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Cần lao vi tiên thủ vô vi thực DB thực CUC

Tự làm đi bạn êi :D

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 ]

\(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)

\(+\left(x^7-x^5+x^4-x^2+x\right)\)

\(+\left(x^6-x^4+x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

\(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

\(=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)

\(x^4+2x^2-24\)

\(=x^4+2x^2+1-25\)

\(=\left(x^2+1\right)^2-5^2\)

\(=\left(x^2+6\right)\left(x^2-4\right)\)

\(=\left(x^2+6\right)\left(x+2\right)\left(x-2\right)\)

\(a,2x-1-3x\left(2x-1\right)=0\)

\(\Leftrightarrow2x-1-6x^2+3x=0\)

\(\Leftrightarrow5x-1-6x^2=0\)

\(\Leftrightarrow6x^2-5x+1=0\)

\(\Leftrightarrow6x^2-2x-3x+1=0\)

\(\Leftrightarrow2x\left(3x-1\right)-\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1\\3x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)

\(b,2x^2+4x=0\)

\(\Leftrightarrow2x\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

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)

1)\(x\left(a-b\right)+a-b=x\left(a-b\right)+\left(a-b\right)=\left(a-b\right)\left(a+x\right)\)

2)\(x\left(a+b\right)-a-b=x\left(a+b\right)-\left(a+b\right)=\left(a+b\right)\left(x-1\right)\)

3)\(10ax-5ay-2x+y=\left(10ax-5ay\right)-\left(2x-y\right)=5a\left(2x-y\right)-\left(2x-y\right)=\left(2x-y\right)\left(5a-1\right)\)

4)\(2a^2x-5by-5a^2y+2bx=\left(2a^2x-5a^2y\right)+\left(2bx-5by\right)=a^2\left(2x-5y\right)+b\left(2x-5y\right)=\left(2x-5y\right)\left(a^2+b\right)\)5)\(2ax^2-bx^2-2ax+bx+4a-2b=\left(2ax^2-bx^2\right)-\left(2ax-bx\right)+\left(4a-2b\right)=x^2\left(2a-b\right)-x\left(2a-b\right)+2\left(2a-b\right)\)\(=\left(2a-b\right)\left(x^2-x+2\right)\)

x²+2x+1+2ax+2a

<=> (x²+2x+1)+(2ax+2a)

<=> (x+1)² + 2a(x+1)

<=> (x+1) (x+1+2a)

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

a) Ta có: \(\left(x+2\right).\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0-2\\x=0+3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-2\\x=3\end{cases}}}\)

Vậy x = { -2 ; 3 }

b) \(2x-x^2=0\)

\(\Rightarrow2.0-0^2=0\)

\(\Rightarrow x=0\)

Không biết câu B đúng hay sai! Chúc bạn học tốt!