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Bài 1:
a) 25\(x^2\) - 0,09
= \(\left(5x\right)^2-0,3^2\)
= (5x - 0,3) (5x +0,3)
Bài 5:
a: \(=\left(2x-3\right)^2\)
b: \(=\left(2x+1\right)^2\)
c: \(=\left(6x+1\right)^2\)
d: \(=\left(3x-4y\right)^2\)
e: \(=\left(\dfrac{1}{2}x-2y\right)^2\)
f: \(=-\left(x-5\right)^2\)
1/ a/ 3x(4x2 - 3xy + 5y)
= (3x.4x2)-[3x.3xy)]+(3x.5y)
= 12x3 - 9x2y + 15xy
b/ (2x - 1).(x2 + 5x + 2)
= 2x3 + 10x2 + 4x - x2 - 5x - 2
= 2x3 + 9x2 - x - 2.
2/ a/ 4x - 8y = 4.(x-2y)
b/ x3 - x2y - x + y
= x2.(x - y) - (x - y)
= (x - y).(x2 - 1)
= (x - y).(x - 1).(x + 1)
3/ a/ x3 - 4x = 0
x..(x2 - 4) = 0
x.(x+2).(x-2) = 0
=> x = 0; x + 2 = 0; x - 2 = 0
hay x = 0; x = -2; x = 2.
b/ (2x+3)2 - x(4x+3) = 18
4x2 + 6x + 9 - 4x2 - 3x - 18 = 0
(4x2 - 4x2) + (6x - 3x) + (9-18) = 0
3x - 9 = 0
=> 3x = 9 => x = 3.
\(x^3-2x+y^3-2y=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)
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: \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-2\)
\(=\left(x^2+3x\right)^2+3\left(x^2+3x\right)+2-2\)
\(=\left(x^2+3x\right)\left(x^2+3x+3\right)\)
\(=x\left(x+3\right)\left(x^2+3x+3\right)\)
1.\(\left(4x-1\right)\left(2x^2-x-1\right)\)
\(=8x^3-4x^2-4x-2x^2+x+1\)
\(=8x^3-6x^2-3x+1\)
2.a)\(\left(x+1\right)^2+\left(x-2\right)\left(x+2\right)-3\left(x+1\right)\)
\(=x^2+2x+1+x^2-4-3x-3\)
\(=2x^2-x-6\)
b)\(\left(3x+1\right)^2+2\left(3x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(3x+1+2x-1\right)^2\)
\(=\left(5x\right)^2=25x^2\)
3.a)\(x^2-2x+x^3\)
\(=x\left(x^2+x-2\right)\)
\(=x\left(x^2-x+2x-2\right)\)
\(=x\left[x\left(x-1\right)+2\left(x-1\right)\right]\)
\(=x\left(x-1\right)\left(x+2\right)\)
b)\(2x^2-5x^3-x\)
\(=-x\left(5x^2-2x+1\right)\)
Bài 1: Sử dụng hằng đẳng thức đáng nhớ:
\(A=(2x+3)[(2x)^2-2x.3+3^2]-2(4x^3-1)\)
\(=(2x)^3+3^3-(8x^3-2)=8x^3+27-8x^3+2=29\)
--------------
\(B=(x-1)^3-4x(x+1)(x-1)+3(x-1)(x^2+x+1)\)
\(=(x-1)[(x-1)^2-4x(x+1)+3(x^2+x+1)]\)
\(=(x-1)(x^2-2x+1-4x^2-4x+3x^2+3x+3)\)
\(=(x-1)(-3x+4)\)
Bài 2:
a)
\(x^2-y^2-3x+3y=(x^2-y^2)-(3x-3y)\)
\(=(x-y)(x+y)-3(x-y)=(x-y)(x+y-3)\)
b)
\((b-a)^2+(a-b)(3a-2b)-a^2+b^2\)
\(=(a-b)^2+(a-b)(3a-2b)-(a^2-b^2)\)
\(=(a-b)^2+(a-b)(3a-2b)-(a-b)(a+b)\)
\(=(a-b)[(a-b)+(3a-2b)-(a+b)]=(a-b)(3a-4b)\)
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
4.a) \(2x^2-10x-3x-2x^2-26=0\)
\(-13x-26=0\Rightarrow-13\left(x+2\right)=0\)
\(\Rightarrow x=-2\)
b) \(2\left(x+5\right)-x^2-5x=0\)
\(2x+10-x^2-5x=0\Leftrightarrow-x^2-3x+10=0\)
\(-\left(x^2+3x-10\right)=0\)
\(-\left(x^2-2x+5x-10\right)=-\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\)
\(-\left(x-2\right)\left(x+5\right)=0\)
\(\left\{{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
c) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\left\{{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
d) \(x^3+x^2-4x-4=0\)
\(x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
g) \(\left(x-1\right)\left(2x+3-x\right)=0\)
\(\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
h) \(x^2-4x+8-2x+1=x^2-6x+9=0\)
\(\left(x-3\right)^2=0\Rightarrow x=3\)