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\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
\(\frac{m^4-m}{2m^2+2m+2}=\frac{m\left(m^3-1\right)}{2m^2+2m+2}=\frac{m\left(m-1\right)\left(m^2+m+1\right)}{2\left(m^2+m+1\right)}=\frac{m\left(m-1\right)}{2}\)
a^3 +c^3 = (a+c). (a^2 -a.c+c^2)
= (a+c)^3 -3 ac.(a+c)
=> a^3+c^3-3abc+b^3 =(a+c)^3-3ac (a+c)-3abc +b^3
=(a+c)^3+b^3 -3ac (b+(a+c))
=(a+c+b). ((a+c)^2-(a+c).b+b^2) -3ac (a+c+b)
=(a+c+b)^3-3(a+c)b. (a+c+b)-3ac (a+c+b)
=(a+c+b)((a+c+b)^2 -3ab-3bc-3ac) (1)
(a-b)^2 + (b-c)^2 +(a-c)^2
= 2a^2 +2b^2+2c^2 -2ab-2bc-2ac
=2 (a^2+b^2+c^2-ac-ab-bc)
=2((a+b)^2-3ab +c^2 -ac-bc)
=2 ((a+b+c)^2-2(ac+bc)-3ab-ac-bc)
=2 (( a+c+b)^2 -3ab-3bc -3ac) (2)
Từ (1),(2) =>(a^3+b^3+c^3-3abc)/((a-b)^2
+(b-c)^2+(c-a)^2)
=(a+b+c)/2
a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)=\left(a+b\right)\left(a-b\right)+c\left(a-b\right)=\)\(\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)
\(c^2+bc-a^2-ab=\left(c-a\right)\left(a+b+c\right)\)
\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Phân tích mẫu thức thành nhân tử :
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+ac^2-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right).\)
Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)
Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)
Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)
Tử:
(b - c)3 + (c - a)3 + (a - b)3
= (b - c + c - a + a - b)3 - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)
= 0 - 3(b - a)(a - c)(c - b)
= 3(a - b)(a - c)(c - b)
Mẫu:
a2(b - c) + b2(c - a) + c2(a - b)
= a2(b - c) + b2c - ab2 + ac2 - bc2
= a2(b - c) - a(b2 - c2) + bc(b - c)
= a2(b - c) - a(b - c)(b + c) + bc(b - c)
= (b - c)(a2 - ab - ac + bc)
= (b - c)[a(a - b) - c(a - b)]
= (b - c)(a - b)(a - c)
\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)
\(=\frac{3\left(c-b\right)}{b-c}\)
Phân tích mẫu \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-c^2b\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]\)
\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)=-\left(b-c\right)\left(a-b\right)\left(c-a\right)\)
Đặt b - c = x, c - a = y, a - b = z
=> x + y + z = b - c + c - a + a - b = 0
Từ x+y+z=0 => x3+y3+z3=3xyz (tự c/m)
=>\(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3\)
a,Ta đặt :
a-b-c=x ; b-c-a=y ; c-a-b=z
Ta có:
\(\text{x+y+z=a-b-c+b-c-a+c-a-b=-(a+b+c)}\)
\(\Rightarrow\left(x+y+z\right)^2=\left(a+b+c\right)^2\)
\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y+z\right)^2+x^2+y^2+z^2\)
\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2\)\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=4\left(a^2+b^2+c^2\right)\)
Ta thấy: \(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}=\frac{\left(b-a\right)\left(d-c\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
\(\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(=\frac{\left(a-b\right)\left(c-d\right)}{\left(b-a\right)\left(b+a\right)\left(d-c\right)\left(d+c\right)}\)
\(\frac{1}{\left(a+b\right)\left(c+d\right)}\)