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\(B=7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
\(E=x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
\(F=x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
\(H=8x^2-2x-1\)
\(=8x^2-4x+2x-1\)
\(=4x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
a. \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x+2\right)\left(x-2\right)\)
b. \(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x+y-2\right)\left(x-y-2\right)\)
c. \(\left(x^2+9\right)^2-36x^2=\left(x^2+6x+9\right)\left(x^2-6x+9\right)=\left(x+3\right)^2\left(x-3\right)^2\)
d. \(25-x^2+2xy-y^2=25-\left(x-y\right)^2=\left(5+x-y\right)\left(5-x+y\right)\)
còn lại làm tương tự
a) \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
b) \(x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
c) \(\left(x^2+9\right)^2-36x^2=\left(x^2+9\right)^2-\left(6x\right)^2=\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)
\(=\left(x-3\right)^2\left(x+3\right)^2\)
d) \(25-x^2+2xy-y^2=5^2-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)
e) \(x^3-4x^2+4x-1=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1-4x\right)=\left(x-1\right)\left(x^2-3x+1\right)\)
f) \(3x-3y-x^2+2xy-y^2=3\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3-x+y\right)\)
g) \(2x^2-9x+10=2x^2-4x-5x+10=2x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(2x-5\right)\)
h) \(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i) \(x^3-3x^2+2=x^3-2x^2-x^2+2=x^2\left(x-1\right)-2\left(x^2-1\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x-2x-2\right)\)
k) \(x^4+4=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
nhiều quá bạn ạ
hay bạn tìm hiểu cách thức chung làm dạng bài tìm GTNN chứ như thế này thì làm lâu lắm
mik chỉ tìm hiểu đc đến câu I còn lại mik k hiểu lắm, bn có lm đc k, giúp mik vs
e) \(\left(x-y\right)^2+4\left(x-y\right)-12\)
\(=\left(x-y\right)\left[\left(x-y\right)+4\right]-12\)
\(=\left(x-y\right)\left(x-y+4\right)-12\)
f) \(x^2-2xy+y^2+3x-3y-10\)
\(=\left(x^2-2xy+y^2\right)+\left(3x-3y\right)-10\)
\(=\left(x-y\right)^2+3\left(x-y\right)-10\)
\(=\left(x-y\right)\left[\left(x-y\right)+3\right]-10\)
\(=\left(x-y\right)\left(x-y+3\right)-10\)
g) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left[\left(x^2+4x+8\right)+3x\right]+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+4x+8+3x\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+7x+8\right)+2x^2\)
1,4x2.(5x3+2x-1)
=4x2.5x3+4x2.2x-4x2.1
20x5+8x3-4x2
2,4x3y2:x2
=4xy2
3,(15x2y3-10x3y3+6xy):5xy
15x2y3:5xy-10x3y3:5xy+6xy:5xy
3xy2-2x2y2+\(\dfrac{6}{5}\)
1: \(=20x^5+8x^3-4x^2\)
2: \(=4xy^2\)
3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)
4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)
6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)
7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)
8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)
9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=4x^2-2x+\dfrac{1}{4}\)
10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)
\(=\dfrac{x^2-7}{2\left(x-1\right)}\)
12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)
15:=x^3-y^3+2
1: \(=20x^5+8x^3-4x^2\)
2: \(=4xy^2\)
3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)
4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)
6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)
7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)
8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)
9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=4x^2-2x+\dfrac{1}{4}\)
10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)
\(=\dfrac{x^2-7}{2\left(x-1\right)}\)
12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)
15:=x^3-y^3+2
Đề là phân tích đa thức thành nhân tử nha các bạn.
a) Ta có: \(\left(x+y\right)^2-8\left(x+y\right)+12\)
\(=\left[\left(x+y\right)^2-8\left(x+y\right)+16\right]-4\)
\(=\left(x+y-4\right)^2-4\)
\(=\left(x+y\right)\left(x+y-8\right)\)
lm hết đc k bạn ơi ????
