rút gọn bt
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(c+b\right)}\)
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rút gọn biểu thức
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ca}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
Afrac{a^2+bc}{b+ac}+frac{b^2+ca}{c+ab}+frac{c^2+ab}{a+bc}frac{3left(a^2+bcright)}{left(a+b+cright)b+3ac}+frac{3left(b^2+caright)}{left(a+b+cright)c+3ab}+frac{3left(c^2+abright)}{left(a+b+cright)a+3bc}gefrac{3left(a^2+bcright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(b^2+caright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(c^2+abright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}3
Đọc tiếp
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
rút gọn a) \(\frac{1}{a\left(a-b\right)\left(a-c\right)}+\frac{1}{b\left(b-a\right)\left(b-c\right)}\)
b) \(A=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}+\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Giải chi tiết vs nói hướng giải bt luôn nha
Rút gọ biểu thức
\(P=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
\(P=\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(b+c\right)\left(b+a\right)}+\frac{c^2-ab}{\left(c+a\right)\left(c+b\right)}\)
\(P=\frac{\left(a^2-bc\right)\left(b+c\right)}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\frac{\left(b^2-ac\right)\left(c+a\right)}{\left(b+c\right)\left(b+a\right)\left(c+a\right)}+\frac{\left(c^2-ab\right)\left(b+a\right)}{\left(c+a\right)\left(c+b\right)\left(b+a\right)}\)
\(P=\frac{a^2b+a^2c-b^2c-bc^2}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}+\frac{b^2a+b^2c-a^2c-ac^2}{\left(b+c\right)\left(b+a\right)\left(c+a\right)}+\frac{c^2a+c^2b-a^2b-b^2a}{\left(c+a\right)\left(c+b\right)\left(b+a\right)}\)
\(P=\frac{0}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)
\(P=0\)
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Xét: \(f\left(x\right)=\frac{x^2-bc}{\left(x+b\right)\left(x+c\right)}+\frac{b^2-xc}{\left(b+c\right)\left(b+x\right)}+\frac{c^2-xb}{\left(c+x\right)\left(c+b\right)}\)
\(\Rightarrow f\left(a\right)=P\)
Ta có: \(f\left(b\right)=\frac{b^2-bc}{2b\left(b+c\right)}+\frac{b^2-bc}{2b\left(b+c\right)}+\frac{c^2-b^2}{\left(c+b\right)\left(c+b\right)}\)
\(\Rightarrow f\left(b\right)=\frac{2b\left(b-c\right)}{2b\left(b+c\right)}+\frac{\left(c-b\right)\left(c+b\right)}{\left(c+b\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-b}{c+b}=0\left(1\right)\)
Chứng minh tương tự ta cũng có: \(f\left(c\right)=0\left(2\right)\)
Từ (1) và (2) suy ra \(f\left(x\right)=0\left(\forall x\right)\Rightarrow f\left(a\right)=0\left(\forall x\right)\)
Vậy A =0
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mình ghi sai rồi \(P=1\)mới đúng chứ kg phải A =1
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Xem thêm câu trả lời
Rút gọn
\(M=\frac{2}{a-b}+\frac{2}{bc}+\frac{2}{c-a}+\frac{\left(a+b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\cdot\left(b-c\right)\cdot\left(c-a\right)}\)
giúp mình với m.n ơi.thanks trước nha!!!
Rút gọn phân thức sau:
Q=\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}\)+\(\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}\)+\(\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)=\left(a+b\right)\left(a-b\right)+c\left(a-b\right)=\)\(\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)
\(c^2+bc-a^2-ab=\left(c-a\right)\left(a+b+c\right)\)
\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
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Rút gọn biểu thức:
A=\(\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-c\right)\left(b-a\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
Giúp mik tick cho.
rút gọn :
\(\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ca}{\left(b-c\right)\left(b-a\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{-bc\left(b-c\right)}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}+\frac{-ca\left(c-a\right)}{\left(b-c\right)\left(a-b\right)\left(c-a\right)}+\frac{-ab\left(a-b\right)}{\left(c-a\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{-b^2c+bc^2-c^2a+ca^2-a^2b+ab^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{b^2\left(a-c\right)+ca\left(a-c\right)-b\left(a-c\right)\left(a+c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(a-c\right)\left(b^2+ca-ba-bc\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(a-c\right)\left[b\left(b-a\right)-c\left(b-a\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(a-c\right)\left(b-c\right)\left(b-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
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frac{a}{b-c}+frac{b}{c-a}+frac{c}{a-b}0Rightarrowfrac{a}{b-c}-left(frac{b}{c-a}+frac{c}{a-b}right)Rightarrowfrac{a}{b-c}-frac{ab-b^2+c^2-ac}{left(c-aright)left(a-bright)}Rightarrowfrac{a}{left(b-cright)^2}frac{b^2-ab-c^2+ac}{left(a-bright)left(b-cright)left(c-aright)}Tương tự:frac{b}{left(c-aright)^2}frac{c^2-bc+ba-a^2}{left(a-bright)left(b-cright)left(c-aright)};frac{c}{left(a-bright)^2}frac{a^2-ac+bc-b^2}{left(a-bright)left(b-cright)left(c-aright)}Cộng lại:frac{a}{left(b-cright)^2}+frac{b}{lef...
Đọc tiếp
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=-\left(\frac{b}{c-a}+\frac{c}{a-b}\right)\)
\(\Rightarrow\frac{a}{b-c}=-\frac{ab-b^2+c^2-ac}{\left(c-a\right)\left(a-b\right)}\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab-c^2+ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Tương tự:
\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)};\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Cộng lại:
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
P/S:Đây nha ! Bài này lớp 7 chắc xem xong key cũng chắc là đang khó hiểu nhỉ ? Đưa giấy bút ra rồi nháp vài cái là hiểu ngay thôi !
Có người nhờ giải ấy @gunny :33
