightarrow\left(x+1\right)^3-105=0$

$\Rightarrow\left(x+1\right)^3=0+105$

$\Rightarrow\left(x+1\right)^3=105$Câu trả lời: c)

$x^3+3x^2+3x=100+4$

$\Rightarrow x^3+3x^2+3x=104$

$\Rightarrow x^3+3x^2+3x-104=0$

$\Rightarrow x^3+3x^2+3x+1-105=0$

$\Rightarrow\left(x^3+3x^2+3x+1\right)-105=0$

$\Rightarrow\left(x+1\right)^3-105=0$

$\Rightarrow\left(x+1\right)^3=0+105$

$\Rightarrow\left(x+1\right)^3=105$

Violympic toán 8