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Bài 1.
a) x2 + 7x +12 = 0
Ta có Δ = 72 - 4.12 = 1> 0 => \(\sqrt{\Delta}=\sqrt{1}=1\)
Phương trình có 2 nghiệm phân biệt:
x1 = \(\frac{-7+1}{2}=-3\)
x2= \(\frac{-7-1}{2}=-4\)
Bài 1
b) 2x2 + 5x - 3=0
Ta có: Δ = 52 + 4.2.3 = 49 > 0 => \(\sqrt{\Delta}=\sqrt{49}=7\)
Phương tình có 2 nghiệm phân biệt:
x1 = \(\frac{-5+7}{2.2}=\frac{1}{2}\)
x2 = \(\frac{-5-7}{2.2}-3\)
c) 3x2 +10x+7 = 0
Ta có: Δ = 102 - 4.3.7= 16> 0 => \(\sqrt{\Delta}=\sqrt{16}=4\)
Phương tình có 2 nghiệm phân biệt:
x1= \(\frac{-10+4}{2.3}=-1\)
x2= \(\frac{-10-4}{2.3}=-\frac{7}{3}\)
Bài 1 :
a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(\)\(=2y^2-10xy\)
Câu b tương tự
Bài 2 :
a ) \(x^2-9+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3+x-3\right)\)
\(=2x\left(x-3\right)\)
b ) \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
c ) \(x^3-4x^2+12x-27\)
\(=x^3-9x^2+5x^2+27x-15x-3^3\)
\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)
\(=\left(x-3\right)^3+5\left(x-3\right)\)
\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)
\(=\left(x-3\right)\left(x^2-6x+14\right)\)
d ) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(3x\left(x+1\right)-10x\left(x+1\right)\)
\(=-7x\left(x+1\right)\)
\(x^3-2x+y^3-2y=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)
theo mình đề câu c là 6x2
\(x^3+6x^2+9x-xz^2=x\left(x^2-6x+9-z^2\right)\)
\(=\left(x-3-z\right)\left(x-3+z\right)\)
\(x^2-11x+30=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)=\left(x-5\right)\left(x-6\right)\)
\(4x^2-3x-1=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+x-1=\left(4x+1\right)\left(x-1\right)\)
\(9x^2-7x-2=9x^2-9x+2x-2\)
\(=9x\left(x-1\right)+2\left(x-1\right)=\left(9x+2\right)\left(x-1\right)\)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-5=\left(x^2+x-1\right)^2-4\)
\(=\left(x^2+x-3\right)\left(x^2+x+1\right)\)
còn lại lát mình làm tiếp
Bài 1:
a, \(x^3-2x-y^3-2y=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
b, \(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
c, \(x^3+6x^2+9x-xz^2=x\left(x^2+6x+9-z^2\right)\)
\(=x\left[\left(x+3\right)^2-z^2\right]=x\left(x+3+z\right)\left(x+3-z\right)\)
Mỗi bài mình sẽ làm một câu mẫu ạ
Bài 1:
a) \(x^3-2x+y^3-2y\)
\(=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
Bài 2:
a) \(x^2-11x+30\)
\(=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-6\right)\left(x-5\right)\)
Bài 3:
a) \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-4x-x+4=0\)
\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Bài 2:
b: \(=4x^2-4x+x-1=\left(x-1\right)\left(4x+1\right)\)
c: \(=9x^2-9x+2x-2=\left(x-1\right)\left(9x+2\right)\)
e: Sửa đề: \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-2\)
\(=\left(x^2+3x\right)^2+3\left(x^2+3x\right)+2-2\)
\(=\left(x^2+3x\right)\left(x^2+3x+3\right)\)
\(=x\left(x+3\right)\left(x^2+3x+3\right)\)
ừ thì mình sẽ giúp bạn mà câu a bạn viết sai đề nha
1/a)\(2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
b)\(4x^2-3x-1=4x^2-4x+x-1=4x\left(x-1\right)+\left(x-1\right)=\left(4x+1\right)\left(x-1\right)\)
c)Sai đề: \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d)Sai đề:\(x^3-2x^2y+xy^2-9x=x\left(x-2xy+y^2-9\right)=x\left[\left(x-y\right)^2-9\right]=x\left(x-y+3\right)\left(x-y-3\right)\)
e)\(2x-2y-x^2+2xy-y^2=2\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(2-x+y\right)\)
f)Hình như sai đề đúng không?
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
2/a.\(7x-6x^2-2=0\)
\(\Leftrightarrow-\left(6x^2-3x-4x+2\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=1\end{matrix}\right.\)
b.\(16x-5x^2-3=0\)
\(\Leftrightarrow-\left(5x^2-15x-x+3\right)=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)
c.\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}=-2,5\\x=1\end{matrix}\right.\)
không có đâu bạn ơi
Đây là phương pháp tách hạng tử hay gì đó quên rồi
như bạn thấy là \(3x=-2x+5x\)
cái đó đúng nhá bạn
bạn ơi cho hỏi nha