phân tích đa thức thành nhân tử
a) x^3+4x^2-29x+24
b) x4 +6x3 +7x2-6x+1
c) (x2-x+2)2+(x-2)2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
PT
1
CM
17 tháng 5 2018
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
5 tháng 10 2021
\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)
\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)
T
13 tháng 1
Bài 1:
\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)
\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)
\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)
T
13 tháng 1
Bài 2:
\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)
Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:
\(\left(y-1\right)\left(y+1\right)=120\)
\(\Leftrightarrow y^2-1=120\)
\(\Leftrightarrow y^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)
+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)
+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)
\(\Leftrightarrow x^2+5x+16=0\)
\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)
Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
\(\Rightarrow\) loại
Vậy \(x\in\left\{1;-6\right\}\).
\(b,\) Đề thiếu vế phải rồi bạn.
PH
2
13 tháng 11 2016
a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0
=> x3 + 4x2 - 29x + 24 = x2(x-1) + 5x(x-1) - 24(x-1)
= (x-1)(x2+5x-24) = (x-1)(x-3)(x+8)
b) ...
RZ
13 tháng 11 2016
a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
1 tháng 10 2021
\(x^3+4x^2-29x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-1\right)-3\left(x-1\right)\right]\)
\(=\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
16 tháng 11 2021
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
16 tháng 11 2021
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
NN
5
13 tháng 9 2017
a,\(=x^3-x^2+5x^2-5x-24x+24\)
\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+5x-24\right)\)
\(=\left(x-1\right)\left(x^2-3x+8x-24\right)\)
\(=\left(x-1\right)\left(x\left(x-3\right)+8\left(x-3\right)\right)\)
\(=\left(x-1\right)\left(x-3\right)\left(x+8\right)\)
