x^3 + 1=
8 + x^3=
27x^3 - 64y^3=
x^3 phần 64 - 1 phần 25
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4 tháng 10 2021
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
NM
Nguyễn Minh Quang
Giáo viên
23 tháng 7 2021
a.\(x^3-8=x^3-2^3=\left(x-2\right)\left(x^2+2x+4\right)\)
b.\(27x^3+125y^3=\left(3x\right)^3+\left(5y\right)^3=\left(3x+5y\right)\left(9x^2-15xy+25y^2\right)\)
c.\(\left(2x-1\right)^3+8=\left(2x-1\right)^3+2^3=\left(2x+1\right)\left[\left(2x-1\right)^2-2\left(2x-1\right)+4\right]\)
d.\(x^6+6^3=\left(x^2+6\right)\left(x^4-6x+36\right)\)
e.\(1-27x^3=1-\left(3x\right)^3=\left(1-3x\right)\left(1+3x+9x^2\right)\)
j.\(\left(x-3\right)^3-27=\left(x-3\right)^3-3^3=\left(x-6\right)\left[\left(x-3\right)^2+3\left(x-3\right)+9\right]\)
g.\(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)
t.\(8x^3-\frac{1}{8}=\left(2x\right)^3-\left(\frac{1}{2}\right)^3=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
u.\(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{x}{3}+\frac{1}{9}\right)\)
10 tháng 9 2017
b ơi minh thấy đề bài nó cứ kì kì
nếu như bn viết đề bài đúng thì mình có thể lm đc cho bn đó
12 tháng 1 2023
a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
b: =(1-2x)(1+2x)
c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
d: =(x+3)^3
e: \(=\left(2x-y\right)^3\)
f: =(x+2y)(x^2-2xy+4y^2)
7 tháng 9 2017
a) \(x^2\)\(+\)\(6x\)\(+\)\(9\)
\(=\left(x+3\right)^2\)
b) \(x^3\)\(+\)\(3x^2\)\(+\)\(3x\)\(+\)\(1\)
\(=\left(x+1\right)^3\)
c) \(8x^3\)\(-\)\(\frac{1}{8}\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
d) \(10x\)\(-\)\(25\)\(-\)\(x^2\)
\(=\)\(-x^2\)\(+\)\(10\)\(-\)\(25\)
\(=-\left(x^2-10+25\right)\)
\(=-\left(x-5\right)^2\)
e) \(\frac{1}{25}x^2\)\(-\)\(64y^2\)
=\(\left(\frac{1}{25}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
\(x^3+1\)
\(=x^3+1^3\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
______
\(8+x^3\)
\(=2^3+x^3\)
\(=\left(2+x\right)\left(4-2x+x^2\right)\)
______
\(27x^3-64y^3\)
\(=\left(3x\right)^3-\left(4y\right)^3\)
\(=\left(3x-4y\right)\left(9x^2+12xy+16y^2\right)\)
______
\(\dfrac{x^3}{64}-\dfrac{1}{125}\)
\(=\left(\dfrac{x}{4}\right)^3-\left(\dfrac{1}{5}\right)^3\)
\(=\left(\dfrac{x}{4}-\dfrac{1}{5}\right)\left(\dfrac{x^2}{16}+\dfrac{x}{20}+\dfrac{1}{25}\right)\)
x^3+1=(x+1)(x^2-x+1)
x^3+8=(x+2)(x^2-2x+4)
27x^3-64y^3=(3x-4y)(9x^2+12xy+16y^2)
\(\dfrac{x^3}{64}-\dfrac{1}{25}=\left(\dfrac{1}{4}x-\sqrt[3]{\dfrac{1}{5}}\right)\left(\dfrac{1}{16}x^2+\dfrac{1}{4\sqrt[3]{5}}\cdot x+\dfrac{1}{\sqrt[3]{25}}\right)\)