Ôn tập phép nhân và phép chia đa thức - Hỏi đáp

\(\left(x+2\right)^3-x^3+6x^2=7\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2=7\)

\(\Leftrightarrow12x^2+12x+1=0\)

\(\Leftrightarrow12\left(x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}+\frac{1}{12}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\frac{1}{6}=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{1}{6}\)

⇔ \(x=\pm\sqrt{\frac{1}{6}}-\frac{1}{2}\)

PT \(\Leftrightarrow12x^2+12x+1=0\)

\(\Leftrightarrow...\)

a) Ta có: \(x^4+3x^3-7x^2-27x-18\)

\(=x^4-3x^3+6x^3-18x^2+11x^2-33x+6x-18\)

\(=x^3\left(x-3\right)+6x^2\left(x-3\right)+11x\left(x-3\right)+6\left(x-3\right)\)

\(=\left(x-3\right)\left(x^3+6x^2+11x+6\right)\)

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

\(=\left(x-3\right)\left[x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\right]\)

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

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

b) Ta có: \(x^3-8x^2+x+42\)

\(=x^3-7x^2-x^2+7x-6x+42\)

\(=x^2\left(x-7\right)-x\left(x-7\right)-6\left(x-7\right)\)

\(=\left(x-7\right)\left(x^2-x-6\right)\)

\(=\left(x-7\right)\left(x-3\right)\left(x+2\right)\)

c) Ta có: \(x^4+5x^3-7x^2-41x-30\)

\(=x^4+5x^3-7x^2-35x-6x-30\)

\(=x^3\left(x+5\right)-7x\left(x+5\right)-6\left(x+5\right)\)

\(=\left(x+5\right)\left(x^3-7x-6\right)\)

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

\(=\left(x+5\right)\left[x\left(x^2-1\right)-6\left(x+1\right)\right]\)

\(=\left(x+5\right)\left[x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\right]\)

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

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

a ) \(==>x^3.\left(x+3\right)-\left(7x^2+27x+18\right)\)

ko xét phần x^3.( x+3 ) nữa mà mik phân tích trong ngoặc nha zo thi ko lm như vậy mà ghi lại phần đó nha

\(7x^2+21x+6x+18\)

\(7x\left(x+3\right)+6\left(x+3\right)\)

\(\left(x+3\right)\left(7x+6\right)\)

==> \(x^3.\left(x+3\right)-\left(x+3\right)\left(7x+6\right)\)

==>\(\left(x+3\right)\left(x^3-7x-6\right)\)

A=\([\)(x+y)(x+4y)\(][\)(x+2y)(x+3y)]+y⁴

=(x²+4xy+xy+4y²)(x²+3xy+2xy+6y²)+y⁴

=(x²+5xy+4y²)(x²+5xy+6y²)+y⁴

=(x²+5xy+5y²-y²)(x²+5xy+5y²+y²)+y⁴

=(x²+5xy+5y²)²-y⁴+y⁴

=(x²+5xy+5y²)²

vậy A là số chính phương vs \(\forall\) x,y\(\in\)R

1) \(x^2-8x+20=\left(x^2-8x+16\right)+4=\left(x-4\right)^2+4>0\forall x\)

(do \(\left(x-4\right)^2\ge0\forall x\)

2) \(4x^2-12x+11=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2>0\forall x\)

(do \(\left(2x-3\right)^2\ge0\forall x\))

3) \(x^2-2x+y^2+4y+6=\left(x-1\right)^2+\left(y+2\right)^2+1>0\forall x;y\)

(do ....)

4) \(\left(15x-1\right)^2+3\left(7x+3\right)\left(x+1\right)-\left(x^2-73\right)\)

\(=225x^2-30x+1+3\left(7x^2+10x+3\right)-x^2+73\)

\(=225x^2-30x+83+21x^2+30x-x^2\)

\(=245x^2+83>0\forall x\)

a)

ta có:

a(2a - 3) - 2a(a + 1)

= 2a² - 3a - 2a² - 2a

= -5a \(⋮\) 5

b)

ta có:

x² + 2x + 2

= x² + 2x + 1 + 1

= (x + 1)² + 1

vì (x + 1)² \(\ge0\forall x\in R\)

\(\Rightarrow\) (x + 1)² +1 \(\ge1\) > 0 \(\forall x\in R\)

Vậy (x + 1)² +1 > 0 \(\forall x\in R\)

Hay x² + 2x + 2 > 0 \(\forall x\in R\)

Lời giải:
$P(x)$ chia $x+1, x-2, x+4$ đều dư $20$ mà lại có bậc 3 nên $P(x)$ có dạng:

$P(x)=a(x+1)(x-2)(x+4)+20$

Vì $P(x)$ triệt tiêu khi $x=1$ nên:

$P(1)=0$

$\Leftrightarrow a.2.(-1).5+20=0$

$\Rightarrow a=2$

Vậy $P(x)=2(x+1(x-2)(x+4)+20=2x^3+6x^2-12x+4$

a) \(x^2-8x+20\)

