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Đề đúng đây chứ ?
\(3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)
\(=3\left[\left(x^2+y^2+2xy\right)-2xy\right]-2\left[\left(x^3+3x^2y+3xy^2+y^3\right)-3xy\left(x+y\right)\right]\)
\(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\)
Thay \(x+y=1;\)có :
\(=3\left(1-2xy\right)-2\left(1-3xy\right)\)
\(=3-6xy-2+6xy=3-2=1\)
Vậy ...
\(\frac{\left(a-b+c\right)^2}{4}+\frac{\left(a+b-c\right)^2}{4}+\frac{\left(-a+b+c\right)^2}{4}\)
\(=\frac{\left(a+b+c-2b\right)^2}{4}+\frac{\left(a+b+c-2c\right)^2}{4}+\frac{\left(a+b+c-2a\right)^2}{4}\)
\(=\frac{\left(4m-2b\right)^2}{4}+\frac{\left(4m-2c\right)^2}{4}+\frac{\left(4m-2a\right)^2}{4}\)
\(=\frac{16m^2+4b^2-16bm}{4}+\frac{16m^2+4c^2-16cm}{4}+\frac{16m^2+4a^2-16am}{4}\)
\(=4m^2+b^2-4bm+4m^2+c^2-4cm+4m^2+a^2-4am\)
\(=12m^2+b^2+c^2+a^2-4m\left(a+b+c\right)\)
\(=12m^2+b^2+c^2+a^2-4m\left(4m\right)\)
\(=a^2+b^2+c^2-4m^2\)
Chắc hết rồi nhỉ :/
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
Ta có
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(a-c\right)\left(b-a\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)
Tương tự
\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ab-a^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)
\(\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{b^2-ab+ac-c^2+c^2-bc+ab-a^2+a^2-ac+bc-b^2}{\left(a-c\right)\left(b-a\right)\left(a-b\right)}\)
=0 ( ĐPCM)
Lời giải:
Ta có: \(a+b+c=4m\Rightarrow a+b-c=4m-2c\)
\(\Rightarrow \frac{a+b-c}{2}=2m-c\)
Hoàn toàn tương tự với các phân thức còn lại, suy ra:
\(\left(\frac{a+b-c}{2}\right)^2+\left(\frac{a-b+c}{2}\right)^2+\left(\frac{-a+b+c}{2}\right)^2=(2m-c)^2+(2m-b)^2+(2m-a)^2\)
\(=4m^2+c^2-4mc+4m^2+b^2-4mb+4m^2+a^2-4ma\)
\(=12m^2+a^2+b^2+c^2-4m(a+b+c)\)
\(=12m^2+a^2+b^2+c^2-4m.4m=a^2+b^2+c^2-4m^2\)
Ta có đpcm
\(\left(\frac{a+b-c}{2}\right)^2+\left(\frac{b+c-a}{2}\right)^2+\left(\frac{c+a-b}{2}\right)^2\)
\(=\left(\frac{a+b+c-2c}{2}\right)^2+\left(\frac{a+b+c-2a}{2}\right)^2+\left(\frac{a+b+c-2b}{2}\right)^2\)
\(=\left(2m-c\right)^2+\left(2m-a\right)^2+\left(2m-b\right)^2\)
\(=4m^2-4mc+c^2+4m^2-4ma+a^2+4m^2-4mb+b^2\)
\(=a^2+b^2+c^2+12m^2-4m\left(a+b+c\right)\)
\(=a^2+b^2+c^2+12m^2-16m^2\)
\(=a^2+b^2+c^2-4m^2\)