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a,(x+2y)3 =x3+3.x2.2y+3.x.(2y)2+(2y)3
= x3+6x2y+12xy2+8y3
b, phần b tương tự dấu thay đổi một tí
c, (5x+1)(5x+1)= (5x+1)2
=25x2+10x+1
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: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
4:
a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2
b: 8(7^2+1)(7^4+1)(7^8+1)
=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^16-1)<7^16-1
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
mik chỉ biết bài 5 thôi !
\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)
\(2,=-\left(16-8y+y^2\right)=-16+8y-y^2\)
\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)
\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)
\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)
.... mấy ý còn lại bn tự lm nhé, tương tự thhooi
1) \(-\left(y+6\right)^2=-y^2-12y-36\)
2) \(-\left(4-y\right)^2=-y^2+8y-16\)
3) \(-\left(x+\dfrac{2}{3}\right)^2=-x^2-\dfrac{4}{3}x-\dfrac{4}{9}\)
4) \(-\left(x-\dfrac{3}{2}\right)^2=-x^2+3x-\dfrac{9}{4}\)
5) \(-\left(3y+2\right)^2=-9y^2-12y-4\)
6) \(-\left(2y-3\right)^2=-4y^2+12y-9\)
7) \(-\left(5x+2y\right)^2=-25x^2-20xy-4y^2\)
8) \(-\left(2x-\dfrac{3}{2}\right)^2=-4x^2+6x-\dfrac{9}{4}\)
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) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
b) \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)
c) \(\left(-6x-\frac{2}{5}\right)^2=36x^2+\frac{24}{5}x+\frac{4}{25}\)
d) \(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)
e) \(\left(x-y\right)^2\left(x+y\right)^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)
f) \(\left(\frac{1}{2}x-\frac{1}{3}y-1\right)^2=\frac{1}{4}x^2+\frac{1}{9}y^2+1-\frac{1}{3}xy-x+\frac{2}{3}y\)
\(a,\left(x+2\right)^2=x^2+4x+4\\ b,\left(x-1\right)^2=x^2-2x+1\\ c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)
\(a\text{) }pt\Leftrightarrow\left(y^2+2y+1\right)+\left[\left(2^x\right)^2-2.2^x+1\right]=0\)
\(\Leftrightarrow\left(y+1\right)^2+\left(2^x-1\right)^2=0\)
\(\Leftrightarrow y+1=0\text{ và }2^x-1=0\)
\(\Leftrightarrow y=-1\text{ và }x=0\)
\(b\text{) }pt\Leftrightarrow\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow x+y=0\text{ và }x-1=0\text{ và }y+1=0\)
\(\Leftrightarrow x=1\text{ và }y=-1\)
a) `(3/5 x^2 -1/2 y)^2 = (3/5 x^2)^2 - 2. 3/5 x^2 .1/2 y + (1/2 y)^2`
`= 9/25 x^4 - 3/5 x^2y + 1/4 y^2`