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14 tháng 10 2016
nhiều quá, các bn ngại làm, chia nhỏ ra,mk làm cho 2 câu
a) x2 +5x -6 = x2 -x +x + 5x -6
= x2 -x +6x -6
= x( x-1) + 6(x-1) = (x-1)(x+6)
b) 5x2 +5xy -x-y = 5x(x+y) -(x+y)
= (x+y)(5x-1)
14 tháng 10 2016
c) 7x -6x2 -2 = 6( x+2/3)(x+1/2)
d) x2 +4x +3 = x2 +x +3x +3
= x(x+1) + 3(x+1)
= (x+1)(x+3)
26 tháng 6 2023
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
15 tháng 10 2017
Tìm x:
\(5x\left(x-1\right)=x-1\)
\(5x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(5x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\)\(\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\)\(\Rightarrow\)\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
Vậy x=\(\dfrac{1}{5}\)hoặc x=1
\(2\left(x+5\right)-x^2-5x=0\)
\(2\left(x+5\right)-x\left(x+5\right)=0\)
\(\left(2-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy...
15 tháng 10 2017
A)\(x^2+5x-6=x^2-x+6x-6\\ =\left(x-1\right)\left(x+6\right)\)
B)\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(5x-1\right)\)
C)\(7x-6x^2-2=-6x^2+3x+4x-2\\ =-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2x-1\right)\left(2-3x\right)\)
D)\(x^2+4x+3=x^2+x+3x+3=\left(x+1\right)\left(x+3\right)\)
E)\(2x+3x-5=5x-5=5\left(x-1\right)\)
F)\(16x-5x^3=x\left(16-5x^2\right)\)
MM
2
DT
20 tháng 12 2021
\(a,3x^2y-6x=3x\left(xy-2\right)\)
\(b,x^2-4=x^2-2^2=\left(x-2\right)\left(x+2\right)\)
\(c,5x^2+5xy-x^2-2xy-y^2\)
\(=5x\left(x+y\right)-\left(x+y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-x+y\right)\)
\(=\left(x+y\right)\left(4x+y\right)\)
3 tháng 10 2021
\(a,=x^2-6x-x+6=\left(x-6\right)\left(x-1\right)\\ b,=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\\ c,=x^3-x^2+2x^2-2x+2x-2=\left(x-1\right)\left(x^2+2x+2\right)\\ d,=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\\ e,=x^2\left(x-6\right)+\left(x-6\right)=\left(x^2+1\right)\left(x-6\right)\\ f,=x^3\left(x^2+1\right)+\left(x^2+1\right)=\left(x^3+1\right)\left(x^2+1\right)\\ =\left(x+1\right)\left(x^2-x+1\right)\left(x^2+1\right)\)
BA
1
30 tháng 7 2023
a: =x^2+2x+3x+6
=(x+2)(x+3)
b: =5x(x+y)-(x+y)
=(x+y)(5x-1)
c: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
11 tháng 7 2023
13: =x(a^3-1)-b(a-1)
=x(a-1)(a^2+a+1)-b(a-1)
=(a-1)(a^2x+a*x+x-b)
12: =(x-y)(x+y)-(x-y)
=(x-y)(x+y-1)
10: =3(x^2-4y^2)
=3(x-2y)*(x+2y)
7: =x^2-x-5x+5=(x-1)(x-5)
8: =x^2+3x+4x+12=(x+3)(x+4)
9: =2x^2-6x-x+3=(x-3)(2x-1)
14 tháng 1 2023
a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)
=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)
=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22
=>-x^2+59x+14-8x^2+5x+22=0
=>-9x^2+54x+36=0
=>x^2-6x-4=0
=>\(x=3\pm\sqrt{13}\)
b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)
=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)
=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32
=>x^2+6x+19=x^2+4x-32
=>2x=-51
=>x=-51/2
13 tháng 2 2022
1.
a.\(\Leftrightarrow7x-5x=3+12\)
\(\Leftrightarrow2x=15\Leftrightarrow x=\dfrac{15}{2}\)
b.\(\Leftrightarrow6x-10-7x-7=2\)
\(\Leftrightarrow x=-19\)
c.\(\Leftrightarrow1-3x=4x-3\)
\(\Leftrightarrow7x=2\Leftrightarrow x=\dfrac{2}{7}\)
d.\(\Leftrightarrow8x^2-4x+12x-6-8x^2-8x-2=12\)
\(\Leftrightarrow-2=12\left(voli\right)\)
\(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=-\left(6x^2-7x+2\right)=-\left[\left(6x^2-3x\right)-\left(4x+2\right)\right]=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left[\left(3x-2\right)\left(2x-1\right)\right]\)
d) \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)