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Bài 1:
a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)
b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)
c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)
d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)
e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)
g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)
f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)
Bài 2 :
a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)
c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)
e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)
1) \(B=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)+2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+\left(4x-4\right)\cdot\left(x+3\right)+2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+12x-4x-12+50-60+18x^2\)
\(=42x^2-72x+43\)
2) \(C=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a+1\right)^2\)
\(=4a^4-4a^3+2a^2+4a^3-4a^2+2a+2a^2-2a+1-\left(4a^2+4a+1\right)\)
\(=4a^4+2a^2-4a^2+2a^2+1-4a^2-4a-1\)
\(=4a^4-4a^2-4a\)
3) Sky Sơn Tùng làm đúng rồi nhé.
4) \(E=\left(x^2-5x+1\right)^2+2\left(5x-1\right)\left(x^2-5x+1\right)\left(5x-1\right)^2\)
\(=x^4+27x^2+1-10x^3+250x^5-1400x^4+1030x^3-302x^2+40x-2\)
\(=-1399x^4-275x^2-1+1020x^3+250x^5+40x\)
5) \(F=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2\)
\(=\left[a^2+b^2-c^2-\left(a^2-b^2+c^2\right)\right]\cdot\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\)
\(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\cdot2a^2\)
\(=\left(2b^2-2c^2\right)\cdot2a^2\)
\(=2\left(b^2-c^2\right)\cdot2a^2\)
\(=2\left(b-c\right)\left(b+c\right)\cdot2a^2\)
\(=2\cdot2a^2\cdot\left(b-c\right)\left(b+c\right)\)
\(=4a^2\cdot\left(b-c\right)\left(b+c\right)\)
6) \(G=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+\left(-c\right)^2+2ab-2ac-2bc-2\left(a^2+2ab+b^2\right)\)
\(=a^2+b^2+c^2+2ab+a^2+b^2+\left(-c\right)^2+2ab-2a^2-4ab-2b^2\)
\(=0+0+c^2+0+c^2\)
\(=2c^2\)
7) \(H=\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2x\right)\)
\(=a^2-c^2-\left[\left(a-b\right)^2-c^2\right]+b^2-2bx\)
\(=a^2-c^2-\left(a^2-2ab+b^2-c^2\right)+b^2-2bx\)
\(=a^2-b^2-a^2+2ab-b^2+c^2+b^2-2bx\)
\(=2ab-2bx\)
\(D=\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)=\left(9x-1+1-5x\right)^2=\left(4x\right)^2=16x^2\)
a) Biến đổi VT ta có :
(a2-b2)2 + (2ab)2
= a4 -2a2+b4+4a2b2
= a4+2a2b2 +b4
= (a2b2)2 = VP (đpcm)
b) Biến đổi vế trái ta có :
(ax+b)2 + (a-bx)2+cx2+c2
= a2x2+2axb+b2 +a2 - 2axb+b2x2 +c2x2+ c2
= (a2+b2+c2) + x2(a2+b2+c2)
= (a2+b2+c2) (x2+1) = VP (đpcm)
Cho a-3b=1, 2ab=-4. Tính:
A=2a+(7ab)/2-6b+2
B= (2a+6b)2-2
C+ 3a2+27b2-ab-1
D=a3-27b3+a2+9b2+2
E=a4+81b4-1
Cho a-3b=1, 2ab=-4. Tính:
A=2a+(7ab)/2-6b+2
B=(2a+6b)2-2
C= 3a2+27b2-ab-1
D= a3-27b3+a2+9b2+2
E=a4+81b4-1
a. x2y2+1-x2-y2
=x2.(y2-1)-(y2-1)
=(y2-1).(x2-1)
=(y-1)(y+1)(x-1)(y-1)
c. x^3+3x^2-3x-1
= (x-1).(x^2+x+1)+3x.(x-1)
=(x-1)(x^2+x+1+x-1)
=(x-1)(x^2+2x)
=x(x-1)(x+2)
a) (3+\(xy^2\))\(^2\)= \(3^2\)+2*3*\(xy^2\)+\(\left(xy^2\right)\)\(^2\)
=9+6\(xy^2\)+\(x^2\)\(y^4\)
b) (a+\(b^2\))(a-\(b^2\))
=a(a-\(b^2\))+\(b^2\)(a-\(b^2\))
=a*a+a*-\(b^2\)+\(b^2\)*a+\(b^2\)*-\(b^2\)
=\(a^2\)-a\(b^2\)+\(b^2\)a-\(b^4\)
=\(a^2\)+(-a\(b^2\)+a\(b^2\))-\(b^4\)
=\(a^2\)-\(b^4\)
c)(2y-1)\(^2\)=(2y)\(^2\)-2*2y*1+1\(^{^2}\)
=4y\(^2\)-4y+1
d)(10-m\(^2\))\(^2\)=10\(^2\)-2*10*m\(^2\)+(m\(^2\))\(^2\)
=100-20m\(^2\)+m\(^4\)
câu e bạn tự làm nha tương tự như câu b
chúc bạn học tốt