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\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)
c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)
a)\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
=\(2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
b) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24=-11x+24\)
c) \(\dfrac{1}{2}x^2\left(6x-3\right)-x\left(x^2+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+4\right)\)
\(=3x^3-\dfrac{3}{2}x^2-x^3-\dfrac{1}{2}x+\dfrac{1}{2}x+2=2x^3-\dfrac{3}{2}x^2+2\)
a) \(\left(1+x\right)^2+\left(1-x\right)^2\)
\(=1+2x+x^2+1-2x+x^2\)
\(=2x^2+2\)
b) \(\left(x+2\right)^2+\left(1+x\right)\left(1-x\right)\)
\(=x^2+4x+4+1-x^2\)
\(=4x+5\)
c) \(\left(x-3\right)^2+3\left(x+1\right)^2\)
\(=x^2-6x+9+3x^2+6x+3\)
\(=4x^2+12\)
d)\(\left(2+3x\right)\left(3x-2\right)-\left(3x+1\right)^2\)
\(=9x^2-4-9x^2-6x-1\)
\(=-6x-5\)
e) \(\left(x+5\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=x^2-2x+5x-10-x^2-4x-4\)
\(=-x-14\)
f) \(\left(x+3\right)\left(2x-5\right)-2\left(1+x\right)^2\)
\(=2x^2-5x+6x-15-2-4x-2x^2\)
\(=-3x-17\)
g) \(\left(4x-1\right)\left(4x+1\right)-4\left(1-2x\right)^2\)
\(=16x^2-1-4+16x-16x^2\)
\(=16x-5\)
#Học tốt!
a: \(=3x+5-3x+\dfrac{5}{3}-3x-1=3x+\dfrac{17}{3}\)
b: \(=\left(3a+2-3a+2\right)^2=4^2=16\)
câu a là hằng đẳng thức luôn
A=(2x+4)^2
B khai triển tung tóe ra thì phần sau triệt tiêu hết còn 4(a^2+b^2+c^2)
câu c cảm giác sai đề vì mấy câu này phải là (3x)^ ms ra hdt chứ nhỉ
cau a : (3x^2y-6xy+9x)(-4/3xy)
=-4/3xy.3x^2y+4/3xy.6xy-4/3xy.9x
=-4x+8-8y
cau b : (1/3x+2y)(1/9x^2-2/3xy+4y^2)
=(1/3)^3-2/9x^2y+8y^3+4/3xy^2+2/9x^2y-4/3xy^2+8y^3
=(1/3)^3 + (2y)^3x-2
cau c : (x-2)(x^2-5x+1)+x(x^2+11)
=x^3-5x^2+x-2x^2+10x-2+x^3+11x
=2x^3-7x^2+22x-2
cau d := x^3 + 6xy^2 -27y^3
cau e := x^3 + 3x^2 -5x - 3x^2y - 9xy = 15y
cau f := x^2-2x+2x -4-2x-1
= x(x-2)-5
MK ĐANG CẦN GẤP Ạ AI NHANH MK SẼ VOTE Ạ
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)