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a) \(3x^2-3y^2-12x+12y\)
\(=\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3\left(x^2-y^2\right)-12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-3y-12\right)\)
\(=\left(x-y\right).3.\left(x-y-4\right)\)
b) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
c) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=\left(x^4-x^2\right)-\left(4x^2-4\right)\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
a) 2x3 + 8x2 - 8x
= 2x(x2 + 4x - 4)
= 2x(x2 + 4x + 4 - 8)
= 2x[(x + 2)2 - 8]
= \(2x\left(x+2-\sqrt{8}\right)\left(x+2+\sqrt{8}\right)\)
b) a2 - b2 + 4a + 4b
= (a - b)(a + b) + 4(a + b)
= (a + b)(a - b + 4)
c) x2 - 2x - 3
= x2 + x - 3x - 3
= x(x + 1) - 3(x + 1)
= (x + 1)(x - 3)
d) x2 - 4x - 3
= x2 - 4x + 4 - 7
= (x + 2)2 - 7
= \(\left(x+2-\sqrt{7}\right)\left(x+2+\sqrt{7}\right)\)
giải
a.(2x-3)(4x^2+6x+9)-2x(4x^2-1)
=8x^3+12x^2+18x-12x^2-18x-27-8x^3+2x
=2x-27
bài 1
b.(x+y)2+2(x+y)(x-y)+(x-y)2
= [(x+y)+(x-y)]2
= (x+y-x+y)2
= (2y)2
= 4y2
bài 2
a. 2x2y+4xy+2y
=2y(x2+2x+1)
=2y(x+1)2
b.9x2+6xy-4z2+y2
= (9x2+6xy+y2)-4z2
= (3x+y)2-(2z)2
= (3x+y-2z)(3x+y+2z)
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)\(x^3-x^2-x+1=\left(x^3-x\right)-\left(x^2-1\right)=x\left(x^2-1\right)-\left(x^2-1\right)=\left(x-1\right)^2.\left(x+1\right)\)
b)\(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)=\left(x+2\right)\left(x-2\right)\left(x+1\right)\)
c)\(a^5+27a^2=a^2\left(a^3+27\right)=a^2\left(a+3\right)\left(a^2-3a+9\right)\)
d)\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
e)\(x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)
f)\(2x^4-32=2\left(x^4-16\right)=2\left(x^2+4\right)\left(x^2-4\right)=2\left(x^2+4\right)\left(x+2\right)\left(x-2\right)\)
a) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x^2-1\right)\left(x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x+1\right)\)
b) \(x^3+x^2-4x-4\)
\(=x^2\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x^2-4\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)
c) \(a^5+27a^2=a^2\left(a^3+27\right)\)
\(=a^2\left(a+3\right)\left(a^2-3a+9\right)\)
d) \(x^4-8x=x\left(x^3-8\right)\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
e) \(x^4-4x^3+4x^2\)
\(=\left(x^2\right)^2-2\cdot x^2\cdot2x+\left(2x\right)^2\)
\(=\left(x^2+2x\right)^2\)\(=\left[x\left(x+2\right)\right]^2=x^2\left(x+2\right)^2\)
f) \(2x^4-32=2\left(x^4-16\right)\)
\(=2\left(x^2-4\right)\left(x^2+4\right)\)
\(=2\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)
Bài 1:
a) 25\(x^2\) - 0,09
= \(\left(5x\right)^2-0,3^2\)
= (5x - 0,3) (5x +0,3)
Bài 5:
a: \(=\left(2x-3\right)^2\)
b: \(=\left(2x+1\right)^2\)
c: \(=\left(6x+1\right)^2\)
d: \(=\left(3x-4y\right)^2\)
e: \(=\left(\dfrac{1}{2}x-2y\right)^2\)
f: \(=-\left(x-5\right)^2\)
3)
\(A=\dfrac{5}{x^2-2x+5}\)
ta có x2-2x+5
=x2-2x+1+4
=(x2-2x+1)+4
=(x-1)2+4
=> A=\(\dfrac{5}{\left(x-1\right)^2+4}\)
do \(\left(x-1\right)^2\ge0\forall x\)
=> \(\left(x-1\right)^2+4\ge4\)
=> \(\dfrac{5}{\left(x-1\right)^2+4}\le\dfrac{5}{4}\)
=> A\(\le\dfrac{5}{4}\)
GTLN của A =\(\dfrac{5}{4}\)
khi x-1=0
=> x=1
vậy GTLN của A=\(\dfrac{5}{4}\) khi x=1
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
bằng phương pháp nào zậy bn????
547675675675678768768789980957457346242645657
a, 2x(x+3) +(3+x) = (x + 3) (2x + 1)
b. 4xn+2 + 8xn ( n thuộc N*)
= 4xn (x2 + 2)
c. ( x - 1)2 + y(x - 1) = (x - 1) (x - 1 + y)
d. 4x(x - 2) - (2 - x)2 = 4x(x - 2) - (x - 2)2 = (x - 2) (4x - x + 2) = (x - 2) (3x + 2)
e. (x - 2)2 - (2 - x)3 = (2 - x)2 - (2 - x)3 = (2 - x)2 (1 - 2 + x) = (2 - x)2 (x - 1)