Bài 1 : Phân tích đa thức thành nhân tử
a) 3x2 -3y2-12x+12x
b) 4x3 + 4xy2 + 8x2y - 16x
c) x4 - 5x2 + 4
phân tích đa thức thành nhân tử
a) xy+y2-x-y
b) 25-x2+4xy-4y2
c) xy+yz-2y-2z
d) x2-6xy+9y2-25z2
e) 3x2-3y2-12x+12y
f) 4x3+4xy2+8x2y-16x
g) x2-5x+4
h) x4+5x2+4
i) 2x2+3x-5
k) x3-2x2+6x-5
l) x2-4x+3
Mong mọi người giúp đỡ em cảm ơn ạ
Bài làm
a) xy + y2 - x - y
= ( xy + y2 ) - ( x + y )
= y( x + y ) - ( x + y )
= ( x + y )( y - 1 )
b) 25 - x2 + 4xy - 4y2
= 25 - ( x2 - 4xy + 4y2 )
= 25 - ( x - 2y )2
= ( 5 - x + 2y )( 5 + x - 2y )
c) xy + xz - 2y - 2z
= ( xy + xz ) - ( 2y + 2z )
= x( y + z ) - 2( y + z )
= ( y + z )( x - 2 )
d) x2 - 6xy + 9y2 - 25z2
= ( x2 - 6xy + 9y2 ) - 25z2
= ( x - 3y )2 - 25z2
= ( x - 3y - 5z )( z - 3y + 5z )
e) 3x2 - 3y2 - 12x + 12y
= 3( x - y )( x + y ) - 12( x - y )
= ( x - y )[ 3( x + y ) - 12 ]
f) 4x3 + 4xy2 + 8x2y - 16x
= 4x( x2 + y2 + 2xy - 4 )
= 4x[ ( x + y)2 - 4 ]
= 4x( x + y - 2 )( x + y + 2 )
g) x2 - 5x + 4
= x2 - x - 4x + 4
= x( x - 1 ) - 4( x - 1 )
= ( x - 1 )( x - 4 )
h) x4 + 5x2 + 4
= x4 + x2 + 4x2 + 4
= x2( x2 + 1 ) + 4( x2 + 1 )
= ( x2 + 1 )( x2 + 4 )
i) 2x2 + 3x - 5
= 2x2 - 5x + 2x - 5
= 2x( x + 1 ) - 5( x + 1 )
= ( x + 1 )( 2x - 5 )
k) x3 - 2x2 + 6x - 5 ( không biết làm )
l) x2 - 4x + 3
= ( x2 - 4x + 4 ) - 1
= ( x - 2 )2 - 1
= ( x - 3 )( x - 1 )
# Học tốt #
Phân tích đa thức thành nhân tử:
a) xy + y2 – x – y
b) 25 – x2 + 4xy – 4y2
c) 4x3 + 4xy2 + 8x2y – 16x
d) (x2 + x)2 + 4(x2 + x) – 12
e) (x + 1) (x + 2) (x + 3) (x + 4) - 24 g)
h) x2 – 5x + 4
i) x4 – 5x2 + 4
j) x3 – 2x2 + 6x – 5
k) x2 – 4x + 3
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
Phân tích đa thức thành nhân tử :
a) 3x2-6xy+3y2-12x2
b) 3x2y2-6x2y3+12x2y2
c) 3x2-3y2+12x-12y
a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)
b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)
c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)
a: \(3x^2-6xy+3y^2-12x^2\)
\(=3\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\left[\left(x-y\right)^2-4x^2\right]\)
\(=3\left(x-y-2x\right)\left(x-y+2x\right)\)
\(=3\left(-x-y\right)\left(3x-y\right)\)
b: \(3x^2y^2-6x^2y^3+12x^2y^2\)
\(=3x^2y^2\left(1-2y+4\right)\)
\(=3x^2y^2\left(-2y+5\right)\)
c: Ta có: \(3x^2-3y^2+12x-12y\)
\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
Phân tích đa thức thành nhân tử :
a.x4 - 4x3 + 11x2 - 16x + 16
b.x4 + 6x3 + 13x2 + 12x + 4
c.x4 + x3 - 4x2 + x + 1
d.x4 + x3 - 4x2 + x + 1
c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
Bài 9: Phân tích đa thức thành nhân tử
1, 5x2 – 10xy + 5y2 – 20z2 2, 16x – 5x2 – 3 3, x2 – 5x + 5y – y2 | 4, 3x2 – 6xy + 3y2 – 12z2 5, x2 + 4x + 3 6, (x2 + 1)2 – 4x2 7, x2 – 4x – 5
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1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)
2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
BÀI TẬP PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ BẰNG PHƯƠNG PHÁP ĐẶT NHÂN TỬ CHUNG
1) xy3 – x3y
2) 15xy + 20x2 – 30x
3)6x – 3xy
4)x3 + 2x2 + x
5)4x3 – 12x2 + 9x
6)2x2y + 4xy2 – 10 x3y2
7)x4 + 2x3 + x2
11)x(x – 1) – y(1 – x)
1, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)
2, \(5x\left(3y+4x-6\right)\)
3, \(3x\left(2-y\right)\)
4, \(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)
6, \(2xy\left(x+2y-5x^2y\right)\)
7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
11, \(\left(x+y\right)\left(x-1\right)\)
\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)
bài 1 phân tích các đa thức sau thành nhân tử
a) x2 + 4x +3 b) 16x - 5x2 - 3 c) 2x2 + 7x + 5
d) 2x2 + 3x -5 e) x3 - 3x2 + 1 - 3x f ) x2 - 4x - 5
g) (a2 + 1 )2 - 4a2 h) x3 - 3x2 - 4x + 12 i) x4 + x3 + x + 1
k) x4 - x3 - x2 + 1 l ) (2x + 1 )2 - ( x - 1 )
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
Phân tích đa thức thành nhân tử
a)6x2-15x+9
b)5x2+12x+7
làm bừa thui,ai tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
trời cái này mà không biết đây là toán lơp 5
bài 1:phân tích đa thức thành nhân tử
a,x4 +5x2 +9
b,x4 + 3x2 +4
c,2x4 - x2 -1
Bài 2:tìm x biết
a,(x+1) (x+2)(x+3)(x+4)= 120
b,(x-4x+3)(x2+6x +8) +24
Bài 1:
\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)
\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)
\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)
Bài 2:
\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)
Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:
\(\left(y-1\right)\left(y+1\right)=120\)
\(\Leftrightarrow y^2-1=120\)
\(\Leftrightarrow y^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)
+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)
+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)
\(\Leftrightarrow x^2+5x+16=0\)
\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)
Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
\(\Rightarrow\) loại
Vậy \(x\in\left\{1;-6\right\}\).
\(b,\) Đề thiếu vế phải rồi bạn.
Phân tích các đa thức sau thành nhân tử
a,x4+2x3+3x2+2x+1
b,x4-4x3+2x2+4x+1
c,x4+x3+2x2+2x+4