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a/\(\left(x-1\right)\left(x^5+x^4+x^3+x^2+x+1\right).\)
\(=\left(x-1\right)\left[\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left[x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Câu b/ quên làm ạ :> Bù nè
b/ \(2\left(3x-1\right)\left(2x+5\right)-\left(4x-1\right)\left(3x-2\right)\)
\(=2\left(6x^2+15x-2x-5\right)-\left(12x^2-8x-3x+2\right)\)
\(=2\left(6x^2+13x-5\right)-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-12x^2+11x-2\)
\(=37x-12\)
a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)
b. =\(\frac{2.\left(x+3\right)}{x.\left(3x-1\right)}.\frac{-\left(3x-1\right)}{x.\left(x+3\right)}=\frac{-2}{x^2}\)
Chắc chắn đúng, mik nhaaaaaa
a) ( 3x + 2y - 1 )( x - 5 ) - ( x - 2 )2y
= 3x(x - 5) + 2y(x - 5) - 1(x - 5) - ( 2xy - 4y )
= 3x2 - 15x + 2xy - 10y - x + 5 - 2xy + 4y
= 3x2 - 16x - 6y + 5
b) ( 3x - 2 )( 3x + 2 ) - ( 2x + 1 )( 4x + 3 )
= [ ( 3x )2 - 22 ] - ( 8x2 + 10x + 3 )
= 9x2 - 4 - 8x2 - 10x - 3
= x2 - 10 - 7
a) \(\left(x^2-1\right)\left(x^2+2x\right)=x^4+2x^3-x^2-2x\)
b) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)=6x^2-3x+4x-2\left(3-x\right)\)
\(=6x^2-3x+4x-6+2x\)
\(=6x^2+3x-6\)
c) \(\left(x+3\right)\left(x^2+3x-5\right)=x^3+3x^2+3x^2+9x-5x-15\)
\(=x^3+6x^2+4x-15\)
d) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+x^2-x^2-x+x+1\)
\(=x^3+1\)
e) \(\left(2x^3-3x-1\right)\left(5x+2\right)=10x^4-15x^2-5x+4x^3-6x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
f) \(\left(x^2-2x+3\right)\left(x-4\right)=x^3-2x^2+3x-4x^2+8x-12\)
\(=x^3-6x^2+11x-12\)
Bài 1:
a) \(3x\left(5x^2-2x+1\right)\)
\(=15x^3-6x^2+3x\)
b) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^2\left(x^2-1\right)+2x\left(x^2-1\right)\)
\(=x^4-x^2+2x^3-2x\)
\(=x^4+2x^3-x^2-2x\)
Bài 2:
a) \(3x^2=2x\)
\(\Leftrightarrow3x^2-2x=0\)
\(\Leftrightarrow x\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
b)\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow14x=4\Leftrightarrow x=\frac{2}{7}\)
\(\frac{x}{3x+1}+\frac{2x+1}{3x+1}=\frac{x+2x+1}{3x+1}=\frac{3x+1}{3x+1}=1\)
\(\text{CHÚC BẠN HOK TOT}\)
\(\frac{x}{3x+1}+\frac{2x+1}{3x+1}=\frac{x+2x+1}{3x+1}=\frac{\left(x+2x\right)+1}{3x+1}=\frac{3x+1}{3x+1}=1\)