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a) \(\left(2x+1\right)^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)
\(=8x^3+12x^2+6x+1\)
b) \(\left(x-3\right)^3\)
\(=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
Bài 2:
a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)
b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)
c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)
a. Đề đúng phải là \(\frac{1}{4}a^2+2ab^2+4b^4\)hoặc \(\frac{1}{4}a^2+2ab+4b^2\)
Ở đây mình giải trường hợp 2, bạn dựa theo để giải trường hợp 1 nhé :))
\(\frac{1}{4}a^2+2ab+4b^2\)
\(=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b+\left(2b\right)^2\)
\(=\left(\frac{1}{2}a+2b\right)^2\)
b. \(25+10x+x^2\)
\(=x^2+2.x.5+5^2\)
\(=\left(x+5\right)^2\)
c. \(\frac{1}{9}-\frac{2}{3}y^4+y^8\)
\(=\left(y^4\right)^2-2.y^4.\frac{1}{3}+\left(\frac{1}{3}\right)^2\)
\(=\left(y^4-\frac{1}{3}\right)^2\)
Bài 1:
c: \(\left(-5x-y\right)^3=-125x^3-75x^2y-15xy^2-y^3\)
h: \(\left(3y-2x^2\right)^3=27y^3-54y^2x^2+36yx^4-8x^6\)
\(a,5\left(x-y\right)-3x\left(y-x\right)=5\left(x-y\right)+3x\left(x-y\right)=\left(5+3x\right)\left(x-y\right)\\ b,x^2-4xy+4y^2=\left(x-2y\right)^2\\ c,\left(x+1\right)^2+x\left(5-x\right)=0\\ \Rightarrow x^2+2x+1+5x-x^2=0\\ \Rightarrow7x+1=0\\ \Rightarrow7x=-1\\ \Rightarrow x=-\dfrac{1}{7}\)
a: =(x-y)(5+3x)
c: \(\Leftrightarrow x^2-2x+1+5x-x^2=0\)
hay x=-1/3
2:
-8x^6-12x^4y-6x^2y^2-y^3
=-(8x^6+12x^4y+6x^2y^2+y^3)
=-(2x^2+y)^3
3:
=(1/3)^2-(2x-y)^2
=(1/3-2x+y)(1/3+2x-y)
1, \(x^2+2xy+y^2=\left(x+y\right)^2\)
2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)
3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)
5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
1: =(x+y)^2
2: =(2x+3)^2
3: =(x+5/2)^2
4: =(4x-1)^2
5: =(x+1/2)^2
6: =(x-3/2)^2
7: =(x+1)^3
8: =(1/2x+1)^2
9: =(3y-1/3)^3
10: =(2x+y)^3
`a,(x+y)^2`
`b, x^2-25`
`=x^2-5^2`
`=(x-5)(x+5)`
a) (x+y)^{2}=x^2+2xy+y^2
b) x^2 -25=x^2 - 5^2=(x-5)(x+5)