Bài 1: Phân tích đa thức thành nhân tử:
a) x(y+z) + 3(y+z)
\(=\left(y+z\right)\left(x+3\right)\)
b) 2x2 - 6x
\(=2x\left(x+3\right)\)
c) x2 - y2 - 3x - 3y
\(=\left(x^2-y^2\right)-\left(3x+3y\right)\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-3\right)\)
d) 2x2 - 5x - 3
\(=2x^2-6x+x-3\)
\(=\left(2x^2-6x\right)+\left(x-3\right)\)
\(=2x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(2x+1\right)\)
e) x4 - y4
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
f) mx - my + nx - ny
\(=\left(mx-my\right)++\left(nx-ny\right)\)
\(=m\left(x-y\right)+n\left(x-y\right)\)
\(=\left(x-y\right)\left(m+n\right)\)
Bài 1:
a,\(x\left(y+z\right)+3\left(y+z\right)\)
\(=\left(x+3\right)\left(y+z\right)\)
b,\(2x^2-6x\)
\(=2x\left(x-6\right)\)
c,\(x^2-y^2-3x-3y\)
\(=\left(x^2-y^2\right)+\left(-3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x-y-3\right)\left(x+y\right)\)
d,\(2x^2-5x-3\)
\(=2x^2-6x+1x-3\)
\(=\left(2x^2-6x\right)+\left(1x-3\right)\)
\(=2x\left(x-3\right)+1\left(x-3\right)\)
\(=\left(2x+1\right)\left(x-3\right)\)
e,\(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
f,\(mx-my+nx-ny\)
\(=\left(mx-my\right)+\left(nx-ny\right)\)
\(=m\left(x-y\right)+n\left(x-y\right)\)
\(=\left(m+n\right)\left(x-y\right)\)
Bài 2: Tìm x, biết:
a) x2(2-x) - x + 2 = 0
\(\Rightarrow x^2\left(2-x\right)+\left(2-x\right)=0\)
\(\Rightarrow\left(2-x\right)\left(x^2+1\right)=0\)
\(\Rightarrow2-x=0\)
\(\Rightarrow x=2\)
b) 2(x+5) = x2+5x
\(\Rightarrow2\left(x+5\right)=x\left(x+5\right)\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
c) x(5-2x) + 2x2 = 15
\(\Rightarrow5x-2x^2+2x^2=15\)
\(\Rightarrow5x-15=0\)
\(\Rightarrow5\left(x-3\right)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
d) x(x-2) + x - 2 = 0
\(\Rightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
e) x2 + y2 - 4x + 6y + 13 = 0
\(\Rightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=0\)
\(\Rightarrow\left(x-2\right)^2+\left(y+3\right)^2=0\)
Ta có : \(\left(x-2\right)^2+\left(y+3\right)^2\ge0\)với mọi x
Để\(\left(x-2\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
Bài 2:
a,\(x^2\left(2-x\right)-x+2=0\)
\(\Leftrightarrow x^2\left(2-x\right)+\left(-x+2\right)=0\)
\(\Leftrightarrow x^2\left(2-x\right)+\left(2-x\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2-x\right)=0\)
Vì \(x^2+1\) luôn lớn hơn 0 \(\Rightarrow\) loại.
\(\Leftrightarrow2-x=0\)
\(\Leftrightarrow x=2\)
b,\(2\left(x+5\right)=x^2+5x\)
\(\Leftrightarrow2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy...
c,\(x\left(5-2x\right)+2x^2=15\)
\(\Leftrightarrow5x-2x^2+2x^2-15=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow5\left(x-3\right)=0\)
Vì 5\(\ne0\) \(\Rightarrow\) loại.
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
d,\(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy...
Bài 1:
a x(y+z)+3(y+z)
=> (x+3)(y+z)
b 2x2-6x
=> 2x(x-3)
c x2-y2-3x-3y
=> (x-y)(x+y)-3(x+y)
=> (x+y)(x-y-3)
d 2x2 -5x-3
=> 2x2-6x+x-3
=> 2x(x-3)+(x-3)
=> (2x+1)(x-3)
e x4-y4
=> (x2-y2)(x2+y2)
=> (x-y)(x+y)(x2+y2)
f mx-my+nx-ny
=> m(x-y)+n(x-y)
=> (m+n)(x-y)
2)
a) \(x^2\left(2-x\right)-x+2=0\)
\(\Leftrightarrow x^2\left(2-x\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=2 hoặc x=1 hoặc x=-1
b) \(2\left(x+5\right)=x^2+5x\)
\(\Leftrightarrow2\left(x+5\right)-\left(x^2+5x\right)=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy x=-5 hoặc x=2
c) \(x\left(5-2x\right)+2x^2=15\)
\(\Leftrightarrow5x-2x^2+2x^2=15\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=\dfrac{15}{5}=3\)
Vậy x=3
d) \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy x=2 hoặc x=-1
e) \(x^2+y^2-4x+6y+13=0\)
\(\Leftrightarrow x^2+y^2-4x+6y+4+9=0\)
\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
Vậy x=2 và y=-3
