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A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bài 1.
a) -2x( -3x + 2 ) - ( x + 2 )2
= 6x2 - 4x - ( x2 + 4x + 4 )
= 6x2 - 4x - x2 - 4x - 4
= 5x2 - 8x - 4
b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2
= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1
= 4
d) x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
Bài 2.
a) 4x2 - 4xy + y2 = ( 2x - y )2
b) 9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )
c) x3 + 2 + 3( x3 - 2 )
= x3 + 2 + 3x3 - 6
= 4x3 - 4
= 4( x3 - 1 )
= 4( x - 1 )( x2 + x + 1 )
Bài 3.
2( x - 2 ) = x2 - 4x + 4
⇔ ( x - 2 )2 - 2( x - 2 ) = 0
⇔ ( x - 2 )( x - 2 - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ x = 2 hoặc x = 4
câu 1:
a,x2+2x-4z2+1
=x2+2x.1+12-(2z)2
=(x+1)2-(2z)2
=(x+1-2z)(x+1+2z)
a. y4 - 14y2 + 49
Gọi y2 là t, ta có:
t2 - 14t + 49
<=> t2 - 14t + 72
<=> (t - 7)2
Thay x2 = t
<=> (x2 - 7)2
b. \(\dfrac{1}{4}-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
c. x4 - 16
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
d. x2 - 9
<=> x2 - 32
<=> (x - 3)(x + 3)
Bài 1:
a: \(y^2-14y^2+49=\left(y-7\right)^2\)
b: \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
c: \(x^4-16=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)
d: \(x^2-9=\left(x-3\right)\left(x+3\right)\)
a, y4-14y2+49= (y2)2- 2.y2.7+ 72= (y2- 7)2
b, \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}\right)^2-x^2=\left(\dfrac{1}{2}-x\right).\left(\dfrac{1}{2}+x\right)\)
c, x4- 16= (x2)2 - 42= (x2-4).(x2+4)
Bài 2
a, \(4x^2-2x+\dfrac{1}{4}=\left(2x\right)^2-2.2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(2x-\dfrac{1}{2}\right)^2\)
+) x=3 ta có: \(\left(2.3-\dfrac{1}{2}\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\)
+) x= 19 ta có: \(\left(2.19-\dfrac{1}{2}\right)^2=\left(\dfrac{75}{2}\right)^2=\dfrac{5625}{4}\)
b, \(25a^2+\dfrac{10}{3}a+\dfrac{1}{9}=\left(5a\right)^2+2.5a.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2=\left(5a+\dfrac{1}{3}\right)^2\)
+) a=9 ta có: \(\left(5.9+\dfrac{1}{3}\right)^2=\left(\dfrac{136}{3}\right)^2\)
+) a=2009 ta có: \(\left(5.2009+\dfrac{1}{3}\right)^2=\left(\dfrac{30136}{3}\right)^2\)
Hoctot
tick cho mik zới
Bài 2:
a. 4x2 - 2x + \(\dfrac{1}{4}\)
<=> (2x)2 - 2 . \(\dfrac{1}{2}\) . 2x + \(\left(\dfrac{1}{2}\right)^2\)
<=> \(\left(2x-\dfrac{1}{2}\right)^2\)
Thay x = 3 vào PT:
\(\left(6-\dfrac{1}{2}\right)^2\)
= 5,52
= 30,25
Thay x = 19 vào PT:
\(\left(38-\dfrac{1}{2}\right)^2\)
= 37,52
= \(\dfrac{5625}{4}\)
b. 25a2 + \(\dfrac{10}{3a}\) + \(\dfrac{1}{9}\)
<=> \(\left(5a+\dfrac{1}{3}\right)^2\)
Thay a = 9 vào PT:
\(\left(45+\dfrac{1}{3}\right)^2\)
= \(\left(\dfrac{136}{3}\right)^2\)
= \(\dfrac{18496}{9}\)
Thay x = 2009 vào PT:
\(\left(10045+\dfrac{1}{3}\right)^2\)
= \(\left(\dfrac{30136}{3}\right)^2\)
= 100908721,8
c. 4x2 + y2 - 4x
<=> 4x2 - 4x + y2
Mik cảm thấy đề câu C có phần sai sai, bạn xem lại rồi nhắc mik
câu c bài 2 mk bị nhầm
sửa lai là: 4x^2-4x+y^2 (2x-y=4)
Câu c: 4xy đâu ra vậy bn, là 4x chứ
Khác gì đâu bn
Câu c đề sai nên mình đã sửa lại thành 4xy đó bạn. Nghe vậy có vẻ hợp lý hơn...
ok bn