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ĐKXĐ : \(x,y\ne0\)\(;\)\(x\ne y\)
\(a)\) \(P=\frac{2}{x}-\left(\frac{x^2}{x^2-xy}+\frac{x^2-y^2}{xy}-\frac{y^2}{y^2-xy}\right):\frac{x^2-xy+y^2}{x-y}\)
\(P=\frac{2}{x}-\left(\frac{x^2y}{xy\left(x-y\right)}+\frac{\left(x-y\right)^2\left(x+y\right)}{xy\left(x-y\right)}+\frac{xy^2}{xy\left(x-y\right)}\right):\frac{x^2-xy+y^2}{x-y}\)
\(P=\frac{2}{x}-\left(\frac{xy\left(x+y\right)+\left(x-y\right)^2\left(x+y\right)}{xy\left(x-y\right)}\right):\frac{x^2-xy+y^2}{x-y}\)
\(P=\frac{2}{x}-\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{xy\left(x-y\right)}.\frac{x-y}{x^2-xy+y^2}\)
\(P=\frac{2y}{xy}-\frac{x+y}{xy}=\frac{y-x}{xy}\)
\(b)\)
+) Với \(\left|2x-1\right|=1\)\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)
Mà \(x\ne0\) ( ĐKXĐ ) nên \(x=1\)
+) Với \(\left|y+1\right|=\frac{1}{2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y+1=\frac{1}{2}\\y+1=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{-1}{2}\\y=\frac{-3}{2}\end{cases}}}\)
Thay \(x=1;y=\frac{-1}{2}\) vào \(A=\frac{y-x}{xy}\) ta được : \(A=\frac{\frac{-1}{2}-1}{1.\frac{-1}{2}}=\frac{\frac{-3}{2}}{\frac{-1}{2}}=3\)
Thay \(x=1;y=\frac{-3}{2}\) vào \(A=\frac{y-x}{xy}\) ta được : \(A=\frac{\frac{-3}{2}-1}{1.\frac{-3}{2}}=\frac{\frac{-5}{2}}{\frac{-3}{2}}=\frac{15}{4}\)
Vậy ...
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\frac{4y^2-\left(x-y\right)^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{x\left(x-2y\right)-2\left(x^2-xy\right)}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{3y^2+2xy-x^2}{y^2\left(x-y\right)}.\frac{y^2-xy}{x-3y}+\frac{-x^2}{2\left(x-2y\right)}.\frac{2x-4y}{xy+y^2}\)
\(=\frac{\left(x+y\right)\left(3y-x\right)}{y^2\left(x-y\right)}.\frac{y\left(y-x\right)}{x-3y}-\frac{x^2}{2\left(x-2y\right)}.\frac{2\left(x-2y\right)}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)}{y}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}=\frac{2xy+y^2}{y\left(x+y\right)}=\frac{2x+y}{x+y}\)
Giờ chỉ cần thế x, y vô nữa là xong nhé.
