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a) Ta có: 10(x-y)-8y(y-x)
\(=10\left(x-y\right)+8y\left(x-y\right)\)
\(=2\left(x-y\right)\left(5+4y\right)\)
d) Ta có: \(x^2y-x^3-9y+9x\)
\(=x^2\left(y-x\right)-9\left(y-x\right)\)
\(=\left(y-x\right)\left(x^2-9\right)\)
\(=\left(y-x\right)\left(x-3\right)\left(x+3\right)\)
e) Ta có: \(2x+2y-x^2-xy\)
\(=2\left(x+y\right)-x\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x\right)\)
f) Ta có: \(x^2-25+y^2+2xy\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
g) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
h) Ta có: \(x^2\left(x-1\right)+16\left(1-x\right)\)
\(=x^2\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-16\right)\)
\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
9) \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=b^2\left[a^2+2ab+b^2+a\left(a-b\right)+b\left(a-b\right)+a^2-2ab+b^2\right]\)
\(=b^2\left(a^2+2ab+b^2+a^2-ab+ab-b^2+a^2-2ab+b^2\right)\)
\(=b^2\left(3a^2+b^2\right)\)
10) \(\left(6x-1\right)^2-\left(3x+2\right)^2\)
\(=\left(6x-1-3x-2\right)\left(6x-1+3x+2\right)\)
\(=\left(3x-3\right)\left(9x+1\right)\)
11) \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
12) \(\left(x^2-25\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-25-x+5\right)\left(x^2-25+x-5\right)\)
\(=\left(x^2-x-20\right)\left(x^2-30+x\right)\)
13) \(x^6-x^4+2x^3+2x^2\)
\(=x^6-x^4+2x^3+2x^2-1+1\)
\(=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left[\left(x^3\right)^2+2x^3.1+1^2\right]-\left[\left(x^2\right)^2-2x^2.1+1^2\right]\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2\)
\(=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)\)
\(=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
1) \(\left(x+y\right)^2-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y-5\right)\left(x+y+5\right)\)
2) \(100-\left(3x-y\right)^2\)
\(=10^2-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
3) \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
4) \(4a^2b^4-c^4d^2\)
\(=\left(2ab^2\right)^2-\left(c^2d\right)^2\)
\(=\left(2ab^2-c^2d\right)\left(2ab^2+c^2d\right)\)
5) Đề đúng ko vậy ạ?
6) \(16x^3+54y^3\)
\(=2\left(8x^3+27y^3\right)\)
\(=2\left[\left(2x\right)^3+\left(3y\right)^3\right]\)
\(=2\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]\)
\(=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
7) \(8x^3-y^3\)
\(=\left(2x\right)^3-y^3\)
\(=\left(2x-y\right)\left[\left(2x\right)^2+2xy+y^2\right]\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
8) \(\left(a+b\right)^2-\left(2ab-b\right)^2\)
\(=\left(a+b-2ab+b\right)\left(a+b+2ab-b\right)\)
\(=\left(a+2b-2ab\right)\left(a+2ab\right)\)
a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1
=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1
=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)
=(x2+x+1)(x5-x4+x3-x+1)
b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)
=(2x2-6x+1)(2x2+6x+1)
c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)
d)3(x4+x2+1)-(x2+x+1)
=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2
=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)
=(x2+x+1)(3x2-3x+2)
e)bạn tự làm nhé
a) x2 - 7x + 5 = ( x2 - 2 . 7/2 . x + 49 / 4 ) + 5 - 49 / 4
= (x - 7/2)^2 - 29/4
= (x - 7/2)^2 - (√ 29 / 2 )^2
= ( x - ( 7 + √ 29 / 2 )). ( x + ( 7 - √ 29 / 2 ))
a) Ta có: \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
b) Ta có: \(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
c) Ta có: \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
d) Ta có: \(2x^2+3x-5\)
\(=2x^2+5x-2x-5\)
\(=x\left(2x+5\right)-\left(2x+5\right)\)
\(=\left(2x+5\right)\left(x-1\right)\)
e) Ta có: \(x^3-3x^2+1-3x\)
\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
f) Ta có: \(x^2-4x-5\)
\(=x^2-4x+4-9\)
\(=\left(x-2\right)^2-3^2\)
\(=\left(x-2-3\right)\left(x-2+3\right)\)
\(=\left(x-5\right)\left(x+1\right)\)
g) Ta có: \(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)
h) Ta có: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
k) Ta có: \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
m) Ta có: \(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
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
a) \(x^4-4x^3+8x^2-16x+16\)
\(=x^4-2x^3-2x^3+4x^2+4x^2-8x-8x+16\)
\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-2\right)\left(x^2+4\right)\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
b) \(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2-4c^2\right)\)
\(=3\left[\left(a-b\right)^2-\left(2c\right)^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
c/ \(a^2+2ab+b^2-ac-bc\)
\(=\left(a+b\right)^2-c\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-c\right)\)
d/ \(ac-bc-a^2+2ab-b^2\)
\(=c\left(a-b\right)-\left(a^2-2ab+b^2\right)\)
\(=c\left(a-b\right)-\left(a-b\right)^2\)
\(=\left(a-b\right)\left(c-a+b\right)\)
e/ \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
\(=\left(x-y+5\right)\left(x-y+5-2\right)+1\)
\(=\left(x-y+5\right)\left(x-y+3\right)+1\)
f/ \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)