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a) 16( 2x - 3 ) + x2( 3x - 2 ) = 0 < xem lại đề bạn ơi -- >
b) x3 - 7x2 = 7 - x
⇔ x2( x - 7 ) + x - 7 = 0
⇔ x2( x - 7 ) + ( x - 7 ) = 0
⇔ ( x - 7 )( x2 + 1 ) = 0
⇔ x - 7 = 0 hoặc x2 + 1 = 0
⇔ x = 7 ( x2 + 1 ≥ 1 > 0 ∀ x )
Với\(\left(x-3\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)\) và x = 1/3
\(\Rightarrow\left(\frac{1}{3}-3\right)\left(\frac{1}{3}+3\right)-\left(\frac{1}{2}+2\right)\left(\frac{1}{2}-1\right)=\left(\frac{1}{3}\right)^2-3^2-\left(\frac{5}{2}\cdot2-\frac{5}{2}\right)\)\(=\frac{1}{3^2}-\frac{3^2}{1}-\left(\frac{5}{1}-\frac{5}{2}\right)=\frac{1}{3^2}-\frac{3^4}{3^2}-\frac{5}{2}\)\(=\frac{-80}{3^2}-\frac{5}{2}=\frac{-160-45}{18}=\frac{-205}{18}\)
( x - 3 )( x + 3 ) - ( x + 2 )( x - 1 )
= x2 - 9 - ( x2 + x - 2 )
= x2 - 9 - x2 - x + 2
= -x - 7
Với x = 1/3 => Gtri bthuc = -1/3 - 7 = -22/3
Ta có : 3x3 + x2 - 13x + 5
= 3x3 + 6x2 - 5x2 - 3x - 10x + 5
= ( 3x3 + 6x2 - 3x ) - ( 5x2 + 10x - 5 )
= 3x( x2 + 2x - 1 ) - 5( x2 + 2x - 1 )
= ( x2 + 2x - 1 )( 3x - 5 )
=> ( 3x3 + x2 - 13x + 5 ) : ( x2 + 2x - 1 ) = 10x - 1
⇔ ( x2 + 2x - 1 )( 3x - 5 ) : ( x2 + 2x - 1 ) = 10x - 1
⇔ 3x - 5 = 10x - 1
⇔ 3x - 10x = -1 + 5
⇔ -7x = 4
⇔ x = -4/7
a) \(9x^2-12x+4\)
\(=9x^2-6x-6x+4\)
\(=3x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(3x-2\right)^2\)
b) \(2xy+16-x^2-y^2\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y\right)^2+16\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
c) \(3x+2x^2-2\)
\(=2x^2+4x-x-2\)
\(=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
\(A=\frac{\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)}{x+1}\)
ĐKXĐ : x ≠ -1
\(=\frac{\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)}{x+1}\)
\(=\frac{\left(x+1\right)\left[3x-2-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)
\(=3x-2-\left(2x^2+3x-5\right)\)
\(=3x-2-2x^2-3x+5\)
\(=-2x^2+3\)
Với x = 1/2 ( tmđk ) => \(A=-2\cdot\left(\frac{1}{2}\right)^2+3=\frac{5}{2}\)
