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a)\(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
b)\(9x^2+42x+49=\left(3x\right)^2+2\cdot3x\cdot7+7^2=\left(3x+7\right)^2\)
c)\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}\right)^2-2\cdot\dfrac{1}{3}\cdot y^4+\left(y^4\right)^2=\left(y^4-\dfrac{1}{3}\right)^2\)
a) \(x^2+2.2x+2^2\)
\(=\left(x+2\right)^2\)
b)\(\left(3x\right)^2+2.3.7x+7^2\)
\(=\left(3x+7\right)^2\)
c) \(\left(\dfrac{1}{3}\right)^2-2.\dfrac{1}{3}.y^4+\left(y^4\right)^2\)
\(=\left(\dfrac{1}{3}-y^4\right)^2\)
a) =(3x+y)2
câu b) sao kì quá mik k hiểu bạn xem lại đề đi có viết thiếu dấu k
3.
a, (2y- 1)3= (2y)3-3.(2y)2.1+3.2y.12-13
= 8y3-12y2+6y-1
b, (3x2+2y)3=(3x2)3+3.(3x2)2.2y+3.3x2.(2y)2+13
=27x6+54x4y+36x2y2+1
c, ( 1/3x-2)3=(1/3x)3-3.(1/3x)2.2+3.1/3x.22-23
=1/27x3-2/3x2+4x-8
4.
a, -x3+3x3-3x+1=1-3x+3x3-x3
=1-3.12.x+3.1.x3-x3
=(1-x)3
b,64-48x+12x2-x3=43-3.42.x+3.4.x2-x3
=(4-x)3
Bài 3 Tính:
\(a\)) \(\left(2y-1\right)^3=2y^3-3.\left(2y\right)^2.1+3.2y.1^2-1^3\)
\(=2y^3-12y^2+6y-1\)
b)\(\left(3x^2+2y\right)^3\)
\(=\left(3x^2\right)^3=3.\left(3x^2\right)^2.2y+3.\left(3x^2\right).\left(2y\right)^2+\left(2y\right)^3\)
\(=27x^8+3.9x^4.2+9x^2.4y+8y^3\)
\(=27x^8+54x^4+36x^2y+8y^3\)
c)\(\left(\dfrac{1}{3}x-2\right)^3\)
\(=\left(\dfrac{1}{3}x\right)^3-3.\left(\dfrac{1}{3}x\right)^2.2+3.\dfrac{1}{3}x.2^2-2^3\)
\(=\dfrac{1}{27}x^3-3.\dfrac{1}{9}x^2.2+x.2^2-8\)
\(=\dfrac{1}{27}x^3-\dfrac{2}{3}x^2+4x-8\)
Câu a : \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Câu b : \(9m^2+n^2-6mn=\left(3m-n\right)^2\)
Câu c : \(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
Câu d : \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
\(a,4x^2+4xy+y^2=\left(2x\right)^2+4xy+y^2=\left(2x+y\right)^2\)
\(b,9m^2+n^2-6mn=\left(3m\right)^2-6mn+n^2=\left(3m-n\right)^2\)
\(c,16a^2+25b^2+40ab=\left(4a\right)^2+40ab+\left(5b\right)^2=\left(4a+5b\right)^2\)
@Yukru ơi! giúp câu D với!
Chúc bạn học tốt!
b)(y-2)^3=y^3-8+12y-6y^2
c)8x^3+y^3=(2x+y)(4x^2+y^2-4xy)
2)
=(xy+2/3)^2
a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
\(A=2^3-3.2^2.x+3.2.x^2-x^3\)
\(A=\left(2-x\right)^3\)
\(B=\left(2x\right)^3-2.\left(2x\right)^2.y+3.2x.y^2-y^3\)
\(B=\left(2x-y\right)^3\)
a)x2+2x+1=x2+2x.1+12=(x+1)2
b)x2-x+\(\frac{1}{4}\)=x2-2.x.\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)=\(\left(x-\frac{1}{2}\right)^2\)
xl chuyển hộ mk "bình phương" thành "lập phương" nha
\(=\left(\dfrac{2}{3}a\right)^3-3.\left(\dfrac{2}{3}\right)^2a^2.2b+3.\dfrac{2}{3}a.4b^2-\left(2b\right)^3=\left(\dfrac{2}{3}a-2b\right)^3\)