CMR nếu
\(c^2+2\left(ab-ac-bc\right)=0,b\ne c,a+b\ne c\) thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
Những câu hỏi liên quan
1.cho x+y+z=xyz và xy+yz+zx≠3
cmr: x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)/xy+yz+zx=xyz
2.cmr nếu c^2+2(ab-ac-bc)=0và b≠c,a+b≠c thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
3. cho a,b,c thỏa mãn abc≠0 và ab+bc+ca=0
tính :P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
1.cho x+y+z=xyz và xy+yz+zx≠3
cmr: x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)/xy+yz+zx=xyz
2.cmr nếu c^2+2(ab-ac-bc)=0và b≠c,a+b≠c thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
3. cho a,b,c thỏa mãn abc≠0 và ab+bc+ca=0
tính :P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Em(mình) thử nhé, ko chắc đâu
3/ Ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=\left[ab\left(a+b\right)+abc\right]+\left[bc\left(b+c\right)+abc\right]+\left[ca\left(c+a\right)+ca\right]-abc\)
\(=\left(a+b+c\right)ab+\left(a+b+c\right)bc+\left(a+b+c\right)ca-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)= -abc
Suy ra \(P=\frac{-abc}{abc}=-1\)
Vậy..
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chứng minh rằng nếu \(c^2+2\left(ab-ac-bc\right)=0;b\ne c;a+b\ne c\) thì:
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
Do \(c^2+2\left(ab-ac-bc\right)=0\Leftrightarrow-c^2=2\left(ab-ac-bc\right)\)
Ta có; \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+c^2-c^2+\left(a-c\right)^2}{b^2+c^2-c^2+\left(b-c\right)^2}=\frac{a^2+c^2+2\left(ab-ac-bc\right)+\left(a-c\right)^2}{b^2+c^2+2\left(ab-ac-bc\right)+\left(b-c\right)^2}\)
\(=\frac{2\left(a-c\right)^2+2\left(ab-bc\right)}{2\left(b-c\right)^2+2\left(ab-ac\right)}=\frac{2\left(a-c\right)^2+2b\left(a-c\right)}{2\left(b-c\right)^2+2a\left(b-c\right)}=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}\)
\(=\frac{a-c}{b-c}\) (đpcm)
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Chứng minh rằng nếu \(c^2+2.\left(ab-ac-bc\right)=0\)và \(b\ne c\), \(a+b\ne c\)thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
Cho \(c^2+2\left(ab-ac-bc\right)=0;b\ne c;a+b\ne c\)thì
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
thế cuối cùng đề bài là gì'.'???????
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Ta có: \(c^2+2\left(ab-ac-bc\right)=0\)
\(\Rightarrow c^2=-2\left(ab-ac-bc\right)\)
Thay vào
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+a^2-2ac-2\left(ab-ac-bc\right)}{b^2+b^2-2bc-2\left(ab-ac-bc\right)}=\frac{2a^2-2ab+2bc}{2b^2-2ab+2ac}=\frac{a^2-ab+bc}{b^2-ab+ac}\)
\(\frac{a-c}{b-c}=\frac{a^2-2ac-2\left(ab-ac-bc\right)}{b^2-2bc-2\left(ab-ac-bc\right)}=\frac{a^2-2ab+2bc}{b^2-2ab+2ac}\)
=> ...
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Chứng minh rằng nếu \(c^2+2\left(ab-ac-bc\right)=0\) với \(b\ne c\) và \(\left(a+b\right)\ne c\) thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
m.n giúp mk nha!!!
Cho ba số thực a, b, c thỏa mãn: b≠c, a+b≠c, c2=2(ac+bc-ab). CMR: \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
Ta có \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+c^2+2ab-2ac-2bc+\left(a-c\right)^2}{b^2+c^2+2ab-2ac-2bc+\left(b-c\right)^2}\)
\(=\frac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}=\frac{\left(a-c\right)\left(a-c+2b+a-c\right)}{\left(b-c\right)\left(b-c+2a+b-c\right)}=\frac{\left(a-c\right)\left(2a+2b-2c\right)}{\left(b-c\right)\left(2a+2b-2c\right)}=\frac{a-c}{b-c}\)
⇒điều phải chứng minh
Biết \(a\ne-b,b\ne-c,c\ne-a\). CMR:
\(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}+\frac{-a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}\)
Bài 13: Biết \(a\ne-b;b\ne-c;c\ne-a\). CMR:
\(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}+\frac{a^2-b^2}{\left(c+a\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}\)
Lời giải:
\(\text{VT}=\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}=\left(\frac{b}{b+c}-\frac{b}{a+b}\right)+\left(\frac{c}{c+a}-\frac{c}{c+b}\right)+\left(\frac{a}{a+b}-\frac{a}{a+c}\right)\)
\(=\frac{b(a-c)}{(b+c)(a+b)}+\frac{c(b-a)}{(c+a)(c+b)}+\frac{a(c-b)}{(a+b)(a+c)}\)
\(=\frac{b(a-c)(a+c)+c(b-a)(b+a)+a(c-b)(c+b)}{(a+b)(b+c)(c+a)}=\frac{b(a^2-c^2)+c(b^2-a^2)+a(c^2-b^2)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)}{(a+b)(b+c)(c+a)}(*)\)
Và:
\(\text{VP}=\frac{(b^2-c^2)(b+c)+(c^2-a^2)(c+a)+(a^2-b^2)(a+b)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)}{(a+b)(b+c)(c+a)}(**)\)
Từ $(*); (**)\Rightarrow $ đpcm