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a )
ĐKXĐ : \(y\ne0\) , \(y\ne-1\)
\(P=\left(\dfrac{2y^2+1}{y^3+1}-\dfrac{y}{y+y^2}\right):\left(1-\dfrac{y^2-2y-1}{y^2-y+1}\right)\)
\(=\left(\dfrac{2y^2+1}{\left(y+1\right)\left(y^2-y+1\right)}-\dfrac{1}{y+1}\right):\left(\dfrac{y^2-y+1-y^2+2y+1}{y^2-y+1}\right)\)
\(=\dfrac{2y^2+1-y^2+y-1}{\left(y+1\right)\left(y^2-y+1\right)}:\dfrac{y+2}{y^2-y+1}\)
\(=\dfrac{y\left(y+1\right)}{\left(y+1\right)\left(y^2-y+1\right)}\times\dfrac{y^2-y+1}{y+2}\)
\(=\dfrac{y}{y+2}\)
Câu b :
\(\left|2y+5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2y+5=3\\-2y-5=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=-4\end{matrix}\right.\)
Thay \(y=-1\) vào P ta được : \(P=\dfrac{-1}{-1+2}=-1\)
Thay \(y=-4\) vào P ta được : \(P=\dfrac{-4}{-4+2}=2\)
Câu c :
T chỉ biết lập luận thôi :
Để P chia hết cho 4 thì \(\dfrac{y}{y+2}\) chia hết cho 4 hay \(\dfrac{y}{y+2}\) phải là bội của 4.
Do \(y< y+2\) nên \(\dfrac{y}{y+2}\) không thể là các số 4 ; 8 ;12 ;.........
Nên \(\dfrac{y}{y+2}=0\) thì sẽ chia hết cho 4 . \(\Leftrightarrow y=0\) ( Loại )
Nên không có giá trị y nào hết .
Câu d :
\(P=3-m>2\)
\(\Leftrightarrow-m>-1\)
\(\Leftrightarrow m< 1\)
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
a: \(B=\left(x^2+y\right)\left(y+\dfrac{1}{4}\right)+x^2y^2+\dfrac{3}{4}\left(y+\dfrac{1}{3}\right)\)
\(=x^2y+\dfrac{1}{4}x^2+y^2+\dfrac{1}{4}y+x^2y^2+\dfrac{3}{4}y+\dfrac{1}{4}\)
\(=x^2y+x^2y^2+y^2+y+\dfrac{1}{4}x^2+\dfrac{1}{4}\)
\(=y\left(x^2+1\right)+y^2\left(x^2+1\right)+\dfrac{1}{4}\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(y+\dfrac{1}{2}\right)^2\)
\(C=x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\)
\(=x^2y^2+1+x^2-x^2y-y+y^2\)
\(=x^2y^2-y+x^2+y^2-x^2y+1\)
\(=y^2\left(x^2+1\right)-y\left(x^2+1\right)+x^2+1\)
\(=\left(x^2+1\right)\left(y^2-y+1\right)\)
=>\(A=\dfrac{y^2+y+\dfrac{1}{4}}{y^2-y+1}\)
b: \(=\dfrac{y^2-y+1+2y-\dfrac{3}{4}}{y^2-y+1}=1+\dfrac{2y-\dfrac{3}{4}}{y^2-y+1}>=1\)
Dấu = xảy ra khi y=3/8
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
a: \(P=\left(\dfrac{2y^2+1}{\left(y+1\right)\left(y^2-y+1\right)}-\dfrac{1}{y+1}\right):\dfrac{y^2-y+1-y^2+2y+1}{y^2-y+1}\)
\(=\dfrac{2y^2+1-y^2+y-1}{\left(y+1\right)\left(y^2-y+1\right)}\cdot\dfrac{y^2-y+1}{y+2}\)
\(=\dfrac{y^2+y}{\left(y+1\right)}\cdot\dfrac{1}{y+2}=\dfrac{y}{y+2}\)
b: |2y+5|=3
=>2y+5=3 hoặc 2y+5=-3
=>2y=-2 hoặc 2y=-8
=>y=-1(loại) hoặc y=-4(nhận)
Thay y=-4 vào P,ta được:
\(P=\dfrac{-4}{-4+2}=\dfrac{-4}{-2}=2\)
c: Để P chia hết cho 4 thì P=4k
=>y=4k(y+2)