Tìm x, y biết:a, \(\left[\dfrac{1}{2}x^2\left(2x-1\right)^m-\dfrac{1}{2}x^{m+2}\right]:\dfrac{1}{2}x^2=0\) (m thuộc N)
b, \(\left(2x-3\right)^6=\left(2x-3\right)^8\)
c, \(4x^2-4x+y^2-\dfrac{2}{3}y+\dfrac{10}{9}=0\)
Bài 2: Tìm x,y,z biết:
a)\(\left(x-1\right)\)\(:\)\(\dfrac{2}{3}\)=\(\dfrac{-2}{5}\)
b) \(\left|x-\dfrac{1}{2}\right|-\dfrac{1}{3}=0\)
c) \(\left|4x+2\right|=\left|6+2x\right|\)
a) (x-1):2/3=-2/5
=>x-1=-4/15
=>x=11/15
b) |x-1/2|-1/3=0
=>|x-1/2|=1/3
=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\)
c) Tương Tự câu B
1) ghpt a)\(\left\{{}\begin{matrix}2x+\dfrac{y}{\sqrt{4x^2+1}+2x}+y^2=0\\4\left(\dfrac{x}{y}\right)^2+2\sqrt{4x^2+1}+y^2=3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(x^2-1\right)y+\left(y^2-1\right)=2\left(xy-1\right)\\4x^2+y^2+2x-y-6=0\end{matrix}\right.\)
2) tìm các số nguyên x,y thỏa mãn \(x^2+y^2-xy=x+y+2\)
3) gpt \(\sqrt{2x^2-x}=2x-x^2\)
bài 1:
b) đề như vầy hả :\(\left\{{}\begin{matrix}\left(x^2-1\right)y+\left(y^2-1\right)x=2\left(xy-1\right)\left(1\right)\\4x^2+y^2+2x-y-6=0\left(2\right)\end{matrix}\right.\)
\(Pt\left(1\right)\Leftrightarrow x^2y+xy^2-x-y-2xy+2=0\)
\(\Leftrightarrow xy\left(x+y\right)-\left(x+y\right)-2\left(xy-1\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(xy-1\right)-2\left(xy-1\right)=0\)
\(\Leftrightarrow\left(xy-1\right)\left(x+y-2\right)=0\Leftrightarrow\left[{}\begin{matrix}xy=1\\x+y=2\end{matrix}\right.\)
*xét \(xy=1\Leftrightarrow x=\dfrac{1}{y}\)thế vào Pt (2):\(\dfrac{4}{y^2}+y^2+\dfrac{2}{y}-y-6=0\)
\(\Leftrightarrow\dfrac{4+2y}{y^2}+\left(y+2\right)\left(y-3\right)=0\)\(\Leftrightarrow\left(y+2\right)\left(\dfrac{2}{y^2}+y-3\right)=0\)
\(\Leftrightarrow\left(y+2\right)\left(y^3-3y^2+2\right)=0\)\(\Leftrightarrow\left(y+2\right)\left(y-1\right)\left(y^2-2y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=1\\y=1-\sqrt{3}\\y=1+\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\\x=-\dfrac{1+\sqrt{3}}{2}\\x=\dfrac{-1+\sqrt{3}}{2}\end{matrix}\right.\)
* xét x+y=2(tương tự thay x=2-y vào Pt (2))
câu 2:
ta đưa về PT ẩn x:\(x^2-x\left(y+1\right)+y^2-y-2=0\)
Pt phải có nghiệm ,xét \(\Delta=\left(y+1\right)^2-4\left(y^2-y-2\right)\ge0\)
\(\Leftrightarrow y^2-2y-3\le0\Leftrightarrow\left(y+1\right)\left(y-3\right)\le0\)
\(\Leftrightarrow-1\le y\le3\).
