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a) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left[3x\left(x+1\right)-5x^2+7\right]\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
\(=-\left(x+1\right)\left(2x^2-3x-7\right)\)
b) \(\left(x+y\right)\left(2x-y\right)-\left(3x-y\right)\left(y-2x\right)\)
\(=\left(x+y\right)\left(2x-y\right)+\left(3x-y\right)\left(2x-y\right)\)
\(=\left(2x-y\right)\left(x+y+3x-y\right)\)
\(=4x\left(2x-y\right)\)
c) \(5u\left(u-v\right)^2+10u^2\left(v-u\right)^2\)
\(=5u\left(u-v\right)^2+10u^2\left(u-v\right)^2\)
\(=5u\left(u-v\right)^2\left(1+2u\right)\)
Trả lời:
a, 3x ( x + 1 )2 - 5x2 ( x + 1 ) + 7 ( x + 1 )
= ( x + 1 )[ 3x ( x + 1 ) - 5x2 + 7 ]
= ( x + 1 )( 3x2 + 3x - 5x2 + 7 )
= ( x + 1 )( - 2x2 + 3x + 7 )
b, ( x + y )( 2x - y ) - ( 3x - y )( y - 2x )
= ( x + y )( 2x - y ) + ( 3x - y )( 2x - y )
= ( 2x - y )( x + y + 3x - y )
= 4x ( 2x - y )
c, 5u ( u - v )2 + 10u2 ( v - u )2
= 5u ( u - v )2 + 10u2 ( u - v )2
= 5u ( u - v )2( 1 + 2u )
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
Bài2: phân tích đa thức thành nhân tử
\(a,x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(y+x-2\right)\)
\(b,x^3-5x^2+x-5\)
\(=x^2\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+x-5\right)\left(x-x-5\right)\)
\(c,x^2-2xy+y^2-9\)
\(=\left(x^2-y^2\right)-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
chúc bạn học tốt !
a) A = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
A = 6x^2 + 33x - 10x - 55 - 6x^2 - 23x - 21
A = -76
b) B = 4x(3x - 2) - 3x(4x + 1)
B = 12x^2 - 8x - 12x^2 - 3x
B = -11x
c) C = (x + 3)(x - 2) - (x - 1)^2
C = x^2 + x - 6 - x^2 + 2x - 1
C = 3x - 7
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
a) x2 - 3x + 2 = x2 - x - 2x + 2 = x( x - 1 ) - 2( x - 1 ) = ( x - 1 )( x - 2 )
b) 2x2 - x - 6 = 2x2 - 4x + 3x - 6 = 2x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 2x + 3 )
c) x2 - 5x - 6 = x2 + x - 6x - 6 = x( x + 1 ) - 6( x + 1 ) = ( x + 1 )( x - 6 )
d) x2 + 8x + 7 = x2 + x + 7x + 7 = x( x + 1 ) + 7( x + 1 ) = ( x + 1 )( x + 7 )
e) 3x2 + 2x - 5 = 3x2 - 3x + 5x - 5 = 3x( x - 1 ) + 5( x - 1 ) = ( x - 1 )( 3x + 5 )
f) 4x2 - 3x - 1 = 4x2 - 4x + x - 1 = 4x( x - 1 ) + ( x - 1 ) = ( x - 1 )( 4x + 1 )
a \(x^2-3x+2=x^2-x-2x+2=\left(x-1\right)\left(x-2\right)\)
b, \(2x^2-x-6=2x^2-4x+3x-6=\left(x-2\right)\left(2x+3\right)\)
c, \(x^2-5x-6=x^2+x-6x-6=\left(x+1\right)\left(x-6\right)\)
d, \(x^2+8x+7=x^2+x+7x+7=\left(x+1\right)\left(x+7\right)\)
e, \(3x^2+2x-5=3x^2-3x+5x-5=\left(x-1\right)\left(3x+5\right)\)
f, \(4x^2-3x-1=4x^2-4x+x-1=\left(x-1\right)\left(4x+1\right)\)
Bài 1.
a) x( 8x - 2 ) - 8x2 + 12 = 0
<=> 8x2 - 2x - 8x2 + 12 = 0
<=> 12 - 2x = 0
<=> 2x = 12
<=> x = 6
b) x( 4x - 5 ) - ( 2x + 1 )2 = 0
<=> 4x2 - 5x - ( 4x2 + 4x + 1 ) = 0
<=> 4x2 - 5x - 4x2 - 4x - 1 = 0
<=> -9x - 1 = 0
<=> -9x = 1
<=> x = -1/9
c) ( 5 - 2x )( 2x + 7 ) = ( 2x - 5 )( 2x + 5 )
<=> -4x2 - 4x + 35 = 4x2 - 25
<=> -4x2 - 4x + 35 - 4x2 + 25 = 0
<=> -8x2 - 4x + 60 = 0
<=> -8x2 + 20x - 24x + 60 = 0
<=> -4x( 2x - 5 ) - 12( 2x - 5 ) = 0
<=> ( 2x - 5 )( -4x - 12 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\-4x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
d) 64x2 - 49 = 0
<=> ( 8x )2 - 72 = 0
<=> ( 8x - 7 )( 8x + 7 ) = 0
<=> \(\orbr{\begin{cases}8x-7=0\\8x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{7}{8}\end{cases}}\)
e) ( x2 + 6x + 9 )( x2 + 8x + 7 ) = 0
<=> ( x + 3 )2( x2 + x + 7x + 7 ) = 0
<=> ( x + 3 )2 [ x( x + 1 ) + 7( x + 1 ) ] = 0
<=> ( x + 3 )2( x + 1 )( x + 7 ) = 0
<=> x = -3 hoặc x = -1 hoặc x = -7
g) ( x2 + 1 )( x2 - 8x + 7 ) = 0
Vì x2 + 1 ≥ 1 > 0 với mọi x
=> x2 - 8x + 7 = 0
=> x2 - x - 7x + 7 = 0
=> x( x - 1 ) - 7( x - 1 ) = 0
=> ( x - 1 )( x - 7 ) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
Bài 2.
