Phép nhân và phép chia các đa thức - Hỏi đáp

\(\text{a) }A=x^2-10x+25\\ A=x^2-2\cdot x\cdot5+5^2\\ A=\left(x-5\right)^2\\ Do\text{ }\left(x-5\right)^2\ge0\forall x\\ \Leftrightarrow A\ge0\forall x\\ \text{Dấu "=" xảy ra khi : }\\ \left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5\\ \text{Vậy }A_{\left(Min\right)}=0\text{ }khi\text{ }x=5\)

\(\text{b) }B=x^2+y^2-x+6y+10\\ B=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{3}{4}\\ B=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\\ Do\text{ }\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \left(y+3\right)^2\ge0\forall y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x;y\\ \text{Dấu "=" xảy ra khi: }\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\\\left(y+3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\\ \text{ Vậy }B_{\left(Min\right)}=\dfrac{3}{4}\text{ }khi\text{ }x=\dfrac{1}{2};y=-3\)

\(\text{c) }C=2x^2-6x+10\\ C=\left(2x^2-6x+\dfrac{9}{2}\right)+\dfrac{11}{2}\\ C=2\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{2}\\ C=2\left[x^2-2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{11}{2}\\ C=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\\ Do\text{ }\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{2}\ge\dfrac{11}{2}\\ \text{Dấu "=" xảy ra khi: }\\ \left(x-\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{3}{2}=0\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }C_{\left(Min\right)}=\dfrac{11}{2}khi\text{ }x=\dfrac{3}{2}\)

\(\)

a, Ta có:

A= x²-10x+25=(x+5)²

Vì (x+5)²\(\ge\)0\(\forall\)x

\(\Rightarrow\)Min A = 0

Khi đó :

(x+5)²=0

\(\Leftrightarrow\)x+5=0

\(\Rightarrow\)x=5

Vậy min A = 0 khi x=5

\(Q=2x^2-6x\)

\(=2.\left(x^2-3x\right)\)

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

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

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

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

Dấu = xảy ra khi:

\(2.\left(x-\frac{3}{2}\right)^2=0\)

\(\Rightarrow\left(x-\frac{3}{2}\right)^2=0\)

\(\Rightarrow x-\frac{3}{2}=0\)

\(\Rightarrow x=\frac{3}{2}\)

Vậy:..............

B1 : a, M = x³-3xy(x-y)-y³-x²+2xy-y²

= ( x³-y³)-3xy(x-y) -(x²-2xy+y²)

= (x-y)(x²+xy+y²)-3xy(x-y)-(x-y)²

= (x-y) [(x²+xy+y²-3xy-(x-y)]

= (x-y)[(x²-2xy+y²)-(x-y)

= (x-y)[(x-y)²-(x-y)]

= (x-y)(x-y)(x-y-1)

= (x-y)²(x-y-1)

= 7²(7-1) = 49 . 6= 294

N = x²(x+1)-y²(y-1)+xy-3xy(x-y+1)-95

= x³+x²-(y³-y²)+xy-(3x²y-3xy²+3xy)-95

= x³+x²-y³+y²+xy-3x²y+3xy²-3xy-95

= (x³-y³)+(x²-2xy+y²)-(3x²y+y²)-(3x²y-3xy²)-95

=(x-y)(x²+xy+y²)+(x-y)²-3xy(x-y)-95

= (x-y)(x²+xy+y²+x-y-3xy)-95

= (x-y)[(x²-2xy+y²)+(x-y)]-95

= (x-y)[(x-y)²+(x-y)]-95

=(x-y)(x-y)(x-y+1)-95

= (x-y)²(x-y+1)-95

= 7²(7+1)-95=297

a) \(x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+2x-2y+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2.7+34\)

\(=100\)

b) \(x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)-95\)

\(=x^3+x^2-y^3+y^2+xy-3xy-3xy\left(x-y\right)-95\)

