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a) 4x3 - 13x2 + 9x - 18
= 4x3 - 12x2 - x2 + 3x + 6x - 18
= 4x2( x - 3) - x( x - 3) + 6( x - 3)
= ( x - 3)( 4x2 - x + 6)
b) - x3 - 6x2 + 6x + 1
= 6x( 1 - x) + 1 - x3
= 6x( 1 - x) + ( 1 - x )( x2 + x + 1)
= ( 1 - x)( x2 + 7x + 1)
c) x3 + 3x2 + 3x + 2
= x3 + 2x2 + x2 + 2x + x + 2
= x2( x + 2) + x( x + 2) + x + 2
= ( x + 2)( x2 + x + 1)
a) \(4x^3-13x^2+9x-18\)
\(=4x^3-12x^2-x^2+3x+6x-18\)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)
Đặt \(x^2+3x+1=t\)
\(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
\(=t\left(t-4\right)-5\)
\(=t^2-4t-5\)
tự làm nốt ý này nhé.
những ý kia lát nx mình làm.
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
a, (x4-2x3+2x-1):(x2-1) = \(\frac{\left(x^4-1\right)-\left(2x^3-2x\right)}{x^2-1}\)
= \(\frac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\) =\(\frac{\left(x^2-1\right)\left(x^2+1-2x\right)}{x^2-1}\)
= \(x^2+1-2x\)= \(\left(x-1\right)^2\)
b, (8x3-6x2-5x+3):((4x+3)
a) ( x - 2 )3 - x( x + 1 )( x - 1 ) + 6x( x - 3 )
= x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 - 18x
= x3 - 6x - 8 - x3 + x
= -5x - 8
b) ( x + 1 )3 - ( x - 1 )3 - 6( x - 1 )2
= x3 + 3x2 + 3x + 1 - ( x3 - 3x2 + 3x - 1 ) - 6( x2 - 2x + 1 )
= x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 + 12x - 6
= 12x - 4
c) ( 2x + 1 )( 4x2 - 2x + 1 ) + ( 2 - 3x )( 4 + 6x + 9x2 ) - 9
= ( 2x )3 + 13 + 23 - ( 3x )3 - 9
= 8x3 + 1 + 8 - 27x3 - 9
= -19x3
d) ( x + 1 )3 + ( x - 1 )3 + x3 - 3x( x - 1 )( x + 1 )
= x3 + 3x2 + 3x + 1 + x3 - 3x2 + 3x - 1 + x3 - 3x( x2 - 1 )
= 3x3 + 6x - 3x2 + 3x
= 9x
Bài 1:
a: =>9x^2-6x+1=9x^2-2x
=>-4x=-1
=>x=1/4
b: \(\Leftrightarrow x^2+6x+9-x^2-2x-3=14\)
=>4x+6=14
=>4x=8
=>x=2
Bài 2:
a: \(=2x^2-6x+x-3-x^2+5x+3x=x^2+3x-3\)
b: =x^3-6x^2+12x-8-x^3+6x^2
=12x-8
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
a,(x + 3)2 - (x - 3)2 = 6x + 18
=> x^2 + 6x + 9 - x^2 + 6x - 9 = 6x + 19
=> 12x - 6x = 19
=> 6x = 19
=> x = 19/6
\(\left(x+3\right)^2-\left(x-3\right)^2=6x+1\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x+19\)
\(\Leftrightarrow12x-6x=19\)
\(\Leftrightarrow6x=19\)
\(\Leftrightarrow x=\frac{19}{6}\)
Chúc bạn học tốt~~~