a) ( x - 1 )2 - ( x - 2 )( x + 2 )
= x2 - 2x + 1 - ( x2 - 4 )
= x2 - 2x + 1 - x2 + 4
= -2x + 5
b) ( 3x + 5 )2 + ( 26x + 10 )( 2 - 3x ) + ( 2 - 3x )2
= 9x2 + 30x + 25 - 78x2 + 22x + 20 + 9x2 - 12x + 4
= ( 9x2 - 78x2 + 9x2 ) + ( 30x + 22x - 12x ) + ( 25 + 20 + 4 )
= -60x2 + 40x2 + 49
d) ( x + y )2 - ( x + y - 2 )2
= [ x + y - ( x + y - 2 ) ][ x + y + ( x + y - 2 ) ]
= ( x + y - x - y + 2 )( x + y + x + y - 2 )
= 2( 2x + 2y - 2 )
= 4x + 4y - 4
Bài 3.
A = 3x2 + 18x + 33
= 3( x2 + 6x + 9 ) + 6
= 3( x + 3 )2 + 6 ≥ 6 ∀ x
Đẳng thức xảy ra <=> x + 3 = 0 => x = -3
=> MinA = 6 <=> x = -3
B = x2 - 6x + 10 + y2
= ( x2 - 6x + 9 ) + y2 + 1
= ( x - 3 )2 + y2 + 1 ≥ 1 ∀ x,y
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)
=> MinB = 1 <=> x = 3 ; y = 0
C = ( 2x - 1 )2 + ( x + 2 )2
= 4x2 - 4x + 1 + x2 + 4x + 4
= 5x2 + 5 ≥ 5 ∀ x
Đẳng thức xảy ra <=> 5x2 = 0 => x = 0
=> MinC = 5 <=> x = 0
D = -2/7x2 - 8x + 7 ( sửa thành tìm Max )
Để D đạt GTLN => 7x2 - 8x + 7 đạt GTNN
7x2 - 8x + 7
= 7( x2 - 8/7x + 16/49 ) + 33/7
= 7( x - 4/7 )2 + 33/7 ≥ 33/7 ∀ x
Đẳng thức xảy ra <=> x - 4/7 = 0 => x = 4/7
=> MaxC = \(\frac{-2}{\frac{33}{7}}=-\frac{14}{33}\)<=> x = 4/7
Bài 2:
a) \(x^2+y^2-9-2xy\)
\(=\left(x^2-2xy+y^2\right)-3^2\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
b) \(4x^2-5x-9\)
\(=4x^2+4x-9x-9\)
\(=4x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x+1\right)\left(4x-9\right)\)
\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)
\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)
\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)
\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)
\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)
\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)
Bài 1:
a, x2-3xy-10y2
=x2+2xy-5xy-10y2
=(x2+2xy)-(5xy+10y2)
=x(x+2y)-5y(x+2y)
=(x+2y)(x-5y)
b, 2x2-5x-7
=2x2+2x-7x-7
=(2x2+2x)-(7x+7)
=2x(x+1)-7(x+1)
=(x+1)(2x-7)
Bài 2:
a, x(x-2)-x+2=0
<=>x(x-2)-(x-2)=0
<=>(x-2)(x-1)=0
<=>\(\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
b, x2(x2+1)-x2-1=0
<=>x2(x2+1)-(x2+1)=0
<=>(x2+1)(x2-1)=0
<=>x2+1=0 hoặc x2-1=0
1, x2+1=0 2, x2-1=0
<=>x2= -1(loại) <=>x2=1
<=>x=1 hoặc x= -1
c, 5x(x-3)2-5(x-1)3+15(x+2)(x-2)=5
<=>5x(x-3)2-5(x-1)3+15(x2-4)=5
<=>5x(x2-6x+9)-5(x3-3x2+3x-1)+15x2-60=5
<=>5x3-30x2+45x-5x3+15x2-15x+5+15x2-60=5
<=>30x-55=5
<=>30x=55+5
<=>30x=60
<=>x=2
d, (x+2)(3-4x)=x2+4x+4
<=>(x+2)(3-4x)=(x+2)2
<=>(x+2)(3-4x)-(x+2)2=0
<=>(x+2)(3-4x-x-2)=0
<=>(x+2)(1-5x)=0
<=>\(\orbr{\begin{cases}x+2=0\\1-5x=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\-5x=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{-1}{-5}\end{cases}}\)<=>\(\orbr{\begin{cases}x=-2\\x=\frac{1}{5}\end{cases}}\)
Bài 3:
a, Sắp xếp lại: x3+4x2-5x-20
Thực hiện phép chia ta được kết quả là x2-5 dư 0
b, Sau khi thực hiện phép chia ta được :
Để đa thức x3-3x2+5x+a chia hết cho đa thức x-3 thì a+15=0
=>a= -15
Câu a :
\(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
Câu b :
\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
Câu c :
\(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left(3x+15\right)^2-\left(x-7\right)^2\)
\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)
\(=\left(2x+22\right)\left(4x+8\right)\)
\(=8\left(x+11\right)\left(x+2\right)\)