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a: Sửa đề: \(a^2\left(a+1\right)+b^2\left(b-1\right)-a^2b^2\left(a+b\right)\)
\(=a^3+a^2+b^3-b^2-a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a+b\right)-a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)\)
b: \(=a^m\cdot a^3+2\cdot a^m\cdot a^2+a^m\)
\(=a^m\left(a^3+2a^2+1\right)\)
\(A=2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4\)
\(=4a^2b^2-\left(2a^2b^2-2b^2c^2-2a^2c^2+a^4+b^4+c^4\right)\)
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)
\(=\left[c^2-\left(a-b\right)^2\right]\left[\left(a+b\right)^2-c^2\right]\)
\(=\left(c+a-b\right)\left(c-a+b\right)\left(a+b-c\right)\left(a+b+c\right)\)
Nếu a,b,c là độ dài 3 cạnh thì ta có:
c + a > b (bất đẳng thức tam giác)
a + b > c (bất đẳng thức tam giác)
b + c > a (bất đẳng thức tam giác)
mà a,b,c > 0
=> a + b + c dương
a + c - b dương
a + b - c dương
b + c - a dương
=> A dương
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
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) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)
bn chép lại đề nhé
a/ \(=\left(x+y\right)^2-4x^2y^2=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
b/ \(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=\left[\left(b+c\right)^2-a^2\right]\left[-\left(b+c\right)^2+a^2\right]\)
\(=\left(b+c-a\right)\left(b+c+a\right)^2\left(a-b-c\right)\)
c/ \(=2a^2+2b^2-2c^2+4ab=2\left[\left(a^2+b^2+2ab\right)-c^2\right]\)
\(=2\left(a+b-c\right)\left(a+b+c\right)\)
d/ \(=\left(4x^2-25\right)^2-9\left(4x^2-20x+25\right)\)
\(=\left(4x^2-25\right)^2-9\left(4x^2+25\right)+180x\)
tới đây bạn đặt a= 4x^2 -25 rồi làm típ nha, mình lười quá ><
e/ tương tự câu d nha bạn
f/ \(=a^4\left(a^2-1\right)+2a^2\left(a+1\right)\)
\(=a^4\left(a-1\right)\left(a+1\right)+2a^2\left(a+1\right)\)
\(=a^2\left(a+1\right)\left(a^2+2\right)\)
g/ đặt \(a=3x^2+3x+2\) khi đó biểu thức trở thành
\(a^2-\left(a+4\right)^2=a^2-a^2-8a-16\)
\(=-8a-16=-8\left(3x^2+3x+2-8\right)=-8\left(3x^2+3x-6\right)\)
\(=-24\left(x^2+x-2\right)=-24\left(x-1\right)\left(x+2\right)\)
xong rùi nha bn. Chúc bn hc tốt (xin lỗi tại có mấy câu mình lười nha)
\(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2\)
\(=\left(x-2xy+y\right)\left(x+2xy+y\right)\)
c: \(5\left(a+b\right)+x\left(a+b\right)\)
=(a+b)(x+5)
d: \(\left(a-b\right)^2-\left(b-a\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\)
=(a-b)(a-b+1)
e: \(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=2y\cdot6x\cdot\left(2x+1\right)=12xy\left(2x+1\right)\)
Câu 1: \(a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=\left(a^4+b^4+c^4-2a^2b^2-2c^2a^2+2b^2c^2\right)-4b^2c^2=\left(a^2-b^2-c^2\right)^2-4b^2c^2=\left(a^2-b^2-c^2-2bc\right)\left(a^2-b^2-c^2+2bc\right)=\left[a^2-\left(b+c\right)^2\right]\left[a^2-\left(b-c\right)^2\right]=\left(a-b-c\right)\left(a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)\)Câu 2: \(a^3+a^2-ab^2-b^2=a^2\left(a+1\right)-b^2\left(a+1\right)=\left(a^2-b^2\right)\left(a+1\right)=\left(a+b\right)\left(a-b\right)\left(a+1\right)\)
Câu 3: \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=a\left(b^3-c^3\right)-b\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
Câu 1.
a4 + b4 + c4 - 2a2b2 - 2b2c2 - 2a2c2
= [ ( a4 - 2a2b2 + b4 ) - 2a2c2 + 2b2c2 + c4 ] - 4b2c2
= [ ( a2 - b2 )2 - 2( a2 - b2 )c2 + ( c2 )2 ] - ( 2bc )2
= ( a2 - b2 - c2 ) - ( 2bc )2
= ( a2 - b2 - c2 - 2bc )( a2 - b2 - c2 + 2bc )
= [ a2 - ( b2 + 2bc + c2 ) ][ a2 - ( b2 - 2bc + c2 ) ]
= [ a2 - ( b + c )2 ][ a2 - ( b - c )2 ]
= ( a - b - c )( a + b + c )( a - b + c )( a + b - c )
Câu 2.
a3 + a2 - ab2 - b2
= a2( a + 1 ) - b2( a + 1 )
= ( a + 1 )( a2 - b2 )
= ( a + 1 )( a - b )( a + b )