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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)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 3 2020
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 \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 3 2020
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 \(x=-1\)
CM
18 tháng 12 2017
Ta có
Phần dư của phép chia f(x) cho g(x) là R = (a – 3)x + b + 4. Để phép chia trên là phép chia hết thì R = 0, Ɐx
ó (a – 3)x + b + 4 = 0, Ɐx ó a - 3 = 0 b + 4 = 0
ó a = 3 b = - 4 => ab = -12
Đáp án cần chọn là: A
NT
0
17 tháng 9 2021
\(1,A⋮B\Leftrightarrow x^3-3x^2-ax+3=\left(x-1\right)\cdot a\left(x\right)\)
Thay \(x=1\)
\(\Leftrightarrow1-3-a+3=0\\ \Leftrightarrow a=1\)
\(2,A⋮B\Leftrightarrow3x^3-16x^2+25x+a=\left(x^2-4x+3\right)\cdot b\left(x\right)\\ \Leftrightarrow3x^3-16x^2+25x+a=\left(x-3\right)\left(x-1\right)\cdot b\left(x\right)\)
Thay \(x=1\)
\(\Leftrightarrow3-16+25+a=0\\ \Leftrightarrow a=-12\)
Thay \(x=3\)
\(\Leftrightarrow3\cdot27-16\cdot9+25\cdot3+a=0\\ \Leftrightarrow81-144+75+a=0\\ \Leftrightarrow12+a=0\Leftrightarrow a=-12\)
Vậy \(a=-12\)
28 tháng 12 2021
\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)
\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)
24 tháng 10 2021
\(f\left(x\right)⋮g\left(x\right)\)
\(\Leftrightarrow x^4-3x^3+4x^2-x^2+3x-4+\left(a-3\right)x+\left(b+4\right)⋮x^2-3x+4\)
\(\Leftrightarrow\left(a,b\right)=\left(3;-4\right)\)
2 tháng 9 2021
Bài 2:
a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)
b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)
d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)
f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)
g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)
i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)
2 tháng 9 2021
a: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=\left(x+1\right)\left(3x-10\right)\)
b: \(x^2+6x+9-4y^2\)
\(=\left(x+3\right)^2-4y^2\)
\(=\left(x+3-2y\right)\left(x+3+2y\right)\)
c: \(x^2-2xy+y^2-5x+5y\)
\(=\left(x-y\right)^2-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-5\right)\)
\(5x-3x^2+3x^3-x^4=\left(x+1\right)^2\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4=x^2+2x+1\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4-1=x^2+2x+1-1\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4-1=x^2+2x\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4-1-2x=x^2+2x-2x\)
\(\Leftrightarrow-x^4+3x^3-3x^2+3x-1=x^2\)
\(\Leftrightarrow-x^4+3x^3-3x^2-1-x^2=x^2-x^2\)
\(\Leftrightarrow-x^4+3x^3-4x^2+3x-1=0\)
\(\Leftrightarrow-\left(x-1\right)^2\left(x^2-x+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=1\\x=\frac{1}{2}+i\frac{\sqrt{3}}{2}\\x=\frac{1}{2}-i\frac{\sqrt{3}}{2}\end{cases}}\)
Mình ko chắc :(