vì x,y thuộc Z ,lần luợt thay các giá trị của y vừa tìm được vào PT ban đầu ta được các cặp (x,y) t/m là (0;-1);(-1;0);(2;0);(0;2);(3;2);(2;3)
bài 3:
DKXĐ:\(\left\{{}\begin{matrix}2x^2-x\ge0\\2x-x^2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le0\end{matrix}\right.\\0\le x\le2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}\le x\le2\end{matrix}\right.\)
bình phương , self study
\(\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\)
\(\dfrac{1}{x+2}+\dfrac{3}{x^2-4}+\dfrac{x-14}{\left(x^2+4x+4\right).\left(x-2\right)}\)
\(\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\dfrac{1}{x+3}+\dfrac{1}{\left(x+3\right).\left(x+2\right)}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
M=\(\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
a) tìm ĐKXĐ của x
b) rút gọn M
c) tìm x để M≥-3
a: ĐKXĐ: x<>2; x<>0
b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
c: M>=-3
=>(x+1+6x)/2x>=0
=>(7x+1)/x>=0
=>x>0 hoặc x<=-1/7
Rút gọn các biểu thức sau :
a)\(\dfrac{25xy^3\left(2x-y\right)^2}{75xy^2\left(y-2x\right)}\)
b)\(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}\)
c)\(\dfrac{\left(2x+3\right)-x^2}{x^2-1}\)
d)\(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)
`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`
`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`
`=(y(y-2x))/3`
`b,(x^2-y^2)/(x^2-y^2+xz-yz)`
`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`
`=(x+y)/(x+y+z)`
`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`
`=(-(x^2-3x+x-3))/((x-1)(x+1))`
`=(-x(x-3)+x-3)/((x-1)(x+1))`
`=((x-3)(1-x))/((x-1)(x+1))`
`=(3-x)/(1+x)`
`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`
`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`
`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`
`=(3x^2-4x+1)/(2x^2+x-3)`
`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`
`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`
`=(3x-1)/(2x+3)`
a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)
\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)
\(=\dfrac{y\left(y-2x\right)}{3}\)
Rút gọn:
a) \(\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
b) \(\dfrac{6x^2y^2}{8xy^5}\)
c) \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}\)
d) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
e) \(\dfrac{x^2-2x+1}{x^2-1}\)
f) \(\dfrac{8x-4}{8x^3-1}\)
g) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
k) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)
a: \(=\dfrac{x-z}{2}\)
b: \(=\dfrac{3x}{4y^3}\)
Thực hiện các phép tính sau :
a) \(\left[\dfrac{1}{\left(2x-y\right)^2}+\dfrac{2}{4x^2-y^2}+\dfrac{1}{\left(2x+y\right)^2}\right]\)\(\cdot\dfrac{4x^2+4xy+y^2}{16x}\)
b) \(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)
HELP MÌNH 2 câu này vớiiii !!!
a: \(\left(\dfrac{1}{\left(2x-y\right)^2}+\dfrac{2}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{1}{\left(2x+y\right)^2}\right)\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{4x^2+4xy+y^2+2\left(4x^2-y^2\right)+4x^2-4xy+y^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{8x^2+2y^2+8x^2-2y^2}{\left(2x-y\right)^2}\cdot\dfrac{1}{16x}\)
\(=\dfrac{16x^2}{16x}\cdot\dfrac{1}{\left(2x-y\right)^2}=\dfrac{x}{\left(2x-y\right)^2}\)
b: \(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)
\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)
\(=\dfrac{2x}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{-x}=\dfrac{-2\left(x-2\right)}{x+2}\)
thuc hien phep tinh
a.\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
b.\(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+1-2\right)\)
c.\(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.\left(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2}\right)\)
d.\(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\right)\)
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: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
Thực hiện các phép tính sau:
a).\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
b). \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
c). \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.\left(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2}\right)\)
d). \(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\right)\)
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{5}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x^2+1}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{x^2-2x+1}{x}\right)\)
\(=\dfrac{\left(x-1\right)^2}{x^2+1}.\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
c) d) Tự làm đi mình làm biếng quass >.< ^^