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7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)
9, ĐK x >= 0
\(x-2\sqrt{x}-3=x-3\sqrt{x}+\sqrt{x}-3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
10, \(-4x^2-4x+10=-\left(4x^2+4x+1\right)+11\)
\(=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)
11;12 xem lại đề
13, \(-x^3+6xy^2-12xy^2+8y^3=-\left(x^3-6xy^2+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
Trả lời:
7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)
9, \(x-2\sqrt{x}-3\left(ĐK:x\ge0\right)\)
\(=x-3\sqrt{x}+\sqrt{x}-3=\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
10, \(10-4x-4x^2=-\left(4x^2+4x-10\right)=-\left(4x^2+4x+1-11\right)=-\left[\left(2x+1\right)^2-11\right]\)
\(=-\left(2x+1\right)^2+11=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)
11,sửa đề: \(15x\left(x-3y\right)+20y\left(3y-x\right)=15x\left(x-3y\right)-20y\left(x-3y\right)=5\left(x-3y\right)\left(3x-4y\right)\)
12, \(25x^2-2=\left(5x-\sqrt{2}\right)\left(5x+\sqrt{2}\right)\)
13, sửa đề: \(-x^3+6x^2y-12xy^2+8y^3=-\left(x^3-6x^2y+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
a,(2x-y)2+(2x+y)2=(2x2-2*2xy+y2)+(2x2+2*2xy+y2)
=2x2-4xy+y2+2x2+4xy+y2
=4x2+2y2
Bài 1 :
a) 3x2 . ( 5x2 - 7x + 4 ) = 15x4 - 21x3 + 12x2
b) xy2 . ( 2x2y - 5xy + y ) = 2x3y3 - 5x2y3 + xy3
c) ( 2x2 - 5x ) . ( 3x2 - 2x + 1 ) = 6x4 - 4x3 + 2x2 - 15x3 + 10x2 - 5x
= 6x4 - 19x3 + 12x2 - 5x
d) ( x - 3y ) . ( 2xy + y2 + x ) = 2x2y + xy2 + x2 - 6xy2 - 3y3 - 3xy
Bài 2 :
a) A = x2 + 9y2 - 6xy
=> A = x2 - 2 . x . 3y + ( 3y )2
=> A = ( x - 3y )2
Thay x = 19 và y = 13 vào biểu thức A ta có :
A = ( 19 - 3 . 13 )2
=> A = ( 19 - 39 )2
=> A = ( -20 )2
=> A = 400
b) B = x3 - 6x2y + 12xy2 - 8y3
=> B = ( x - 2y )3
Thay x = 12 và y = -4 vào biểu thức B ta có :
B = [ 12 - 2 . ( -4 ) ]3
=> B = ( 12 + 8 )3
=> B = 203
=> B = 8000
= -3y3 + 2x2y - 5xy2 + x2 - 3xy
ta có:
a) (x2 - 3x + xy - 3y) : (x + y)
= [x(x - 3) + y(x - 3)] : (x + y)
= (x + y)(x - 3) : (x + y)
= x - 3
b) (x2 - y2 + 6x + 9) : (x + y + 3)
= [(x2 + 6x + 9) - y2] : (x + y + 3)
= [(x + 3)2 - y2] : (x + y + 3)
= (x + y + 3)(x - y + 3) : (x + y + 3)
= x - y + 3
1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)
2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)
4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
5, \(2x^3+3x^2+2x+3\)
\(=x^2\left(2x+3\right)+2x+3\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
6, \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xy^2\)
\(=xz\left(x^2-xz\right)+xz\left(xy-yz\right)\)
\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)
\(=xz\left(x+y\right)\left(x-z\right)\)
8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)
9, \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-2\right)\left(x-4\right)\)
10, \(x^2-8x+12\)
\(=x^2-6x-2x+12\)
\(=x\left(x-6\right)-2\left(x-6\right)\)
\(=\left(x-2\right)\left(x-6\right)\)
Chỗ còn lại mai làm nốt nha.
Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha
11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)
\(=ab\left(a-b\right)-ac\left(a-b\right)-bc\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
12, \(x^3-7x-6\)
\(=x^3-3x^2+3x^2-9x+2x-6\)
\(=x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)\)
\(=\left(x-3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
13, \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
14, \(a^4+64\)
\(=a^4+16a^2+64-16a^2\)
\(=\left(a^2+8\right)^2-16a^2\)
\(=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
15, \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+x^2+x+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)
16, \(x^5+x-1\)
\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)
\(=x^3\left(x^2-x+1\right)-x^2\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x^2-1\right)\)
17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)
19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)
Đặt \(x^2+8x+7=a\) ta có:
(*) \(\Leftrightarrow a\left(a+8\right)+15\)
\(\Leftrightarrow a^2+8a+15\)
\(\Leftrightarrow a^2+3a+5a+15\)
\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)
\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)
Đặt \(x^2+3x+1=a\) ta có:
(*) \(\Leftrightarrow a\left(a+1\right)-6\)
\(\Leftrightarrow a^2+a-6\)
\(\Leftrightarrow a^2+3a-2a-6\)
\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)
Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)
Bài 2: Ta có :\(x^3-6x^2y+12xy^2-8y^3=-8\)
\(\Leftrightarrow\left(x-2y\right)^3=\left(-2\right)^3\)
\(\Rightarrow x-2y=-2\) (*)
\(3x^2-12xy+12y^2=3.\left(x^2-4xy+4y^2\right)=3.\left(x-2y\right)^2\)
Thay (*) vào bt ta được: \(3.\left(-2\right)^2=12\)
Bài 3: Ta có: a+b=13
=> (a+b)3=2197
<=> a3 + b3 + 3ab.(a+b)=2197
<=> a3 + b3 +3.9.13=2197
=> a3 + b3 =1846