F=(3x-2)^2+(3x+2)^2+2(9x^2-4) tại x = -1/3
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NC
23 tháng 8 2020
1) \(\left(3x^2-1\right)\left(9x^4+3x^2+1\right)\)
\(=27x^6+9x^4+3x^2-9x^4-3x^2-1\)
\(=27x^6-1\) (hằng đẳng thức dạng a3 - b3)
2) \(\left(x^2-4\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left[\left(x-2\right)\left(x^2+2x+4\right)\right].\left[\left(x+2\right)\left(x^2-2x+4\right)\right]\)
\(=\left(x^3-8\right)\left(x^3+8\right)\)
\(=x^6-64\)
G
23 tháng 8 2020
a) \(\left(3x^2-1\right)\left(9x^4+3x^2+1\right)=\left(3x^2-1\right)\left[\left(3x^2\right)^2+3x^2.1+1^2\right]=\left(3x^2\right)^3-1^3=3x^6-1\)
b) \(\left(x^2-4\right).\left(x^2+2x+4\right).\left(x^2-2x+4\right)=\left(x^2-2^2\right).\left(x+2\right)^2.\left(x-2\right)^2=\left(x+2\right).\left(x-2\right).\left(x+2\right)^2.\left(x-2\right)^2=\left(x+2\right)^3.\left(x-2\right)^3\)
YN
2 tháng 10 2021
a) \(\left(3x-2\right)^2=\left(3x\right)^2-2.3x.2+2^2=9x^2-12x+4\)
b) \(\left(\dfrac{x}{3}+y^3\right)^2=\left(\dfrac{x}{3}\right)^2+2\dfrac{x}{3}y^3+\left(y^3\right)^2=\dfrac{x^2}{9}+\dfrac{2}{3}xy^3+y^6\)
c) \(9x^2-225=\left(3x\right)^2-\left(15\right)^2=\left(3x-15\right)\left(3x+15\right)\)
YN
2 tháng 10 2021
d) \(\left(2x-3y\right)^3=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^2-\left(3y\right)^3=8x^3-3.4x^2.3y+6x.9y^2-27y^3=8x^3-36x^2y+54xy^2-27y^3\)
e) \(\left(2x^2+\dfrac{3}{2}\right)^3=\left(2x^2\right)^3+3\left(2x^2\right)^2\dfrac{3}{2}+3.2x^2\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3=8x^6+3.4x^4.\dfrac{3}{2}+6x^2.\dfrac{9}{4}+\dfrac{27}{8}=8x^6+18x^4+\dfrac{27}{2}x^2+\dfrac{27}{8}\)
f) \(\left(-2xy^2+\dfrac{1}{2}x^3y\right)^3=\left(-2xy^2\right)+3\left(-2xy^2\right)^2\dfrac{1}{2}x^3y+3\left(-2xy^2\right)\left(\dfrac{1}{2}x^3y\right)^2+\left(\dfrac{1}{2}x^3y\right)^3=-8x^3y^6+3.4x^2y^4.\dfrac{1}{2}x^3y-6xy^2.\dfrac{1}{4}x^6y^2+\dfrac{1}{8}x^9y^3=-8x^3y^6+6x^5y^5-\dfrac{3}{2}x^7y^4+\dfrac{1}{8}x^9y^3\)
16 tháng 9 2021
\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)
16 tháng 9 2021
\(a,\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)
\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)
\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)
\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)
\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)
\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)
10 tháng 7 2016
xin lỗi vì ko giúp đc zì !!! Tại ....... e ms lớp 6 à !!!!
10 tháng 7 2016
a) \(100x^2-\left(x^2+25\right)^2\)
\(=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)( Áp dụng hằng đẳng thức số 3 )
b) ko khai phân tích dc bạn ạ
c)
Q
20 tháng 6 2017
a) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)
\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
d) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=\left(x^2-3\right)\left(4x^2+9\right)\)
\(=4x^4+9x^2-12x^2-27\)
\(=4x^4-3x^2-27\)
f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=\left(2x\right)^3-1^3\)
\(=8x^3-1\)
TN
20 tháng 6 2017
\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)
2 tháng 10 2021
a: \(\left(3x-2\right)^2=9x^2-12x+4\)
c: \(9x^2-225=9\left(x^2-25\right)=9\left(x-5\right)\left(x+5\right)\)
2 tháng 10 2021
a: \(\left(3x-2\right)^2=9x^2-12x+4\)
c: \(9x^2-225=\left(3x-15\right)\left(3x+15\right)\)
d: \(\left(2x-3y\right)^3=8x^3-36x^2y+54xy^2-27y^3\)
\(F=\left(3x-2\right)^2+\left(3x+2\right)^2+2\left(9x^2-4\right)\\=\left[\left(3x+2\right)^2+2.\left(3x+2\right)\left(3x-2\right)+\left(3x-2\right)^2\right]\\ =\left[\left(3x+2\right)+\left(3x-2\right)\right]^2\\ =\left(6x\right)^2=36x^2\\ Thay.x=-\dfrac{1}{3}.vào.F.thu.gọn:\\ F=36x^2=36.\left(-\dfrac{1}{3}\right)^2=36.\left(\dfrac{1}{9}\right)=4\)