a: \(A\left(2\right)=2^5-2\cdot2^4+5\cdot2-3=32-32+10-3=7\)

\(B\left(-1\right)=-\left(-1\right)^5+3\cdot\left(-1\right)^3+5\cdot\left(-1\right)+11=1-3-5+11=4\)

b: Ta có: A(x)+B(x)

\(=x^5-2x^4+5x-3-x^5+3x^3+5x+11\)

\(=-2x^4+3x^3+10x+8\)

Ta có: A(x)-B(x)

\(=x^5-2x^4+5x-3+x^5-3x^3-5x-11\)

\(=2x^5-2x^4-3x^3-14\)

1.

a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)

\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)

\(\Leftrightarrow x^3+3x^2+2x-6=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)

b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)

\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)

1c/

\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)

\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)

\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn

Vậy pt có nghiệm duy nhất \(x=-1\)

2.

a. \(x^4-x^3+x^2+x^2-x+1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\x^2-x+1=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

b.

\(x^4+x^3+x^2+x+1=0\)

\(\Leftrightarrow x\left(x^3+1\right)+x^3+1+x^2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)+x^2=0\)

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

Mà \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(x^2-x+1\right)\ge0\\x^2\ge0\end{matrix}\right.\)

Nên dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\) ko tồn tại x thỏa mãn

\(a,A\left(x\right)=-3x^3+2x^2-6+5x+4x^3-2x^2-4-4x\\ =\left(-3x^3+4x^3\right)+\left(2x^2-2x^2\right)+\left(5x-4x\right)+\left(-6-4\right)\\ =x^3+0+x-10\\ =x^3+x-10\)

Bậc của đa thức : \(3\)

Hệ số cao nhất ứng với hệ số của số mũ cao nhất : \(1\)

b, \(B\left(x\right)=A\left(x\right).\left(x-1\right)\\ =\left(x^3+x-10\right)\left(x-1\right)\\ =x^3.x+x.x-10x-x^3-x+10\\ =x^4+x^2-x^3-10x-x+10\\ =x^4-x^3+x^2-11x+10\)

\(B\left(2\right)=2^4-2^3+2^2-11.2+10=0\)

\(a,A\left(x\right)=-3x^3+2x^2-6+5x+4x^3-2x^2-4-4x\\ =\left(-3x^3+4x^3\right)+\left(2x^2-2x^2\right)+\left(5x-4x\right)+\left(-6-4\right)\\ =x^3+0+x-10\\ =x^3+x-10\)

Bậc của đa thức \(3\)

Hệ số cao nhất là \(1\)

\(b,B\left(x\right)=A\left(x\right).\left(x-1\right)=\left(x^3+x-10\right)\left(x-1\right)\\ =x^3.x+x.x-10x-x^3-x+10\\ =x^4+x^2-x^3-x-10x+10\\ =x^4-x^3+x^2-11x+10\)

Thay \(x=2\) vào \(B\left(x\right)\)

\(=2^4-2^3+2^2-11.2+10\\ =0\) 

Vậy tại \(x=2\) thì \(B\left(x\right)=0\)

a)

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

Bậc của A(x) là 3

Hệ số tự do A(x) là -1

Hệ số cao nhất của A(x) là 3

Tại A(-2)

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

\(=-17\)

b)

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

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

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

c)

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

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

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

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

d)

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

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

\(C\left(x\right)=3x^3+3x^2+2x-1-4x^2+2x+8\)

\(C\left(x\right)=3x^3+\left(3x^2-4x^2\right)+\left(2x+2x\right)+\left(-1+8\right)\)

\(C\left(x\right)=3x^3-x^2+4x+7\)

chúc bạn học giỏi

A(x)+B(x)=2x-3x³+2x²+1+4x³+2x²-5

               = x³+4x²+2x-4

thay x=1 vào B(x) ta được

B(x)=4.1³+2.1³-5

      =4+2-5

     =1

\(A\left(x\right)+B\left(x\right)=\left(x+2\right)\left(x^2+2x-2\right)\)

thay x=1 \(=>A\left(1\right)+B\left(1\right)=3\left(1+2-2\right)=3\)

Ta có:

A(x)+B(x)=2x-3x³+2x²+1+4x³+2x²-5

                 =(-3x³+4x³)+(2x²+2x²)+2x+(1-5)

                 =x³+4x²+2x-4

Thay x=1 vào B(x) ta có:

B(1)=4*1³+2*1²-5

       =4+2-5=1

`#Namnam041005`

`a)`

`A(x) =`\(x^5+ x^3- 4x - x^5 + 3x - x^2 + 7\)

`= (x^5 - x^5) + x^3 - x^2 + (-4x + 3x) + 7`

`= x^3 - x^2 - x + 7`

`B(x) = `\(3x^2 - x^5 + 5x - 2x^2 - 9\)

`= (3x^2 - 2x^2) - x^5 + 5x - 9`

`= -x^5 + x^2 + 5x - 9`

`b)`

`A(x)``= x^3 - x^2 - x + 7`

Bậc của đa thức: `3`

Hệ số cao nhất: `1`

Hệ số tự do: `7`

`c)`

`A(x) + B(x) = x^3 - x^2 - x + 7 -x^5 + x^2 + 5x - 9`

`= -x^5 + x^3 + (-x^2 + x^2) + (-x+5x) + (7-9)`

`= -x^5 + x^3 + 4x - 2`

`A(x) - B(x) = x^3 - x^2 - x + 7 - (-x^5 + x^2 + 5x - 9)`

`= x^3 - x^2 - x + 7 +x^5 - x^2 - 5x + 9`

`= x^5 + x^3 + (-x^2 - x^2) + (-x-5x) + (7+9)`

`= x^5 + x^3 - 2x^2 - 6x + 16`

___

`A(x) + B(x) = -x^5 + x^3 + 4x - 2=0`

Bạn xem lại đề

`d)`

`H(x) - B(x) = x^3 + x^2 - x + 1`

`=> H(x) = (x^3 + x^2 - x + 1) + B(x)`

`=> H(x) = x^3 + x^2 - x + 1 -x^5 + x^2 + 5x - 9`

`= -x^5 + x^3 + (x^2 + x^2) + (-x+5x) + (1 - 9)`

`= -x^5 + x^3 + 2x^2 + 4x - 8`

a: A(x)=x^5-x^5+x^3-x^2-4x+3x+7

=x^3-x^2-x+7

B(x)=-x^5+3x^2-2x^2+5x-9

=-x^5+x^2+5x-9

b: Bậc: 3

Hệ số cao nhất: 1

hệ số tự do: 7

c: A(x)+B(x)

=x^3-x^2-x+7-x^5+x^2+5x-9

=-x^5+x^3+4x-2

A(x)-B(x)

=x^3-x^2-x+7+x^5-x^2-5x+9

=x^5+x^3-2x^2-6x+16

d: H(x)=x^3+x^2-x+1+B(x)

=x^3+x^2-x+1-x^5+x^2+5x-9

=-x^5+x^3+2x^2+4x-8

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

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

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

d) \(x^2-2x-15=x^2+3x-5x-15=\left(x+3\right)\left(x-5\right)\)