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a, x2-7x-14y+2x
=x(x+2)-7(x-2y)
b, x3-4x2y+4xy2-25x
=x3-4x2y+4xy2-y3-25x+y3
=(x-y)3-25x+y3
a ) = x(x+2) - 7(x+2y)
b) = -4 xy ( x-y) + (x^3-25x) [ câu này mk , chaqcs là làm đúng đâu ]
\(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)
\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)
\(+\left(x^7-x^5+x^4-x^2+x\right)\)
\(+\left(x^6-x^4+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
\(a,2x-1-3x\left(2x-1\right)=0\)
\(\Leftrightarrow2x-1-6x^2+3x=0\)
\(\Leftrightarrow5x-1-6x^2=0\)
\(\Leftrightarrow6x^2-5x+1=0\)
\(\Leftrightarrow6x^2-2x-3x+1=0\)
\(\Leftrightarrow2x\left(3x-1\right)-\left(3x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\3x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=1\\3x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)
\(b,2x^2+4x=0\)
\(\Leftrightarrow2x\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
a, x(a - b) + (a - b)
= (x + 1)(a - b)
b, x(a + b) - a - b
= x(a + b) - (a + b)
= (x - 1)(a + b)
c, 10ax - 5ay - 2x + y
= 5a(2x - y) - (2x - y)
= (5a - 1)(2x - y)
d, 2a^2x - 5by - 5a^2y + 2bx
= 2x(a^2 + b) - 5y(b + a^2)
= (2a - 5y)(a^2 + b)
làm tiếp:
2ax2 - bx2 - 2ax +bx +4a-2b
= x2(2a-b) - x(2a-b) +2(2a-b)
=(2a-b)(x2-x+2)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
a) (x+2)(x-3)=0
<=> x+2=0
x-3=0
<=> x=-2
x= 3
b) 2x-x2=0
<=> x(2-x) =0
<=> x=0
2-x=0
<=> x=0
x=2
a)(x+2)(x-3)=0
=>\(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)=>\(\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
Vậy x=-2 hoặc x=3
b) 2x-x2=0
=> x(2-x)=0
=>\(\orbr{\begin{cases}x=0\\2-x=0\end{cases}}\)=>\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy x=0 hoặc x=2
