khai triển hằng đẳng thức sau
a.\(\left(x-3\right)^3\)
b.\(\left(2x-3\right)^3\)
c.\(\left(x-\frac{1}{2}\right)^3\)
d.\(\left(x^2-2\right)^3\)
e.\(\left(2x-3y\right)^3\)
f.\(\left(\frac{1}{2}x-y^2\right)^3\)
K
Khách
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Những câu hỏi liên quan
29 tháng 7 2020
a) Ta có: \(\left(x+1\right)^3\)
\(=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\)
\(=x^3+3x^2+3x+1\)
b) Ta có: \(\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+3\cdot4x^2\cdot3+27\cdot2x+27\)
\(=8x^3+36x^2+54x+27\)
c) Ta có: \(\left(x+\frac{1}{2}\right)^3\)
\(=x^3+2\cdot x^2\cdot\frac{1}{2}+2\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=x^3+x^2+\frac{1}{2}x+\frac{1}{8}\)
d) Ta có: \(\left(x^2+2\right)^3\)
\(=\left(x^2\right)^3+3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2+2^3\)
\(=x^6+6x^4+12x^2+8\)
e) Ta có: \(\left(2x+3y\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3y+3\cdot2x\cdot\left(3y\right)^2+\left(3y\right)^3\)
\(=8x^3+36x^2y+54xy^2+27y^3\)
f) Ta có: \(\left(\frac{1}{2}x+y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3+3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2+\left(y^2\right)^3\)
\(=\frac{1}{8}x^3+\frac{3}{4}x^2y^2+\frac{3}{2}xy^4+y^6\)
HQ
Hà Quang Minh
Giáo viên
28 tháng 9 2023
a) \({\left( {2x + 1} \right)^4} = {\left( {2x} \right)^4} + 4.{\left( {2x} \right)^3}{.1^1} + 6.{\left( {2x} \right)^2}{.1^2} + 4.\left( {2x} \right){.1^3} + {1^4} = 16{x^4} + 32{x^3} + 24{x^2} + 8x + 1\)
b) \(\begin{array}{l}{\left( {3y - 4} \right)^4} = {\left[ {3y + \left( { - 4} \right)} \right]^4} = {\left( {3y} \right)^4} + 4.{\left( {3y} \right)^3}.\left( { - 4} \right) + 6.{\left( {3y} \right)^2}.{\left( { - 4} \right)^2} + 4.{\left( {3y} \right)^1}{\left( { - 4} \right)^3} + {\left( { - 4} \right)^4}\\ = 81{y^4} - 432{y^3} + 864{y^2} - 768y + 256\end{array}\)
c) \({\left( {x + \frac{1}{2}} \right)^4} = {x^4} + 4.{x^3}.{\left( {\frac{1}{2}} \right)^1} + 6.{x^2}.{\left( {\frac{1}{2}} \right)^2} + 4.x.{\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^4} = {x^4} + 2{x^3} + \frac{3}{2}{x^2} + \frac{1}{2}x + \frac{1}{{16}}\)
d) \(\begin{array}{l}{\left( {x - \frac{1}{3}} \right)^4} = {\left[ {x + \left( { - \frac{1}{3}} \right)} \right]^4} = {x^4} + 4.{x^3}.{\left( { - \frac{1}{3}} \right)^1} + 6.{x^2}.{\left( { - \frac{1}{3}} \right)^2} + 4.x.{\left( { - \frac{1}{3}} \right)^3} + {\left( { - \frac{1}{3}} \right)^4}\\ = {x^4} - \frac{4}{3}{x^3} + \frac{2}{3}{x^2} - \frac{4}{27}x + \frac{1}{{81}}\end{array}\)
HN
20 tháng 6 2018
a) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y\right)^3-3\left(2x^3y\right)^20,5x^2+3.2x^3y\left(0,5x^2\right)^2-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+1,5x^7y-0,125x^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=x^3-3^3\)
\(=x^3-27.\)
19 tháng 6 2018
a,\(\left(2x^3y-0,5x^2\right)^3=\left(2x^3y\right)^3-3.\left(2x^3y\right)^2.\left(0,5x^2\right)+3.\left(0,5x^2\right)^2.\left(2x^3y\right)-\left(0,5x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\frac{3}{2}x^7y-\frac{1}{8}x^6\)
b,\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)
\(=x^3-\left(3y\right)^3=x^3-27y^3\)
\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=\left(x^2-3\right)\left[\left(x^2\right)^2+3.x^2+3^2\right]\)
\(=\left(x^2\right)^3-3^3=x^6-27\)
TN
2
5 tháng 7 2016
a\(=\frac{1}{4}x^2+2.\frac{1}{2}x.1+1=\frac{1}{4}x^2+x+1\)
b\(=4x^2-2.2x.\frac{1}{3}+\frac{1}{9}=4x^2-\frac{4}{3}x+\frac{1}{9}\)
Bạn học tốt nha >>>>>>
nha
5 tháng 7 2016
a/\(\left(\frac{1}{2}x+1\right)^2=\frac{1}{4}x^2+x+1^2\)
b/\(\left(2x-\frac{1}{3}\right)^3=8x^3-2x+\frac{2}{3}x-\frac{1}{27}\)
k nha
a) Ta có: \(\left(x-3\right)^3\)
\(=x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\)
\(=x^3-9x^2+27x^2-27\)
b) Ta có: \(\left(2x-3\right)^3\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2-3^3\)
\(=8x^3-36x^2+54x-27\)
c) Ta có: \(\left(x-\frac{1}{2}\right)^3\)
\(=x^3-3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^3\)
\(=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)
d) Ta có: \(\left(x^2-2\right)^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot2+3\cdot x^2\cdot2^2-2^3\)
\(=x^6-6x^4+12x^2-8\)
e) Ta có: \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-2\cdot\left(2x\right)^2\cdot3y+2\cdot2x\cdot\left(3y\right)^2-\left(3y\right)^3\)
\(=8x^3-24x^2y+36xy^2-27y^3\)
f) Ta có: \(\left(\frac{1}{2}x-y^2\right)^3\)
\(=\left(\frac{1}{2}x\right)^3-3\cdot\left(\frac{1}{2}x\right)^2\cdot y^2+3\cdot\frac{1}{2}x\cdot\left(y^2\right)^2-\left(y^2\right)^3\)
\(=\frac{1}{8}x^3-\frac{3}{4}x^2y^2+\frac{3}{2}xy^4-y^6\)