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Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
=> \(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
=> \(\frac{a+b}{ab}=\frac{-\left(a+b\right)}{\left(a+b+c\right).c}\)
Khi a + b = 0
=> (a + b)(b + c)(c + a) = 0 (2)
Nếu a + b \(\ne0\)
=> ab = -(a + b + c).c
=> ab + (a + b + c).c = 0
=> ab + ac + bc + c2 = 0
=> (a + c)(b + c) = 0
=> (a + b)(b + c)(a + c) = 0 (1)
Từ (2)(1) => (a + b)(b + c)(a + c) = 0 \(\forall a;b;c\)
=> a = -b hoặc b = -c hoặc = c = -a
Nếu a = -b => a11 = -b11 => a11 + b11 = 0
=> P = 0 (3)
Nếu b = -c => b9 = - c9 => b9 + c9 = 0
=>P = 0 (4)
Nếu c = -a => c2001 = -a2001 => c2001 + a2001 = 0
=> P = 0 (5)
Từ (3);(4);(5) => P = 0 trong cả 3 trường hợp
Vạy P = 0
a) 9x2 - 36
=(3x)2-62
=(3x-6)(3x+6)
=4(x-3)(x+3)
b) 2x3y-4x2y2+2xy3
=2xy(x2-2xy+y2)
=2xy(x-y)2
c) ab - b2-a+b
=ab-a-b2+b
=(ab-a)-(b2-b)
=a(b-1)-b(b-1)
=(b-1)(a-b)
P/s đùng để ý đến câu trả lời của mình
dễ!Ta có:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\)
Chứng minh tương tự,Ta được:
\(\hept{\begin{cases}\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{a-b}+\frac{1}{b-c}\\\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}+\frac{1}{b-c}\end{cases}}\)
\(\Rightarrow\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)\(\Rightarrow\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}=\frac{2013}{2}\)
Xong!
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
\(\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)
\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{a}{b}=1;\frac{b}{c}=1;\frac{c}{a}=1\)
\(\Rightarrow M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)
Vì a+b+c=1 nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)
\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{a}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)=2+\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}\)
Do đó
\(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\left(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{ab}\right)+\left(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{bc}\right)+\left(\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}\right)+\frac{3}{4}\)
\(\ge2\sqrt{\frac{ab}{a^2+b^2}\cdot\frac{a^2+b^2}{ab}}+2\sqrt{\frac{bc}{c^2+b^2}\cdot\frac{c^2+b^2}{bc}}+2\sqrt{\frac{ca}{a^2+c^2}+\frac{c^2+a^2}{ca}}+\frac{3}{4}\)
\(=2\cdot\frac{1}{2}+2\cdot\frac{1}{2}+\frac{2}{3}=\frac{15}{4}\)
Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{c-a-b-c}{c\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-\left(a+b\right)}{c\left(a+b+c\right)}\)
\(\Leftrightarrow c\left(a+b\right)\left(a+b+c\right)=-ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2\right)+ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
TH1 : \(a+b=0\Leftrightarrow a=-b\)
\(M=\left(-b^{15}+b^{15}\right)\left(b^4+c^4\right)\left(c^{2016}+a^{2016}\right)\)
\(M=0\left(b^4+c^4\right)\left(c^{2016}+a^{2016}\right)=0\)
TH2 : \(b+c=0\Leftrightarrow b=-c\)
Đến đây tịt :) bác nào biết giải tiếp giúp Nghị Hồng Vân Anh
đề cho a,b trái dấu rồi nên có một trường hợp thôi nha Trần Thanh Phương, cảm ơn bạn