Violympic toán 8
Các câu hỏi tương tự
Chứng minh đẳng thức:
\(\dfrac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\dfrac{1}{\left(c-a\right)\left(b^2+ba-c^2-ca\right)}+\dfrac{1}{\left(a-b\right)\left(c^2+cb-a^2-ab\right)}=0\)
21 tháng 10 2018 lúc 12:10
Xét:
dfrac{c}{a-b}.left(dfrac{a-b}{c}+dfrac{b-c}{a}+dfrac{c-a}{b}right)1+dfrac{c}{a-b}left(dfrac{b-c}{a}+dfrac{c-a}{b}right)1+dfrac{c}{a-b}.dfrac{b^2-bc+ac-a^2}{ab}1+dfrac{c}{a-b}.dfrac{cleft(a-bright)-left(a^2-b^2right)}{ab}1+dfrac{c}{a-b}.dfrac{left(c-a-bright)left(a-bright)}{ab}1+dfrac{c^2-cleft(a+bright)}{ab}1+dfrac{2c^2}{ab}1+dfrac{2c^3}{abc}
CMTT cộng theo vế:
BTCCM3+dfrac{2left(a^3+b^3+c^3right)}{abc}dfrac{6left(a^3+b^3+c^3right)}{3abc}
Mà Khi a+b+c0 thì a^3+b^3+c^33abc ( tự cm,ez)...
Đọc tiếp
Xét:
\(\dfrac{c}{a-b}.\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}\left(\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}=1+\dfrac{c}{a-b}.\dfrac{c\left(a-b\right)-\left(a^2-b^2\right)}{ab}=1+\dfrac{c}{a-b}.\dfrac{\left(c-a-b\right)\left(a-b\right)}{ab}=1+\dfrac{c^2-c\left(a+b\right)}{ab}=1+\dfrac{2c^2}{ab}=1+\dfrac{2c^3}{abc}\)
CMTT cộng theo vế:
\(BTCCM=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}=\dfrac{6\left(a^3+b^3+c^3\right)}{3abc}\)
Mà Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\) ( tự cm,ez)
Vậy \(BTCCM=3+6=9\left(đpcm\right)\)
CMR nếu \(\left(a^2-bc\right).\left(b-abc\right)=\left(b^2-ac\right).\left(a-abc\right)\) và các số a, b, c, a-b khác 0 thì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
Rút gọn biểu thức: \(A=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}+\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)
Cho a + b + c = 1 (a,b,c khác 1,2). Chứng minh
\(\dfrac{c+ab}{a^2+b^2+abc-1}+\dfrac{a+bc}{b^2+c^2+abc-1}+\dfrac{b+ac}{a^2+c^2+abc-1}=\dfrac{bc+ac+ab+8}{\left(a-2\right)\left(b-2\right)\left(a-2\right)}\)
rút gon biểu thức:\(A=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ca}{\left(b-c\right)\left(b-a\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
Rút gọn biểu thức: \(B=\left(ab+bc+ca\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
rút gọn
\(\dfrac{-a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\dfrac{\left(b-c\right)\left(1+a\right)^2}{x+a^2}+\dfrac{\left(c-a\right)\left(1+b\right)^2}{x+b^2}+\dfrac{\left(a-b\right)\left(1+c\right)^2}{x+c^2}=0\)