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1,
\(\left(\frac{2}{3}x+y\right)^2=\left(\frac{2}{3}x\right)^2+2.\frac{2}{3}x.y+\left(y\right)^2=\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
\(\left(3a+\frac{1}{2}b\right)^2=\left(3a\right)^2+2.3a.\frac{1}{2}b+\left(\frac{1}{2}b\right)^2=9a^2+3ab+\frac{1}{4}b^2\)
2,
\(25a^2+4b^2+20ab=\left(5a\right)^2+\left(2b\right)^2+2.5a.2b=\left(5a+2b\right)^2\)
\(x^2+2x+1=\left(x\right)^2+2.x.1+\left(1\right)^2=\left(x+1\right)^2\)
\(9x^2+6x+1=\left(3x\right)^2+2.3x.1+\left(1\right)^2=\left(3x+1\right)^2\)
\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
1.
a)\(\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
b)\(9a^2+3ab+\frac{1}{4}a^2\)
2.
a)\(\left(5x+2b\right)^2\)
b)\(\left(x+1\right)^2\)
c)\(\left(3x+1\right)^2\)
d)\(\left[\left(2x+3y\right)+1\right]^2\)
\(A=x^2-2x+1-x^2+4=5-2x\)
\(B=27x^3+8-x^2+9=27x^3-x^2+17\)
\(C=3x^2y-6xy^2-2x\left(x^2-2x^2y+x^2y^2\right)=3x^2y-6xy^2-2x^3+4x^3y-2x^3y^2\)
Em chỉ cần nhớ hằng đẳng thức và áp dụng là biến đổi được ^^
2X.(3X-5)=10-6X
<=>2X.(3X-5)+6X-10=0
<=>2X.(3X-5)+2.(3X-5)=0
<=>(3X-5).(2X+2)=0
<=>3X-5=0 và 2X+2=0
=> X=
a, \(\left(4x+5\right)^2=\left(4x+5\right)\left(4x+5\right)=\left[\left(4x+5\right)4x\right]+\left[\left(4x+5\right)5\right]=4x^2+20x+25\)
b, \(\left(5x-2\right)^2=\left(5x-2\right)\left(5x-2\right)=\left[\left(5x-2\right)5x-\left(5x-2\right)2\right]=5x^2-10x+25\)
b, \(8^2-12x^2=\left(8^2-12x^2\right)\left(8^2+12x^2\right)\)
đúng ko :)
@No name: Bị sai rồi nhé, a,b,c sai hết :>
a) ( 4x + 5 )2
= ( 4x )2 + 2.4x.5 + 52
= 16x2 + 40x + 25
b) ( 5x - 2 )2
= ( 5x )2 - 2.5x.2 + 22
= 25x2 - 20x + 4
c) 82 - 12x2
= 64 - 12x2
= ( V8 - V12x )( V8 + V12x )
\(x^3-8+12x-6x^2\)
\(=x^3-3.x^2.2+3.x.2^2-2^3\)
\(=\left(x-2\right)^3\)
(1)\(\left(3x\right)^2+2×3x+y^2=\left(3x+y\right)^2\)
(2)\(-x^2+6x-9=-\left(x^2-6x+9\right)=-\left(x^2-6x+3^2\right)=-\left[\left(x-3\right)^2\right]\)
(3)\(x^2+4xy+4y^2=x^2+2×x×2y^2+\left(2y\right)^2=\left(x+2y\right)^2\)
a) \(x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\)
\(\left(x^2+4x+4\right)\div\left(x+2\right)=x+2\)
b) \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(\left(x^3-1\right)\div\left(x-1\right)=x^2+x+1\)
c) \(x^3+6x^2+12x+8=x^3+3.x^2.2+3.x.2^2+2^3=\left(x+2\right)^3\)
\(\left(x^3+6x^2+12x+8\right)\div\left(x+2\right)=\left(x+2\right)^2\)
= x^3-3x^2-3x^2+9x
=x^2(x-3)-3x(x-3)
=x(x-3)(x-3)
=x(x-3)^2
Đầu tiên nhé
x^3 -6x^2+9x
= x(x^2-6x+9)
= x(x-3)^2
Chúc bạn học tốt.