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HM
16 tháng 7 2017
\(^{x^3+2.3x^2y+3.x\left(2y\right)^2+\left(2y\right)^3=\left(x+2y\right)^3}\)
C
29 tháng 6 2017
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: \(3.\left(-2\right)^2=12\)
16 tháng 8 2020
Bài 1: Phân tích đa thức thành nhân tử
a) Ta có: \(3x\left(x-a\right)+5a^2-5ax\)
\(=3x\left(x-a\right)+5a\left(a-x\right)\)
\(=3x\left(x-a\right)-5a\left(x-a\right)\)
\(=\left(x-a\right)\left(3x-5a\right)\)
b) Ta có: \(x^3+8y^3+6x^2y+12xy^2\)
\(=x^3+3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)
\(=\left(x+2y\right)^3\)
c) Ta có: \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7x+7\)
\(=3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left[3x\left(x+1\right)-5x^2+7\right]\)
\(=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)\)
\(=\left(x+1\right)\left(-2x^2+3x+7\right)\)
f) Ta có: \(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=a^2b-ab^2+b^2c-bc^2+c^2a-ca^2\)
\(=abc+a^2b-ab^2+b^2c-bc^2+c^2a-ca^2-abc\)
\(=\left(a^2b-abc\right)-\left(ab^2-b^2c\right)-\left(bc^2-ac^2\right)-\left(a^2c-abc\right)\)
\(=ab\left(a-c\right)-b^2\left(a-c\right)-c^2\left(b-a\right)-ac\left(a-b\right)\)
\(=\left(a-c\right)\left(ab-b^2\right)-c^2\left(b-a\right)+ac\left(b-a\right)\)
\(=b\left(a-c\right)\left(a-b\right)-\left(b-a\right)\left(c^2-ac\right)\)
\(=b\left(a-c\right)\left(a-b\right)+\left(a-b\right)\cdot c\cdot\left(c-a\right)\)
\(=b\left(a-c\right)\left(a-b\right)-c\left(a-b\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
g) Ta có: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
17 tháng 10 2017
$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^{7-5}\)
\(=\left(-x\right)^2\)
\(=x^2\)
$c)$ \(5x^2y^4:10x^2y\)
\(=\dfrac{1}{2}y^3\)
$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^{14-7}\)
\(=\left(-xy\right)^7\)
Các câu còn lại tương tự nha bạn!
5 tháng 8 2018
\(a,x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
\(b,27-8y^3=\left(3-2y\right)\left(9+6y+4y^2\right)\)
\(c,y^6+1=\left(y^2\right)^3+1=\left(y^2+1\right)\left(y^4-y^2+1\right)\)
\(d,64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)
\(e,125x^6-27y^9=\left(5x^2\right)^3-\left(3y^3\right)^3=\left(5x^2-3y^3\right)\left(25x^4+15x^2y^3+9y^9\right)\)
\(g,16x^2\left(4x-y\right)-8y^2\left(x+y\right)+xy\left(16+8y\right)\)
\(=8\left[2x^2\left(4x-y\right)-y^2\left(x+y\right)\right]+8xy\left(2+y\right)\)
\(=8\left(8x^3-2x^2y-xy^2-y^3+2xy+xy^2\right)\)
\(f,-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left[\left(\dfrac{x^2}{5}\right)^3+\dfrac{y^3}{4^3}\right]=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)
13 tháng 7 2019
Duy Nguyễn, Duy Nguyễn, Duy Nguyễn, Duy Nguyễn, Duy nguyễn, Duy Nguyễn, Duy Nguyễn, Duy Nguyễn, Duy Nguyễn
giúp bn ấy vs
12 tháng 8 2021
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, \(-x^3+6xy^2-12xy^2+8y^3=-\left(x^3-6xy^2+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
QA
12 tháng 8 2021
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: \(-x^3+6x^2y-12xy^2+8y^3=-\left(x^3-6x^2y+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
CC
0
NH
1
BK
6 tháng 10 2018
b)5x^2+9y^2-12xy-6x+9=0
=>4x^2-12xy+9y^2+x^2-6x+9=0
=>(2x-3y)^2+(x-3)^2=0
=>2x-3y=0 và x-3=0
=>y=2 và x=3
Áp dụng: \(\left(A-B\right)^3=A^3-3A^2B+3AB^2-B^3\)
\(8y^3-12xy^2+6x^2y-x^3\)
\(=\left(2y\right)^3-3\left(2y\right)^2.x+3.2y.x^2-x^3\)
\(=\left(2y-x\right)^3\)