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a) ( x2 - 25 )2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) ]2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) - ( x - 5 ) ][ ( x - 5 )( x + 5 ) + ( x - 5 ) ]
= ( x - 5 )( x + 5 - 1 )( x - 5 )( x + 5 + 1 )
= ( x - 5 )2( x + 4 )( x + 6 )
b) ( 4x2 - 25 )2 - 9( 2x - 5 )2
= ( 4x2 - 25 )2 - 32( 2x - 5 )2
= ( 4x2 - 25 )2 - ( 6x - 15 )2
= [ ( 4x2 - 25 ) - ( 6x - 15 ) ][ ( 4x2 - 25 ) + ( 6x - 15 ) ]
= ( 4x2 - 25 - 6x + 15 )( 4x2 - 25 + 6x - 15 )
= ( 4x2 - 6x - 10 )( 4x2 + 6x - 40 )
= ( 4x2 + 4x - 10x - 10 )( 4x2 + 16x - 10x - 40 )
= [ 4x( x + 1 ) - 10( x + 1 ) ][ 4x( x + 4 ) - 10( x + 4 ) ]
= ( x + 1 )( 4x - 10 )( x + 4 )( 4x - 10 )
= ( 4x - 10 )2( x + 1 )( x + 4 )
c) 4( 2x - 3 )2 - 9( 4x2 - 9 )2
= 22( 2x - 3 )2 - 32( 4x2 - 9 )2
= ( 4x - 6 )2 - ( 12x2 - 27 )2
= [ ( 4x - 6 ) - ( 12x2 - 27 ) ][ ( 4x - 6 ) + ( 12x2 - 27 ) ]
= ( 4x - 6 - 12x2 + 27 )( 4x - 6 + 12x2 - 27 )
= ( -12x2 + 4x + 21 )( 12x2 + 4x - 33 )
= ( -12x2 + 18x - 14x + 21 )( 12x2 - 18x + 22x - 33 )
= [ -12x( x - 3/2 ) - 14( x - 3/2 ) ][ 12x( x - 3/2 ) + 22( x - 3/2 ) ]
= ( x - 3/2 )( -12x - 14 )( x - 3/2 )( 12x + 22 )
= ( x - 3/2 )2( -12x - 14 )( 12x + 22 )
d) a6 - a4 + 2a3 + 2a2
= a2( a4 - a2 + 2a + 2 )
= a2( a4 - 2a3 + 2a3 + 2a2 - 4a2 + a2 + 4a - 2a + 2 )
= a2[ ( a4 - 2a3 + 2a2 ) + ( 2a3 - 4a2 + 4a ) + ( a2 - 2a + 2 ) ]
= a2[ a2( a2 - 2a + 2 ) + 2a( a2 - 2a + 2 ) + 1( a2 - 2a + 2 ) ]
= a2( a2 + 2a + 1 )( a2 - 2a + 2 )
= a2( a + 1 )2( a2 - 2a + 2 )
e) ( 3x2 + 3x + 2 )2 - ( 3x2 + 3x - 2 )2
= [ ( 3x2 + 3x + 2 ) - ( 3x2 + 3x - 2 ) ][ ( 3x2 + 3x + 2 ) + ( 3x2 + 3x - 2 ) ]
= ( 3x2 + 3x + 2 - 3x2 - 3x + 2 )( 3x2 + 3x + 2 + 3x2 + 3x - 2 )
= 4( 6x2 + 6x )
= 4.6x( x + 1 )
= 24( x + 1 )
- =>((x-5)(x+5))2-(x-5)2 => (x-5)2(x+5)2-(x-5)2 => (x-5)2 ((x+5)2-1) => (x2+10x+25)(x+6)(x+4)
bn chép lại đề nhé
a/ \(=\left(x+y\right)^2-4x^2y^2=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
b/ \(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=\left[\left(b+c\right)^2-a^2\right]\left[-\left(b+c\right)^2+a^2\right]\)
\(=\left(b+c-a\right)\left(b+c+a\right)^2\left(a-b-c\right)\)
c/ \(=2a^2+2b^2-2c^2+4ab=2\left[\left(a^2+b^2+2ab\right)-c^2\right]\)
\(=2\left(a+b-c\right)\left(a+b+c\right)\)
d/ \(=\left(4x^2-25\right)^2-9\left(4x^2-20x+25\right)\)
\(=\left(4x^2-25\right)^2-9\left(4x^2+25\right)+180x\)
tới đây bạn đặt a= 4x^2 -25 rồi làm típ nha, mình lười quá ><
e/ tương tự câu d nha bạn
f/ \(=a^4\left(a^2-1\right)+2a^2\left(a+1\right)\)
\(=a^4\left(a-1\right)\left(a+1\right)+2a^2\left(a+1\right)\)
\(=a^2\left(a+1\right)\left(a^2+2\right)\)
g/ đặt \(a=3x^2+3x+2\) khi đó biểu thức trở thành
\(a^2-\left(a+4\right)^2=a^2-a^2-8a-16\)
\(=-8a-16=-8\left(3x^2+3x+2-8\right)=-8\left(3x^2+3x-6\right)\)
\(=-24\left(x^2+x-2\right)=-24\left(x-1\right)\left(x+2\right)\)
xong rùi nha bn. Chúc bn hc tốt (xin lỗi tại có mấy câu mình lười nha)
\(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2\)
\(=\left(x-2xy+y\right)\left(x+2xy+y\right)\)
a ) \(x^3+3x^2-3x+1\)
\(=x^3-3x+3x^2-1\)
\(=\left(x-1\right)^3\)
a) 4(2x-3)^2-9(4x^2-9)^2
=[2(2x-3)]^2-[3(4x^2-9)]^2
=(4x-6)^2-(12x^2-27)^2
=(4x-6+12x^2-27)(4x-6-12x^2+27)
=(12x^2+4x-33)(4x-12x^2+21)
b) a^6-a^4+2a^3+2a^2
=a^4(a^2-1)+2a^2(a+1)
=a^4(a+1)(a-1)+2a^2(a+1)
=(a+1)[(a^4)(a-1)+2a^2]
=(a+1)(a^5+a^4+2a^2)
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
Bài 2;
\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)
\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)