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\(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)
\(=\left[\left(\frac{1}{2}a+b\right)+\left(\frac{1}{2}a-b\right)\right].\left[\left(\frac{1}{2}a+b\right)^2-\left(\frac{1}{2}a+b\right).\left(\frac{1}{2}a-b\right)+\left(\frac{1}{2}a-b\right)^2\right]\)
Tới đấy phân tích ra là được nhé . Mình đang bận .
\(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)
\(=\left(\frac{1}{2}a+b+\frac{1}{2}a-b\right)\left[\left(\frac{1}{2}a+b\right)^2-\left(\frac{1}{2}a+b\right)\left(\frac{1}{2}a-b\right)+\left(\frac{1}{2}a-b\right)^2\right]\)
\(=a\left(\frac{1}{4}a^2+ab+b^2-\frac{1}{4}a^2+b^2+\frac{1}{4a^2}-ab+b^2\right)\)
\(=a\left(\frac{1}{4}a^2+3b^2\right)\)
\(=\frac{1}{4}a^3+3ab^2\)
b. Sử dụng các hằng đẳng thức
\(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Do (a - b) + (b - c) + (c - a) = 0 nên áp dụng hđt \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:
\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)
Bài 1 :
\(b,ax^2+3ax+9=a^2\)
\(\Leftrightarrow a^2x+3ax+9-a^2=0\)
\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)
Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\)
\(\Leftrightarrow ax=a-3\)
Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
c) \(\frac{a\left(a^2-ab+b^2\right)}{b\left(a+b\right)\left(a^2-ab+b^2\right)}\)
=\(\frac{a}{b\left(a+b\right)}\)
a) \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6}{1-x}\)
\(=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{4x^2-3x+17+2x^2-x-2x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=-\frac{12}{x^2+x+1}\)
b) \(\frac{1}{x^2-x+1}-\frac{x^2+2}{x^3+1}+1=\frac{x+1-x^2-2+x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x-x^2+x^3}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x}{x+1}\)
c) \(N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{2017c}{ac+2017c+2017}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac+abc^2+abc}\)
\(N=\frac{1+b}{b+1+bc}+\frac{abc^2}{ac\left(1+bc+b\right)}\)
\(N=\frac{1+b}{b+1+bc}+\frac{bc}{1+bc+b}\)
\(N=\frac{1+b+bc}{b+1+bc}\)
\(N=1.\)
1.
a + b + c = 0 \(\Rightarrow\)a = - ( b + c ) \(\Rightarrow\)a2 = [ -( b + c ) ]2 \(\Rightarrow\)a2 = b2 + c2 + 2bc
Tương tự : b2 = a2 + c2 + 2ac ; c2 = a2 + b2 + 2ab
a + b + c = 0 \(\Rightarrow\)a3 + b3 + c3 = 3abc ( chứng minh )
Ta có : \(A=\frac{a^2}{b^2+c^2+2bc-b^2-c^2}+\frac{b^2}{a^2+c^2+2ac-a^2-c^2}+\frac{c^2}{a^2+b^2+2ab-a^2-b^2}\)
\(A=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\)
\(A=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
2. quy đồng mà giải
a) Ta có:
B = \(\frac{1}{x+3}-\frac{x}{x-1}-\frac{4x}{x^2+2x-3}\)
=> B = \(\frac{x-1}{\left(x+3\right)\left(x-1\right)}-\frac{x\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4x}{\left(x+3\right)\left(x-1\right)}\)
=> B = \(\frac{\left(x-1\right)-x\left(x+3\right)-4x}{\left(x+3\right)\left(x-1\right)}\)
=> B = \(\frac{x-1-x^2-3x-4x}{\left(x+3\right)\left(x-1\right)}\)
=> B = \(\frac{-6x-1-x^2}{\left(x+3\right)\left(x-1\right)}\)
b) xem lại đề
a) ( a + b )^3 - 3ab( a + b )
=(a+b)[(a+b)2-3ab]
=(a+b)[a2+2ab+b2-3ab]
=(a+b)(a2-ab+b2)
=a3+b3
b)
dùng hđt a3+b3=(a+b)(a2-ab+b2)