Cho a > b > c biết : \(2\left(a^2+b^2\right)5ab\)
\(\left(a+b\right)^2=a^2+b^2+2ab\)
\(\left(a-b\right)^2=a^2+b^2-2ab\)
Tính \(\frac{3a-b}{2a+b}\)
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9 tháng 8 2019
\(=\left(a+b-c\right)\left(a-b\right)^2\) nha !
P/S:Ko có mục đích xấu,đăng lên cho bạn thôi.
7 tháng 8 2018
a) \(a^2+b^2=\left(a+b\right)^2-2ab\)
\(VP=\left(a+b\right)^2-2ab=a^2+2ab+b^2-2ab\)\(=a^2+b^2=VT\)
\(\Rightarrowđpcm\)
b)\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2\)
\(VP=a^4+b^4+2a^2b^2-2a^2b^2=a^4+b^4=VT\)\(\Rightarrowđpcm\)
c) \(a^6+b^6=\left(a^2+b^2\right)\left[\left(a^2+b^2\right)^2-3a^2b^2\right]\)
\(VP=\left(a^2+b^2\right)\left(a^4-a^2b^2+b^4\right)=a^6+b^6\)
\(VP=VT\Rightarrowđpcm\)
d)\(a^6-b^6=\left(a^2-b^2\right)[\left(a^2+b^2\right)^2-a^2b^2]\)
\(VP=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)=a^6-b^6=VT\)
\(VP=VT\Rightarrowđpcm\)
26 tháng 1 2017
a) \(A=\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)
\(=\frac{2bc+b^2+c^2-a^2}{2bc}.\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{b^2+c^2-b^2+2bc-c^2}{a+b+c}\)
\(=\frac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}.\frac{a+b+c}{b+c-a}.\frac{2bc}{a+b+c}\)
\(=a+b+c\)
b) \(B=\frac{\frac{3a}{a+b}}{\frac{2a}{a^2-2ab+b^2}}\)\(=\frac{3a}{a+b}.\frac{\left(a-b\right)^2}{2a}=\frac{3\left(a-b\right)^2}{2\left(a+b\right)}\)
c) \(C=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-\frac{b}{a}}:\frac{\frac{a^2}{b^2}-\frac{b^2}{a^2}}{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\frac{\frac{a^2+b^2}{ab}}{\frac{a^2-b^2}{ab}}:\frac{\frac{a^4-b^4}{a^2b^2}}{\frac{\left(a+b\right)^2}{a^2b^2}}\)
\(=\frac{a^2+b^2}{a^2-b^2}.\frac{\left(a+b\right)^2}{a^4-b^4}\)
\(=\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)\left(a-b\right)\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)}\)
\(=\frac{1}{\left(a-b\right)^2}\)
M
1
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2019
\(3a-b=3m\Rightarrow b=3a-3m\)
\(\Rightarrow2a+\left(3a-3m\right)=7m\Rightarrow5a=10m\Rightarrow a=2m\)
\(\Rightarrow b=3a-3m=6m-3m=3m\)
\(\Rightarrow P=\frac{\left(2m\right)^2-2.2m.3m}{\left(2m\right)^2+\left(3m\right)^2}=\frac{4m^2-12m^2}{4m^2+9m^2}=\frac{-8m^2}{13m^2}=\frac{-8}{13}\)
7 tháng 8 2016
biến đổi vế trái : a. \(\left(a+b\right)^2=a^2+2ab+B^2=VP\)
b. \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3=VP\)
c. \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=VP\)
xem 7 hằng đẳng thức đáng nhớ
DD
7 tháng 8 2016
a)\(=\left(a+b\right)^2=\left(a+b\right)\left(a+b\right)=a^2+ab+ab+b^2\)
\(=a^2+2ab+b^2\)
b)\(\left(a-b\right)^3=\left(a-b\right)\left(a-b\right)\left(a-b\right)=\left(a^2-ab-ab+b^2\right)\left(a-b\right)\)
\(=\left(a^2-2ab+b^2\right)\left(a-b\right)\)
\(=a^3-a^2b-2a^2b+2ab^2+ab^2-b^3\)
\(=a^3-3a^2b-3ab^2-b^3\)
c)\(\left(a+b+c\right)^2=\left(a+b+c\right)\left(a+b+c\right)\)
\(=a^2+ab+ac+ab+b^2+bc+ac+cb+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac\)
Ta có:
\(2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow2a^2-4ab-ab+2b^2=0\)
\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)
\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)
\(\Leftrightarrow a=2b\) hay \(b=2a\)
Vì \(a>b>c\Leftrightarrow a=2b\)
\(\Leftrightarrow\frac{3a-b}{2a+b}=\frac{3.2b-b}{2.2b+b}=\frac{5b}{5b}=1\)
Vậy \(\frac{3a-b}{2a+b}=1\)