chia theo cột dọc nhá

bạn chia đa thức 1 biến đã sắp xếp nha

\(P\left(x\right)=5^6-6.5^5+6.5^4-6.5^3+6.5^2-6.5+1=5^6-6\left(5^5-5^4-5^3-5^2-5\right)+1=1556\)

mình quên là k dùng máy tính bỏ túi nha

\(P\left(x\right)=x^6-6x^5+6x^4-6x^3+6x^2-6x+1\)

\(=x^6-5x^5-x^5+5x^4+x^4-5x^3-x^3+5x^2+x^2-5x-x+1\)

\(=x^5\left(x-5\right)-x^4\left(x-5\right)+x^3\left(x-5\right)-x^2\left(x-5\right)+x\left(x-5\right)-x+1\)

-Thay \(x=5\) vào P(x) ta được:

\(P\left(5\right)=5^5\left(5-5\right)-5^4\left(5-5\right)+5^3\left(5-5\right)-5^2\left(5-5\right)+5\left(5-5\right)-5+1\)\(=-5+1=-4\)

2x4 ,4 là mũ hay số vậy

a, \(x^4-4x^3-6x^2-4x+1=0\)(*)

<=> \(x^4+4x^2+1-4x^3-4x+2x^2-12x^2=0\)

<=> \(\left(x^2-2x+1\right)^2=12x^2\)

<=>\(\left(x-1\right)^4=12x^2\) <=> \(\left[{}\begin{matrix}\left(x-1\right)^2=\sqrt{12}x\\\left(x-1\right)^2=-\sqrt{12}x\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x^2-2x+1-\sqrt{12}x=0\left(1\right)\\x^2-2x+1+\sqrt{12}x=0\left(2\right)\end{matrix}\right.\)

Giải (1) có: \(x^2-2x+1-\sqrt{12}x=0\)

<=> \(x^2-2x\left(1+\sqrt{3}\right)+\left(1+\sqrt{3}\right)^2-\left(1+\sqrt{3}\right)^2+1=0\)

<=> \(\left(x-1-\sqrt{3}\right)^2-3-2\sqrt{3}=0\)

<=> \(\left(x-1-\sqrt{3}\right)^2=3+2\sqrt{3}\) <=> \(\left[{}\begin{matrix}x-1-\sqrt{3}=\sqrt{3+2\sqrt{3}}\\x-1-\sqrt{3}=-\sqrt{3+2\sqrt{3}}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(ktm\right)\\x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(tm\right)\end{matrix}\right.\)

=> \(x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)

Giải (2) có: \(x^2-2x+1+\sqrt{12}x=0\)

<=> \(x^2-2x\left(1-\sqrt{3}\right)+\left(1-\sqrt{3}\right)^2-\left(1-\sqrt{3}\right)^2+1=0\)

<=> \(\left(x+\sqrt{3}-1\right)^2=3-2\sqrt{3}\) .Có VP<0 => PT (2) vô nghiệm

Vậy pt (*) có nghiệm x=\(-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)

a: \(=\dfrac{2x\left(3x^2+2\right)+3x^2+2}{3x^2+2}=2x+1\)

b: 

Sửa đề: 6x^4-4x^3+3x-2/3x-2

\(=\dfrac{6x^4-4x^3+3x-2}{3x-2}\)

\(=\dfrac{2x^3\left(3x-2\right)+3x-2}{3x-2}=2x^3+1\)

a) 12x³ : 4x = (12:4) . (x³ : x) = 3.x²

b) (-2x⁴ ) : x⁴ = [(-2) : 1] . (x⁴ : x⁴) = -2

c) 2x⁵ : 5x² = (2:5) . (x⁵ : x²) = \(\frac{2}{5}\)x³

a: \(P\left(x\right)=6x^3+4x^2+2x-4\)

\(Q\left(x\right)=-x^4+6x^3-4x^2+3x-8\)

b: \(P\left(x\right)-Q\left(x\right)=x^4+8x^2-x+4\)

\(P\left(x\right)+Q\left(x\right)=-x^4+12x^3+5x-12\)

a) \(\left(x^5+4x^3-6x^2\right):4x^2\)

\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)

\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

b) 

Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)

c) (-2x⁵ : 2x²) + (3x² : 2x²) + (-4x^3 : 2x^2)

= \(-x^3+\dfrac{3}{2}-2x\)

d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)

\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)

\(=x-4\)

(dùng hẳng đẳng thức thứ 7)

Bài 2 :

a) 3x(x - 2) - 5x(1 - x) - 8(x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= (3x² + 5x² - 8x²) + (-6x - 5x) + 24 

= -11x + 24

b) (x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³ 

c) (x - y)² + (x + y)² - 2(x - y)(x + y)

= (x - y)² - 2(x - y)(x + y) + (x + y)²

= [(x - y) + x + y)² = [x - y + x + y] = (2x)² = 4x²

Bài 1 :

a]=  \(\frac{1}{4}\)x³ + x - \(\frac{3}{2}\).

b] => [x³ + x² -12 ] = [ x² +3 ][x-2] + [-6]

c]= -x³ -2x +\(\frac{3}{2}\).

d] = [ x³ - 64 ]  = [ x² + 4x + 16][ x- 4].