\(A=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y^2-xy}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x^2-xy}{x-2y}\right):\frac{xy+y^2}{2x-4y}\)
\(=\left(\frac{4}{x-y}-\frac{x-y}{y^2}\right).\frac{y\left(y-x\right)}{x-3y}\)\(+\left(\frac{x}{2}-\frac{x\left(x-y\right)}{x-2y}\right):\frac{y\left(x+y\right)}{2\left(x-2y\right)}\)
\(=\frac{4y\left(y-x\right)}{\left(x-y\right)\left(x-3y\right)}-\frac{\left(x-y\right)y\left(y-x\right)}{y^2\left(x-3y\right)}\)\(+\frac{x.2\left(x-2y\right)}{2.y\left(x+y\right)}-\frac{x\left(x-y\right).2\left(x-2y\right)}{\left(x-2y\right).y\left(x+y\right)}\)
\(=\frac{-4y}{x-3y}+\frac{\left(x-y\right)^2}{y\left(x-3y\right)}+\frac{x\left(x-2y\right)}{y\left(x+y\right)}-\frac{2x\left(x-y\right)}{y\left(x+y\right)}\)
\(=\frac{-4y^2+x^2-2xy+y^2}{y\left(x-3y\right)}+\frac{x^2-2xy-2x^2+2xy}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy-3y^2}{y\left(x-3y\right)}+\frac{-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2+xy-3xy-3y^2}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x\left(x+y\right)-3y\left(x+y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(\frac{\left(x+y\right)\left(x-3y\right)}{y\left(x-3y\right)}-\frac{x^2}{y\left(x+y\right)}\)
\(=\frac{x+y}{y}-\frac{x^2}{y\left(x+y\right)}=\frac{\left(x+y\right)^2-x^2}{y\left(x+y\right)}\)
\(=\frac{x^2-2xy+y^2-x^2}{y\left(x+y\right)}=\frac{-2xy+y^2}{y\left(x+y\right)}\)
\(=\frac{y\left(y-2x\right)}{y\left(x+y\right)}=\frac{y-2x}{x+y}\)
Thay \(x=\frac{1}{2};y=\frac{1}{3}\)vào A ta có :
\(A=\frac{\frac{1}{3}-2.\frac{1}{2}}{\frac{1}{2}+\frac{1}{3}}=\frac{\frac{1}{3}-1}{\frac{3}{6}+\frac{2}{6}}=\frac{2}{3}:\frac{5}{6}=\frac{2.6}{3.5}=\frac{4}{5}\)
Vậy \(A=\frac{4}{5}\)tại \(x=\frac{1}{2};y=\frac{1}{3}\)
Đặt \(A=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\)
\(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
\(C=\frac{x+1}{2x^2+y+2}\)
Ta có:
A = \(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-y^2-xy-y^2}=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}\)
=>A=\(\frac{x^2-y^2+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}\)
B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)
=>\(P=\left(A:B\right):C\)
\(=\left[\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}:\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)
\(=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}.\frac{2x^2+y+2}{x+1}\)
\(=\frac{1}{2y-x}\)
=>\(P=\frac{1}{2y-x}\)
Thế x=-1,76 và y=3/25 vào P
=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)
\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2+y+2}\)
\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right):\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)
\(P=\left(\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right)\left(2y-x\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\right):\frac{2x^2+y+2}{x+1}\)
\(P=\left(\frac{2x^2+y-2}{2y-x}.\frac{x+1}{2x^2+y-2}\right).\frac{1}{x+1}\)
\(P=\frac{1}{2y-x}\)
Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
Mình làm mẫu cho 1 câu nha !
a, ĐKXĐ : x khác -3 ; -1 ; 2
Biểu thức = 2/x-2 - 2/(x+1).(x-2) . (1+x) = 2/x-2 - 2/x-2 = 0
=> Với điều kiện xác định thì giá trị biểu thức ko phụ thuộc vào biến
k mk nha
\(\left(xy+\frac{1}{xy}\right)^2-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)
\(=\left(xy+\frac{1}{xy}\right)\left[\left(xy+\frac{1}{xy}\right)-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\right]\)
\(=\left(xy+\frac{1}{xy}\right)\left(xy+\frac{1}{xy}-xy-\frac{x}{y}-\frac{y}{x}-\frac{1}{xy}\right)\)
\(=\left(xy+\frac{1}{xy}\right)\left(-\frac{x}{y}-\frac{y}{x}\right)\)
\(=-\left(xy+\frac{1}{xy}\right)\left(\frac{x}{y}+\frac{y}{x}\right)=-\left(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(-\left(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}\right)+\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(=4\)
Vậy giá trị bt ko phụ thuộc vào biến
bn có thể giải thích rõ hơn tại sao lại bằng 4 được không? Dù gì thì cx cảm ơn bn đã tl câu hỏi của mk