Ôn tập phép nhân và phép chia đa thức

a: \(\Leftrightarrow x^3+3x^2+3x+1-3⋮x+1\)

\(\Leftrightarrow\left(x+1\right)^3-3⋮x+1\)

\(\Leftrightarrow x+1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{0;-2;2;-4\right\}\)

b: \(\Leftrightarrow2x^2-4x+5x-10+3⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

\(x^2-7xy+12y^2\)

\(=x^2-4xy-3xy+12y^2\)

\(=x\left(x-4y\right)-3y\left(x-4y\right)\)

\(=\left(x-3y\right)\left(x-4y\right)\)

\(x^2-7xy+12y^2\)

= \(x^2-4xy-3xy+12y^2\)

\(=\left(x^2-4xy\right)-\left(3xy-12y^2\right)\)

\(=x\left(x-4y\right)-3\left(x-4y\right)\)

\(=\left(x-3\right)\left(x-4y\right)\)

\(x^2-7xy+12y^2\)

= \(x^2-3xy-4xy+12y^2\)

= \(x\left(x-3y\right)-4y\left(x-3y\right)\)

= \(\left(x-3y\right)\left(x-4y\right)\)

Câu 1:

\(a(b^3-c^3)+b(c^3-a^3)+c(a^3-b^3)\)

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

\(=a(b^3-c^3)-b[(b^3-c^3)+(a^3-b^3)]+c(a^3-b^3)\)

\(=(a-b)(b^3-c^3)-(b-c)(a^3-b^3)\)

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

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

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

\(=(a-b)(b-c)[(c^2-a^2)+b(c-a)]\)

\(=(a-b)(b-c)(c-a)(c+a+b)\)

Câu 2:

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

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

\(=a^3(b^2-c^2)-b^3[(b^2-c^2)+(a^2-b^2)]+c^3(a^2-b^2)\)

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

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

\(=(a-b)(b-c)[(b+c)(a^2+ab+b^2)-(a+b)(b^2+bc+c^2)]\)

\(=(a-b)(b-c)(a^2b+a^2c-ac^2-bc^2)\)

\(=(a-b)(b-c)[b(a^2-c^2)+ac(a-c)]\)

\(=(a-b)(b-c)[b(a-c)(a+c)+ac(a-c)]\)

\(=(a-b)(b-c)(a-c)(ab+bc+ac)\)

a: ĐKXĐ: \(x\ne\pm y\)

b: \(A=\dfrac{-4xy}{\left(x-y\right)\left(x+y\right)}:\left(\dfrac{-1}{\left(x-y\right)\left(x+y\right)}+\dfrac{1}{\left(x+y\right)^2}\right)\)

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

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

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

\(6x^4+x^2-15=0\)

\(\Leftrightarrow\left(2x^2-3\right)\left(3x^2+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3=0\\3x^2+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2=3\\3x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2x^2}{2}=\dfrac{3}{2}\\\dfrac{3x^2}{3}=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{3}{2}\\x^2=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{3}{2}}\\x=-\sqrt{\dfrac{3}{2}}\\x=\sqrt{\dfrac{5}{3}}\\x=-\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)

sau khi kiểm tra lại thì ta nhận thấy: \(x\ne\pm\dfrac{5}{3}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{-\sqrt{\dfrac{3}{2}};\sqrt{\dfrac{3}{2}}\right\}\)

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

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

\(\Leftrightarrow\dfrac{3x-1}{3x+1}=\dfrac{2x+6-x+3}{x+3}\)

\(\Leftrightarrow\dfrac{3x-1}{3x+1}=\dfrac{x+9}{x+3}\)

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)=\left(x+9\right)\left(3x+1\right)\)

\(\Leftrightarrow3x^2+8x-3=3x^2+28x+9\)

\(\Leftrightarrow-20x=12\)

\(\Leftrightarrow x=\dfrac{-3}{5}\)

Vậy...

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

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

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

\(=x^5-x^2+x^2+x+1\)

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

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

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

CHÚC BẠN HỌC TỐT...

\(x^5+x+1\)

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

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

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

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

Đặt A = \(1+2+2^2+2^3+...+2^{49}-\left(2^{50}+3\right)\)

A = \(1+2+2^2+2^3+...+2^{49}-2^{50}-3\) (1)

2A = \(2+2^2+2^3+2^4+...+2^{50}-2^{51}-6\) (2)

Lấy (2) trừ (1) theo từng vế, ta được:

A = \(-2^{51}-4\)

Lời giải:

Áp dụng định lý Bê-du về phép chia đa thức, dư khi chia $x^8$ cho $x+\frac{1}{2}$ là \((-\frac{1}{2})^8=\frac{1}{2^8}\)

Do đó: \(x^8=(x+\frac{1}{2})B(x)+\frac{1}{2^8}\)

\(\Rightarrow B(x)=\frac{x^8-\frac{1}{2^8}}{x+\frac{1}{2}}=(x-\frac{1}{2})(x^2+\frac{1}{2^2})(x^4+\frac{1}{2^4})\)

Tiếp tục áp dụng định lý Bê-du, dư khi chia $B(x)$ cho $x+\frac{1}{2}$ là $B(-\frac{1}{2}$

Do đó:

\(r_2=B(\frac{-1}{2})=(\frac{-1}{2}-\frac{1}{2})[(-\frac{1}{2})^2+\frac{1}{2^2}][(-\frac{1}{2})^4+\frac{1}{2^4}]=-\frac{1}{16}\)

Lời giải:

Áp dụng định lý Bê-du về phép chia đa thức, dư khi chia $x^8$ cho $x+\frac{1}{2}$ là \((-\frac{1}{2})^8=\frac{1}{2^8}\)

Do đó: \(x^8=(x+\frac{1}{2})B(x)+\frac{1}{2^8}\)

\(\Rightarrow B(x)=\frac{x^8-\frac{1}{2^8}}{x+\frac{1}{2}}=(x-\frac{1}{2})(x^2+\frac{1}{2^2})(x^4+\frac{1}{2^4})\)

Tiếp tục áp dụng định lý Bê-du, dư khi chia $B(x)$ cho $x+\frac{1}{2}$ là $B(-\frac{1}{2}$

Do đó:

\(r_2=B(\frac{-1}{2})=(\frac{-1}{2}-\frac{1}{2})[(-\frac{1}{2})^2+\frac{1}{2^2}][(-\frac{1}{2})^4+\frac{1}{2^4}]=-\frac{1}{16}\)