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a, ĐK: \(a\ne0,b\ne0,a+b\ne0\)
\(A=\left[\frac{1}{a^2}+\left(\frac{1}{a}+\frac{1}{b}\right):\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{a+b}{ab}:\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\frac{\left(a+b\right)^2}{a^2b^2}.\frac{a^2b^2}{\left(a+b\right)\left(a^2-ab+b^2\right)}.\frac{1}{a+b}\)
\(=\frac{1}{a^2-ab+b^2}\)
b, \(a^2-ab+b^2=\left(a-\frac{1}{2}b\right)^2+\frac{3}{4}b^2>0\left(a,b\ne0\right)\)
\(\Rightarrow A=\frac{1}{a^2-ab+b^2}>0\forall a;b\)
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
\(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)
\(=\left[\left(\frac{1}{2}a+b\right)+\left(\frac{1}{2}a-b\right)\right].\left[\left(\frac{1}{2}a+b\right)^2-\left(\frac{1}{2}a+b\right).\left(\frac{1}{2}a-b\right)+\left(\frac{1}{2}a-b\right)^2\right]\)
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 \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
a) \(B=\frac{1}{x+3}+\frac{x}{x-1}-\frac{4x}{x^2+2x-3}=\frac{x-1}{x^2+2x-3}+\frac{x^2+3x}{x^2+2x-3}-\frac{4x}{x^2+2x-3}\)
\(\Leftrightarrow B=\frac{x-1+x^2+3x-4x}{x^2+2x-3}=\frac{x^2-1}{x^2+2x+1-4}=\frac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2-2^2}\)
\(\Leftrightarrow B=\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{x+1}{x+3}\)
b) \(\frac{A-1}{B}=\frac{\frac{x-1}{x+3}-1}{\frac{x+1}{x+3}}=\frac{\frac{-4}{x+3}}{\frac{x+1}{x+3}}=\frac{-4}{x+1}\le\frac{1}{2}\Leftrightarrow-8\le x+1\Leftrightarrow x\ge-9\)
1. a + b + c = 0 \(\Rightarrow\)a + b = -c \(\Rightarrow\)( a + b )2 = ( -c )2 \(\Rightarrow\)a2 + b2 - c2 = -2ab
Tương tự : b2 + c2 - a2 = -2bc ; c2 + a2 - b2 = -2ac
Ta có : \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)
\(=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{-1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(=\frac{-1}{2}\left(\frac{a+b+c}{abc}\right)=0\)
2. tương tự
3,4 . có ở dưới, câu hỏi của Quyết Tâm chiến thắng
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)