Thực hiện phép tính
C=\(\frac{bc}{\left(a-b\right).\left(a-c\right)}+\frac{ac}{\left(b-a\right).\left(b-c\right)}+\frac{ab}{\left(c-a\right).\left(c-b\right)}\)
Thực hiện phép tính :
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Ta có:
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)\)
\(=\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b+c\right)\)(1)
\(b^2+ab-c^2-ac=\left(b^2-c^2\right)+\left(ab-ac\right)\)
\(=\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\)
\(=\left(b-c\right)\left(a+b+c\right)\)(2)
\(c^2+bc-a^2-ab=\left(c^2-a^2\right)+\left(bc-ab\right)\)
\(=\left(c-a\right)\left(a+c\right)+b\left(c-a\right)\)
\(=\left(c-a\right)\left(a+b+c\right)\)(3)
Ta có : \(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)\(+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)\(+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)(*)
Thế (1),(2),(3) vào (*)
=>\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(\Leftrightarrow\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
Dễ thôi bạn chỉ cần quy đồng thôi
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\)\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
=\(\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
=\(\frac{c-a+a-b+b-c}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}=0\)
Thực hiện phép tính :
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Ta có :\(\left(a-b\right)\left(c^2+bc-a^2-ab\right)=\left(a-b\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)
Tương tự : \(\left(b-c\right)\left(a^2+ac-b^2-bc\right)=\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
\(\left(c-a\right)\left(b^2+ab-c^2-ac\right)=\left(c-a\right)\left(b-c\right)\left(a+b+c\right)\)
\(MTC=\left(a-b\right)\left(b-c\right)\left(c-s\right)\left(a+b+c\right)\)
Kí hiệu biểu thức đã cho bởi \(Q\),ta có :
\(Q=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
A=\(\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\)\(\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
thực hiện phép tính
Lời giải :
\(A=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{-bc\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ac\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=\frac{-bc\left(b-c\right)-ac\left(c-a\right)-ab\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Xét tử số :
\(TS=-b^2c+bc^2-ac^2+a^2c-a^2b+ab^2\)
\(=-ab\left(a-b\right)-c^2\left(a-b\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(-ab-c^2+ac+bc\right)\)
\(=\left(a-b\right)\left[-a\left(b-c\right)+c\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Khi đó \(A=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
Thực hiện phép tính sau:
\(P=\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
Xét \(\left(b-c\right)\left(a^2+ac-b^2-bc\right)=\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(\left(c-a\right)\left(b^2+ab-c^2-ac\right)=\left(c-a\right)\left(b-c\right)\left(a+b+c\right)\)
\(\left(a-b\right)\left(c^2+bc-a^2-ab\right)=\left(a-b\right)\left(c-a\right)\left(a+b+c\right)\)
=> \(P=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(b-c\right)\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}=0\)
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Cho \(a,b,c\)là số thực khác 0 thỏa mãn đẳng thức \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=8abc\)Chứng minh rằng :
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{3}{4}+\frac{ab}{\left(a+b\right)\left(b+c\right)}+\frac{bc}{\left(b+c\right)\left(c+a\right)}+\frac{ac}{\left(c+a\right)\left(a+b\right)}\)
Trần Hữu Ngọc Minh bn tham khảo nha:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{a+b}=\frac{b}{b+c}=\frac{c}{c+a}=\frac{a+b+c}{"b+c"+"a+c"+"a+b"}=\frac{a+b+c}{2."a+b+c"}\)
Xét 2 trường hợp, ta có:
\(\cdot TH1:a+b+c=0\)thì \(\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}}\)
Có: \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-1+-1+-1=-3\)
Không phụ thuộc vào các giá trị a,b,c 1:
\(\cdot TH2:a+b+c\ne0\)thì \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2."a+b+c"}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}2a=b+c\\2b=a+c\\2c=a+b\end{cases}}\)
Có: \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}\)
Không phụ thuộc vào các giá trị a,b,c 2
Từ 1 và 2 \(\Rightarrow\)đpcm
tính
\(B=\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}+\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}+\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}\)
(với a, b, c là số thực và a+b+c=1)
Do a + b + c = 1 nên \(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}=\frac{\sqrt{\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ca\right]}}{\sqrt{c\left(a+b+c\right)+ab}}\)
\(=\frac{\sqrt{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}}{\sqrt{ac+bc+c^2+ab}}=\frac{\sqrt{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\left(a+b\right)^2}=a+b\) (1)
Tương tự \(\hept{\begin{cases}\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}=b+c\text{ }\left(2\right)\\\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=a+c\text{ }\left(3\right)\end{cases}}\)
Cộng vế với vế của (1)(2)(3) lại ta được :
\(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}+\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}+\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=2\left(a+b+c\right)=2\)
Cho a,b,c là các số thực dương thỏa mãn a+b+c=3abc. Chứng minh rằng :
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\left[\frac{a^4}{\left(ab+1\right)\left(ac+1\right)}+\frac{b^4}{\left(bc+1\right)\left(ab+1\right)}+\frac{c^4}{\left(ca+1\right)\left(bc+1\right)}\right]\ge\frac{27}{4}\)
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THỰC HIỆN PHÉP TÍNH
\(B=\frac{1}{a\left(a-b\right)\left(a-c\right)}+\frac{1}{b\left(b-a\right)\left(b-c\right)}+\frac{1}{c\left(c-a\right)\left(c-b\right)}\)
Đặt
\(\Rightarrow\hept{\begin{cases}x=a-b\\y=a-c\\z=b-c\end{cases}}\)
Ta được
\(B=\frac{1}{axy}+\frac{1}{bxz}+\frac{1}{cyz}=\frac{bcz-acy+abx}{abcxyz}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-b+b-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc\left(b-c\right)-ac\left(a-b\right)-ac\left(b-c\right)+ab\left(a-b\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{c\left(b-c\right)\left(b-a\right)+a\left(a-b\right)\left(b-c\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(b-c\right)\left(a-b\right)\left(a-c\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{1)}{abc}\)
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