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Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)
\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)
\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)
\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)
Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)
\(\Leftrightarrow A\ne0\forall x;y\)
Lần sau đăng từng bài thôi bạn nhé :)
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 :/