\(=x^2-2.x.4+16+4\)

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

Có: \(\left(x-4\right)^2\ge0\Rightarrow\left(x-4\right)^2+4>0\)

Hay:.............

b) \(x^2+11\)

Có: \(x^2\ge0\Rightarrow x^2+11>0\)

Hay:.............

c) \(4x^2-12x+11\)

\(=4\left(x^2-3x+\frac{11}{4}\right)\)

\(=4\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{1}{2}\right)\)

\(=4\left(x-\frac{3}{2}\right)^2+2>0\)

d) \(x^2+5y^2+2x+6y+34\)

\(=x^2+2.x.1+1+y^2+4y^2+2.y.3+9+24\)

\(=\left(x^2+2.x.1+1\right)+\left(y^2+2.y.3+9\right)+4y^2+24\)

\(=\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24\)

Ta có: \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\\\left(2y\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24>0\)

f) \(x^2-2x+y^2+4y+6\)

\(=x^2-2.x.1+1+y^2+2.y.2+4+1\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+1>0\)

d)

$(x+2)^3-(x-1)(x^2+x+1)=(x^3+6x^2+12x+8)-(x^3-1)$

$=6x^2+12x+9$

e)

$(x+2)(x^2-2x+1)-(x+3)(x-3)=(x^3-3x+2)-(x^2-9)$

$=x^3-x^2-3x+11$

f)

$(x+3)(x^2-3x+9)-(x^2+2x+4)(x-2)$

$=x^3+3^3-(x^3-2^3)=3^3+2^3=35$

a)

$(2x+1)^2-(2x+1)(2x-1)=(2x+1)[(2x+1)-(2x-1)]$

$=2(2x+1)$

b)

$(4x+3)(x-1)-2x(2x+1)=4x^2-x-3-4x^2-2x=-3x-3=-3(x+1)$

c)

$(2x+3)^2-(4x+1)(x+5)=(4x^2+12x+9)-(4x^2+21x+5)$

$=-9x+4$

Bài này mà để vào câu hỏi hay clone con ạ :))

Lời giải:
\(x^2+y^2-4x+6y+13=0\)

\(\Leftrightarrow (x^2-4x+4)+(y^2+6y+9)=0\)

\(\Leftrightarrow (x-2)^2+(y+3)^2=0\)

Vì \((x-2)^2; (y+3)^2\ge 0, \forall x,y\Rightarrow (x-2)^2+(y+3)^2\geq 0\)

Dấu "=" xảy ra khi \((x-2)^2=(y+3)^2=0\Leftrightarrow \left\{\begin{matrix} x=2\\ y=-3\end{matrix}\right.\)

\(x^2+y^2-4x+6y+13=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(y+3\right)^3=0\)

Vì: \(\left(x-2\right)^2+\left(y+3\right)^3\ge0\forall x;y\)

=> ''='' xảy ra khi x = 2; y = -3

Vậy.........

\(4x^2-4xy+y^2-25=\left(4x^2-4xy+y^2\right)-5^2 \)
\(=\left(2x-y\right)^2-5^{2 } \)\(=\left(2x-y-5\right)\left(2x-y+5\right)\)

\(4x^2-4xy+y^2-25\)

\(=\left(2x-y\right)^2-5^2\)

\(=\left(2x-y-5\right)\left(2x-y+5\right)\)

4 Chứng minh rằng:

a)\(n^2+\left(n+1\right)+2n\left(n+1\right)\) chia hết cho 6

Ta có:

\(n^2\left(n+1\right)+2n\left(n+1\right)\)

\(=n^3+3n^2+2n\)

\(=n\left(n^2+3n+2\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

Ta thấy n , n+1 và n+2 là ba số tự nhiên liên tiếp

=> n(n+1) (n+2)\(⋮\)6

=> đpcm

b)\(\left(2n-1\right)^3-\left(2n-1\right)\) chia hết cho 8

Ta có:

\(\left(2n-1\right)^3-\left(2n-1\right)\)

\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)

\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1^2\right]\)

\(=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)

\(=\left(2n-1\right).2\left(n-1\right).2n\)

\(=4n\left(2n-1\right)\left(n-1\right)\)

=>\(4n\left(2n-1\right)\left(n-1\right)⋮4\left(1\right)\)

Mà(2n-1)(n-1)=(n+n-1)(n-1)

=>\(\left(2n-1\right)\left(n-1\right)⋮2\left(2\right)\)

Từ (1) và (2)=> Đpcm

c)\(\left(n+7\right)^2-\left(n-5\right)^2\) chia hết cho 24

Câu hỏi của Ngoc An Pham - Toán lớp 8 | Học trực tuyến

Chúc bạn học tốt!^^

4.a)n²(n+1)+2n(n+1)=(n+1)(n²+2n)=n(n+1)(n+2)

n,(n+1),(n+2) là ba số nguyên liên tiếp nên chia hết cho 2 và 3

\(\Rightarrow\)n(n+1)(n+2) chia hết cho 6