b. ( x2 + 2x )2 - 2x2 - 4x - 3
= x4 + 4x3 + 4x2 - 2x2 - 4x - 3
= x4 + 4x3 + 2x2 - 4x - 3
= ( x4 - 2x3 - 3x2 ) + ( 2x3 - 4x2 - 6x ) + ( x2 - 2x - 3 )
= x2 ( x2 - 2x - 3 ) + 2x ( x2 - 2x - 3 ) + ( x2 - 2x - 3 )
= ( x2 + 2x + 1 ) ( x2 - 2x - 3 )
= ( x + 1 )2 [ ( x2 + 3x ) - ( x + 3 ) ]
= ( x + 1 )2 [ x ( x + 3 ) - ( x + 3 ) ]
= ( x + 1 )2 ( x - 1 ) ( x + 3 )
Xin lỗi vì giờ mình mới làm được câu này ;-;
Mình không viết lại đề bài nhé
a) Đặt t = x + y
pt <=> t2 - 8t + 12
= t2 - 2t - 6t + 12
= t( t - 2 ) - 6( t - 2 )
= ( t - 2 )( t - 6 )
= ( x + y - 2 )( x + y - 6 )
b) = ( x2 + 2x )2 - 2( x2 + 2x ) - 3
Đặt t = x2 + 2x
pt <=> t2 - 2t - 3
= t2 + t - 3t - 3
= t( t + 1 ) - 3( t + 1 )
= ( t + 1 )( t - 3 )
= ( x2 + 2x + 1 )( x2 + 2x - 3 )
= ( x + 1 )2( x2 - x + 3x - 3 )
= ( x + 1 )2[ x( x - 1 ) + 3( x - 1 ) ]
= ( x + 1 )2( x - 1 )( x + 3 )
c) = ( x2 + x )2 + 4( x2 + x ) - 12
Đặt t = x2 + x
pt <=> t2 + 4t - 12
= t2 - 2t + 6t - 12
= t( t - 2 ) + 6( t - 2 )
= ( t - 2 )( t + 6 )
= ( x2 + x - 2 )( x2 + x + 6 )
= ( x2 - x + 2x - 2 )( x2 + x + 6 )
= [ x( x - 1 ) + 2( x - 1 ) ]( x2 + x + 6 )
= ( x - 1 )( x + 2 )( x2 + x + 6 )
d) Đặt t = x2 + x
pt <=> t2 - 2t - 15
= t2 + 3t - 5t - 15
= t( t + 3 ) - 5( t + 3 )
= ( t + 3 )( t - 5 )
= ( x2 + x + 3 )( x2 + x - 5 )
e) = ( x2 + 2x )2 + 9( x2 + 2x ) + 15
Đặt t = x2 + 2x
pt <=> t2 + 9t + 20
= t2 + 4t + 5t + 20
= t( t + 4 ) + 5( t + 4 )
= ( t + 4 )( t + 5 )
= ( x2 + 2x + 4 )( x2 + 2x + 5 )
f) = ( x2 - 2xy + y2 ) + ( 3x - 3y ) - 10
= ( x - y )2 + 3( x - y ) - 10
Đặt t = x - y
pt <=> t2 + 3t - 10
= t2 - 2t + 5t - 10
= t( t - 2 ) + 5( t - 2 )
= ( t - 2 )( t + 5 )
= ( x - y - 2 )( x - y + 5 )
h) ( x2 +x + 1 ) + 2x( x2 + x + 1 ) + x2
= x2 + x + 1 + 2x3 + 2x2 + 2x + x2
= 2x3 + 4x2 + 3x + 1
= 2x3 + 2x2 + 2x2 + 2x + x + 1
= ( 2x3 + 2x2 ) + ( 2x2 + 2x ) + ( x + 1 )
= 2x2( x + 1 ) + 2x( x + 1 ) + ( x + 1 )
= ( x + 1 )( 2x2 + 2x + 1 )
g) < mình làm riêng vì mình sửa lại thành hai trường hợp >
1. ( x2 - 4x )2 - 2( x2 - 4x ) - 15
Đặt t = x2 - 4x
pt <=> t2 - 2t - 15
= t2 + 3t - 5t - 15
= t( t + 3 ) - 5( t + 3 )
= ( t + 3 )( t - 5 )
= ( x2 - 4x + 3 )( x2 - 4x - 5 )
= ( x2 - x - 3x + 3 )( x2 + x - 5x - 5 )
= [ x( x - 1 ) - 3( x - 1 ) ][ x( x + 1 ) - 5( x + 1 )
= ( x - 1 )( x - 3 )( x + 1 )( x - 5 )
2. ( x2 + 4x )2 - 2( x2 + 4x ) - 15
Đặt t = x2 + 4x
pt <=> t2 - 2t - 15
= t2 + 3t - 5t - 15
= t( t + 3 ) - 5( t + 3 )
= ( t + 3 )( t - 5 )
= ( x2 + 4x + 3 )( x2 + 4x - 5 )
= ( x2 + x + 3x + 3 )( x2 - x + 5x - 5 )
= [ x( x + 1 ) + 3( x + 1 ) ][ x( x - 1 ) + 5( x - 1 ) ]
= ( x + 1 )( x + 3 )( x - 1 )( x + 5 )
Đúng cái nào thì bạn xem nhé ;-;