\(=x^3+x^2-y^3+y^2-2xy-3xy\left(x-y\right)-95\)

\(=\left(x-y\right)^3+\left(x-y\right)^2-95\)

\(=7^3+7^2-95\)

\(=297\)

Ta có: ( n -1 ). ( n + 4 ) - ( n - 4 ). ( n + 1 )

= \(n^2+4n-n-4-n^2-n+4n+4\)

= 8n - 2n = 6n

Vậy đa thức trên luôn chia hết cho 6 với mọi n ϵ Z

Chúc bạn học tốt :))

\(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)=n^2+3n-4-n^2+3n+4\\ =6n\)

vì: \(6n⋮6\left(với\:n\in Z\right)\) nên \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)⋮6\left(với\: n\in Z\right)\)

P = 3x² - 2x + 3y² - 2y + 6xy - 100

= (3x² + 6xy + 3y²) - (2x + 2y) - 100

= 3(x² + 2xy + y²) - 2(x + y) - 100

= 3(x + y)² - 2.5 - 100

= 3. 5² -10 - 100

= 75 - 10 - 100 = -35

Q = x³ + y³ - 2x² - 2y² + 3xy(x + y) - 4xy + 3(x+y) +10

= x³ + y³ - 2x² - 2y² + 3x²y + 3xy² - 4xy + 3.5 + 10

= (x³ + 3x²y + 3xy² + y³) - (2x² + 4xy + 2y²) + 15 + 10

= (x + y)³ - 2(x² + 2xy + y²) + 25

= 5³ - 2(x + y)² +25

= 125 - 2. 5² + 25

= 125 - 50 + 25 = 100

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

\(A=10x^2-15xy+6xy-9y^2-\left(10x^2-6xy+15xy-9y^2\right)\)

\(A=10x^2-15xy+6xy-9y^2-10x^2+6xy-15xy+9y^2\)

\(A=-18xy\)

câu D ạ

D)

a, Ta có: \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)

= \(a^3+b^3+a^3-b^3=a^3+a^3=2a^3\)

\(\xrightarrow[]{}\) đpcm

b, Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(\left(a-b\right)^2+ab\right)\)

\(\xrightarrow[]{}\) đpcm

c, Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(\xrightarrow[]{}\) đpcm

\(a,\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a+ab+b^2\right)=a^2+b^3+a^3-b^3=2a^3\)\(b,\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\Rightarrowđpcm\)\(c,\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+a^2d^2+b^2d^2+b^2c^2=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\Rightarrowđpcm\)

Ta có: \(a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\) \(a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc=2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+b^2\right)+\left(a^2-2ac+c^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\)

\(\Rightarrow a=b=c\) (đpcm)

\(a^2+b^2+c^2=ab+ac+bc\\ a^2+b^2+c^2-ab-ac-bc=0\\ 2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\ a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\\ \left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)

a) Ta có: ( x - 1) ( x^2 +x +1) = x³+x²+x-x²-x-1

= x³-1

b) 8x³-y³ = (2x)³-y³ = (2x-y)(4x²+2xy+y²)

c) Đáp số đúng là: x³+8

a,x^3+1

b,8x^3-y^3=<2x>^3-y^3=<2x-y>.<a^2+ab+b^2>

c< thiếu

a, Ta có : x^3+8=x^3+2^3=(x+2)(x^2-2x+4)

b, Ta có: (x+1)(x^2-x+1)=x^3+1

a. x³ + 8 = x³ + 2³

= (x+2)(x² - 2x + 4)

b. (x+1)(x² - x + 1)

= x³ + 1³ = x³ + 1

a) Ta có: \(x^2+10x+27\)

\(=x^2+10x+25+2\)

\(=\left(x+5\right)^2+2\)

Ta có: \(\left(x+5\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+5\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x+5=0

hay x=-5

Vậy: Giá trị nhỏ nhất của biểu thức \(x^2+10x+27\) là 2 khi x=-5