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TN
7 tháng 8 2018
a) \(\dfrac{1}{8}x^3y^3-27=\left(\dfrac{1}{2}xy\right)^3-3^3=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy+9\right)\)
b)\(\dfrac{8}{125}x^3+27y^3=\left(\dfrac{2}{5}x\right)^3+\left(3y\right)^3=\left(\dfrac{2}{5}x+3y\right)\left(\dfrac{4}{25}x^2-\dfrac{6}{5}xy+9y^2\right)\)
c) \(0.008x^6-27y^3=\left(0.2x^2\right)^3-\left(3y\right)^3=\left(0.2x^2-3y\right)\left(0.04x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d)\(\left(2x+y\right)^3-\left(x-y\right)^3=\left(2x+y-x+y\right)[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2]\\ =\left(x+2y\right)\left(4x^2+4xy+y^2+2x^2-2xy+xy-y^2+x^2-2xy+y^2\right)\\ =\left(x+2y\right)\left(6x^2+xy+y^2\right)\)
Y
7 tháng 8 2018
Bài 1:
a) \(\dfrac{1}{8}x^3y^3-27\)
\(=\left(\dfrac{1}{2}xy\right)^3-3^3\)
\(=\left(\dfrac{1}{2}xy-3\right)\left[\left(\dfrac{1}{2}xy\right)^2+\dfrac{1}{2}xy.3+3^2\right]\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{1}{4}xy+\dfrac{3}{2}xy+9\right)\)
\(=\left(\dfrac{1}{2}xy-3\right)\left(\dfrac{7}{4}xy+9\right)\)
b) \(\dfrac{8}{125}x^3+\dfrac{1}{8}y^3\)
\(=\left(\dfrac{2}{5}x\right)^3+\left(\dfrac{1}{2}y\right)^3\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left[\left(\dfrac{2}{5}x\right)^2-\dfrac{2}{5}x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)
\(=\left(\dfrac{2}{5}x+\dfrac{1}{2}y\right)\left(\dfrac{4}{25}x-\dfrac{1}{5}xy+\dfrac{1}{4}y\right)\)
c) \(0.008x^6-27y^3\)
\(=\left(\dfrac{1}{5}x^2\right)^3-\left(3y\right)^3\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left[\left(\dfrac{1}{5}x^2\right)^2+\dfrac{1}{5}x^2.3y+\left(3y\right)^2\right]\)
\(=\left(\dfrac{1}{5}x^2-3y\right)\left(\dfrac{1}{25}x^4+\dfrac{3}{5}x^2y+9y^2\right)\)
d) \(\left(2x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(2x+y\right)-\left(x-y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(2x+y-x+y\right)\left(4x^2+4xy+y^2+2x^3-2xy+xy-y^2+x^2-2xy+y^2\right)\)
\(=\left(x-2y\right)\left(4x^2+2x^3+xy\right)\)
19 tháng 10 2022
1: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
3: \(=18\left(m^2-2mn+n^2-4p^2\right)\)
\(=18\left(m-n-2p\right)\left(m-n+2p\right)\)
4: \(=9\left(a^2-2ab+b^2-4c^2\right)\)
\(=9\left(a-b-2c\right)\left(a-b+2c\right)\)
5: \(=\left(x-3y\right)\left(5a-8b\right)\)
6: \(=7\left(x^2-2xy+y^2-z^2\right)\)
\(=7\left(x-y-z\right)\left(x-y+z\right)\)
19 tháng 7 2017
câu 20
