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Bài làm:
a) Sửa đề: \(4x^2-y^2\)
\(=\left(2x\right)^2-y^2\)
\(=\left(2x-y\right)\left(2x+y\right)\)
b) \(a^2+2ab+b^2\)
\(=\left(a+b\right)^2\)
c) \(x^2-2xy+y^2\)
\(=\left(x-y\right)^2\)
d) \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
b) \(a^2+2ab+b^2=\left(a+b\right)^2.\)
c) \(x^2-2xy+y^2=\left(x-y\right)^2.\)
d) \(x^2+4xy+4y=\left(x+2y\right)^2\)
câu a chịu
a) \(2xy-ax+x^2-2xy\)
= \(-ax+x^2=x\cdot\left(x-a\right)\)
b) \(x^2-y^2-2x-2y=\left(x-y\right)\left(x+y\right)-2\cdot\left(x+y\right)\)
= \(\left(x+y\right)\left(x-y-2\right)\)
bình phương tổng chứ
b, B= x^2+ 2xy+y^2 +4y+4
= x^2+2xy+y^2+y^2+4y+4
=(x+y)^2+(y+2)^2
c, C= 2x^2+6xy+9y^2+2x+1
= x^2+6xy+9y^2+x^2+2x+1
= (x+3)^2+(x+1)^2
d, D= x(x+2) +(x+1)(x+3) +2
= x^2+2x+x^2+3x+x+3+2
= x^2+2x+1+x^2+4x+4
= (x+1)^2+(x+2)^2
e, E= x^2-2xy+2y^2+2y+1
= x^2-2xy+y^2+y^2+2y+1
= (x-y)^2+(y+1)^2
f, F= 4x^2-12xy+10y^2+4y+4
=4x^2-12xy+9y^2+y^2+4y+4
=(2x-3y)^2+(y+2)^2
g, G=2x^2+4xy+4y^2+4x+4
=x^2+4xy+4y^2+x^2+4x+4
=(x+2y)^2+(x+2)^2
Xong r.... dài quá...mới hè lớp 7 nên có j bỏ qua ak
\(a,\left(x+3\right)^2\)
\(b,\left(x+\frac{1}{2}\right)^2\)
\(c,\left(xy^2+1\right)^2\)
a) \(x^6-4=\left(x^3\right)^2-2^2=\left(x^3-2\right).\left(x^3+2\right)\)
b) \(-9x^2+1=1^2-\left(3x\right)^2=\left(1-3x\right).\left(1+3x\right)\)
c) \(x^{10}-9=\left(x^5\right)^2-3^2=\left(x^5-3\right).\left(x^5+3\right)\)
mk chỉ làm đk bài 1 thui ,thông cảm cho mk nha bạn
\(a;x^6-4=\left(x^3\right)^2-2^2=\left(x^3-2\right)\left(x^3+2\right)\)
\(b;-9x^2+1=1^2-3x^2=\left(1-3x\right).\left(1+3x\right)\)
\(c;x^{10}-9=\left(x^5\right)^2-3^2=\left(x^5-3\right).\left(x^5+3\right)\)
\(#LTH\)
\(a,5\left(x-y\right)-3x\left(y-x\right)=5\left(x-y\right)+3x\left(x-y\right)=\left(5+3x\right)\left(x-y\right)\\ b,x^2-4xy+4y^2=\left(x-2y\right)^2\\ c,\left(x+1\right)^2+x\left(5-x\right)=0\\ \Rightarrow x^2+2x+1+5x-x^2=0\\ \Rightarrow7x+1=0\\ \Rightarrow7x=-1\\ \Rightarrow x=-\dfrac{1}{7}\)
a: =(x-y)(5+3x)
c: \(\Leftrightarrow x^2-2x+1+5x-x^2=0\)
hay x=-1/3
a) \(x^2+6x+9=\left(x+3\right)^2\)
b) \(x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)
c) \(2xy^2+x^2y^4+1=\left(xy^2+1\right)^2\)
a, \(\left(x+3\right)^2\)
b,\(\left(x+\frac{1}{2}\right)^2\)Mik giải thích tí nè, cái này =\(x^2+2.x.\frac{1}{2}+\frac{1}{4}\)=\(x^2+x+\frac{1}{4}\)
c,thì mik chịu.
A=x^2+2(x^2+2x+1)+3(x^2+4x+4)+4(x^2+6x+9)
=x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36
=10x^2+40x+50
=(9x^2+30x+25)+(x^2+10x+25)
=(3x+5)^2+(x+5)^2
\(a,\left(x+2\right)^2-4\left(y+2\right)^2\\ =\left(x+2\right)^2-\left(2y+4\right)^2\\ =\left(x+2-2y-4\right)\left(x+2+2y+4\right)\\ =\left(x-2y-2\right)\left(x+2y+6\right)\\ b,x^2y^2+2xy-z^2+1\\ =\left(x^2y^2+2xy+1\right)-z^2\\ =\left(xy+1\right)^2-z^2\\ =\left(xy-z+1\right)\left(xy+z+1\right)\\ c,4x^2y^2+4xy-\left(z^2-1\right)\\ =\left(4x^2y^2+4xy+1\right)-z^2\\ =\left(2xy+1\right)^2-z^2\\ =\left(2xy-z+1\right)\left(2xy+z+1\right)\)
câu c làm sai rồi