Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng định lý Bezout:
2x3 + 3x2 + ax + b chia hết cho (x+1).(x-1)
\(\Leftrightarrow\hept{\begin{cases}2.1^3+3.1^2+a.1+b=0\\2.\left(-1\right)^3-3.\left(-1\right)^2+a.\left(-1\right)+b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b=-5\\a-b=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-5\\b=0\end{cases}}\)
Áp dụng định lý Bezout:
x3 - 4x2+ ax + b chia hết cho x2 - 3x + 2
hay x3 - 4x2+ ax + b chia hết cho (x-1)(x-2)
\(\Leftrightarrow\hept{\begin{cases}1-4+a+b=0\\8-16+2a+b=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b=3\\2a+b=8\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5\\b=-2\end{cases}}\)
a)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x^3+2^3\right)-x^3-2x=0\)
\(\Leftrightarrow8-2x=0\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
b)\(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
\(x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(x^3-3x^2+3x-1-x^3-27+3x^2-12-2=0\)
\(3x-42=0\)
\(3x=42\)
\(x=14\)
\(x^2+4x+3=0\)
\(x^2+x+3x+3=0\)
\(x\left(x+1\right)+3\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+3\right)=0\)
\(\left[\begin{array}{nghiempt}x+1=0\\x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=-1\\x=-3\end{array}\right.\)
\(4x^2+4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(\left[\begin{array}{nghiempt}2x-1=0\\2x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=1\\2x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{3}{2}\end{array}\right.\)
\(x^2-x-12=0\)
\(x^2-4x+3x-12=0\)
\(x\left(x-4\right)+3\left(x-4\right)=0\)
\(\left(x-4\right)\left(x+3\right)=0\)
\(\left[\begin{array}{nghiempt}x-4=0\\x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)
\(x^2-25-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+5-1\right)=0\)
\(\left(x-5\right)\left(x+4\right)=0\)
\(\left[\begin{array}{nghiempt}x-5=0\\x+4=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)
\(x^2\left(x^2+1\right)-x^2-1=0\)
\(x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)
\(\left(x^2+1\right)\left(x^2-1\right)=0\)
\(\left(x^2+1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+1=0\end{array}\right.\) (vì \(x^2+1\ge1>0\))
\(\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\)
a) x^2 - 11x + 18 = 0
=> x^2 - 2x - 9x + 18 = 0
=> x ( x- 2 ) - 9 ( x- 2 ) = 0
=> ( x- 9 )( x- 2 )= 0
=> x- 9 = 0 hoặc x - 2 = 0
=> x= 9 hoặc x = 2
A= (4x2 + y2).[(2x)2 - y2] = (4x2 +y2)(4x2 - y2) = (4x2)2 _ (y2)2 = 16x4 - y4
a, \(x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
b,\(\left(3x-1\right)^2-16=0\Rightarrow\left(3x-1-4\right)\left(3x-1+4\right)\)
\(\Rightarrow\left(3x-5\right)\left(3x+3\right)=0\Rightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
\(x^2-2x=0\Leftrightarrow x.\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=0+2=2\end{cases}}}.\)
\(\left(3x-1\right)^2-16=0\)
\(\Leftrightarrow\left(3x-1\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=4\\3x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=4+1=5\\3x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}}\)
\(1.\)
\(a.=3\left(x+2\right)\)
\(b.=4\left(x-y\right)+x\left(x-y\right)\)
\(=\left(4+x\right)\left(x-y\right)\)
\(c.=\left(x-6\right)\left(x+6\right)\)
\(d.=\left(x^2-2y^2\right)\left(x^2+2y^2\right)\)
\(2.\)
\(a.ĐKXĐ:\)\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
\(b.A=\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{3}{x+1}với\)\(x\ne\pm1\)
\(c.A=-1\Leftrightarrow\frac{3}{x+1}=-1\)
\(\Rightarrow\left(x+1\right).-1=3\)
\(-x-1=3\)
\(-x=4\)
\(\Rightarrow x=4\left(t/mđk\right)\)
\(d.\)Để \(x\in Z,A\in Z\Leftrightarrow x+1\inƯ\left(3\right)\)
\(Ư\left(3\right)\in\left\{\pm1,\pm3\right\}\)
| x+1 | 1 | -1 | 3 | -3 |
| x | 0 | -2 | 2 | -4 |
Vậy \(x\in\left\{0,-2,2,-4\right\}\)
1a) 3x + 6 = 3 (x + 2)
b) 4x - 4y + x2 - xy = (4x - 4y) + (x2 - xy) = 4 (x - y) + x (x - y) = (4 + x) (x - y)
c) x2 - 36 = x2 - 62 = (x + 6) (x - 6)
2a) phân thức A được xác định khi \(x^2-1\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\ne0\)
\(\Rightarrow x+1\ne0..và..x-1\ne0\)
\(x\ne-1..và..x\ne1\)
b) \(A=\frac{3x-3}{x^2-1}=\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{3}{x+1}\)
c) \(A=-1\Rightarrow\frac{3}{x+1}=-1\)
\(\Rightarrow x+1=-3\)
\(x=-4\left(TM\text{Đ}K\right)\)
Vậy x = -1 thì A = -1
#Học tốt!!!
~NTTH~
1. \(\left(x+1\right)^2-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+1-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x+1=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
Vậy ...
\(x\left(x+2\right)-3\left(-x-2\right)=0\)
\(\Leftrightarrow x^2+2x+3x+6=0\)
\(\Leftrightarrow x^2+5x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)
Vậy ...
Còn cậu nữa chịu rồi !
câu 2 nhé :
\(3x\left(2x-8\right)-\left(2x-8\right)^2=0\)
câu này em phải sử dụng tam thức bậc 2 liệu em đã học chưa z :(????