\(\)\(C_{20}=\left(a^2+1\right)^2-4a^2=\left(a^2+1\right)^2-\left(2a\right)^2=\left[\left(a^2+1\right)-2a\right]\left[\left(a^2+1\right)+2a\right]\)\(C_{20}=\left[a^2-2a+1\right]\left[a^2+2a+1\right]=\left(a-1\right)\left(a-1\right)\left(a+1\right)\left(a+1\right)\)
\(C_{20}=\left(a-1\right)\left(a-1\right)\left(a+1\right)\left(a+1\right)\)
KN
18 tháng 9 2017
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
H
2
KC
5 tháng 8 2017
Áp dụng hằng đẳng thức ( a - b ) ( a + b ) = a2 - b2 ta đc:
\(\left(5x-3y+8z\right)\left(5x-3y-8z\right)=\left(5x-3y\right)^2-\left(8z\right)^2\)
\(=25x^2-30xy+9y^2-64z^2\)
Đề có sai ko vậy bn
6 tháng 8 2017
mk lấy kq của bạn Kia Cerato mk giải típ
tc \(x^2=y^2+4z^2\Leftrightarrow x^2-y^2=4z^2\)
\(\Leftrightarrow25x^2-30xy+9y^2-16.4z^2\)
\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)
\(=25x^2+9y^2-30xy-16x^2+16y^2\)
\(=9x^2-30xy+25y^2=\left(3x-5y\right)^2\)
ok
TV
8
TN
10 tháng 6 2016
Cách 1:x2-y2-z2=0
=>x2=y2+z2
(5x-3y+4z)(5x-3y-4z)
=(5x-3y)2-16z2
=25x2-30xy+9y2-16z2(*)
Vì x2=y2+z2=>z2=x2-y2 nên (*)=25x2-30xy+9y2-16(x2-y2)=(3x-5y)2
Cách 2: cách này dễ hiểu hơn
x2-y2-z2=0
=>x2=y2+z2
(5x-3y+4z).(5x-3y-4z)=(3x-5y)2
<=>(5x-3y)2-16z2=(3x-5y)2
<=>(5x-3y)2-(3x-5y)2=16z2
<=>(8x-8y)(2x+2y)=16z2
<=>16(x2-y2)=16z2
<=>x2=y2+z2 (đúng với gt)
10 tháng 6 2016
Ta có: (5x-3y+4z)(5x-3y-4z)=(5x-3y)^2-16z^2=25x^2-30xy+9y^2-16(x^2-y^2)=25x^2-30xy+9y^2-16x^2+16y^2
=9x^2-30xy+25y^2=(3x-5y)^2 (đpcm)
NN
1
B
29 tháng 8 2020
Bài làm :
Ta có:
\(x^2-y^2-z^2=0\)
\(\Leftrightarrow16x^2-16y^2-16z^2=0\)
\(\Leftrightarrow25x^2-9x^2+9y^2-25y^2-16z^2+30xy-30xy=0\)
\(\Leftrightarrow\left[\left(25x^2-30xy+9y^2\right)-16z^2\right]-\left(9x^2-30xy+25y^2\right)=0\)
\(\Leftrightarrow\left(5x-3y\right)^2-16z^2=\left(3x-5y\right)^2\)
\(\Leftrightarrow\left(5x-3y-4z\right)\left(5x-3y+4z\right)=\left(3x-5y\right)^2\)
=> Điều phải chứng minh
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(P+Q=6\Leftrightarrow x^3-3x^2+5x+y^3-3y^2+5y=6\Leftrightarrow\left(x^3-3x^2+3x-1\right)+\left(y^3-3y^2+3y-1\right)+\left(2x-4+2y\right)=0\)
\(\Leftrightarrow\left(x-1\right)^3+\left(y-1\right)^3+2\left(x+y-2\right)=0\Leftrightarrow\left(x-1+y-1\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(y-1\right)+\left(y-1\right)^2\right]+2\left(x+y-2\right)=0\)
\(\left[\left(x-1\right)^2-\left(x-1\right)\left(y-1\right)+\left(y-1\right)^2+2\right]\left(x+y-2\right)=0\)
=> x+y-2=0 <=> x+y=2
( trong ngoặc là bình phương thiếu của hiệu. có dạng \(a^2-ab+b^2\) luôn >=0 => +2 vào thì luôn khác 0