Rút gọn P
P = \(\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)
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2
30 tháng 6
1.7:
a: \(\left\{{}\begin{matrix}3x+2y=6\\2x-2y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y+2x-2y=6+14\\x-y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x=20\\x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=x-7=4-7=-3\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}0,3x+0,5y=3\\1,5x-2y=1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1,2x+2y=12\\1,5x-2y=1,5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1,2x+2y+1,5x-2y=12+1,5\\0,3x+0,5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2,7x=13,5\\0,5y=3-0,3x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\0,5y=3-0,3\cdot5=1,5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\y=3\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}-2x+6y=8\\3x-9y=-12\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x+3y=4\\x-3y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=0\\x=3y-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in R\\3y=x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\y=\dfrac{x+4}{3}\end{matrix}\right.\)
1.6:
a: \(\left\{{}\begin{matrix}x-y=3\\3x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\3\left(y+3\right)-4y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=y+3\\3y+9-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+3\\9-y=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7\\x=7+3=10\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}7x-3y=13\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-3y=13\\y=2-4x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x-3\left(2-4x\right)=13\\y=2-4x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x-6=13\\y=2-4x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}19x=19\\y=2-4x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2-4\cdot1=-2\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}0,5x-1,5y=1\\-x+3y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0,5x-1,5y=1\\x=3y-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0,5\left(3y-2\right)-1,5y=1\\x=3y-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1,5y-1-1,5y=1\\x=3y-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-1=1\\x=3y-2\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\varnothing\)
30 tháng 6
Xét ΔIDC có AB//DC
nên \(\dfrac{IA}{AD}=\dfrac{IB}{BC}\)
mà AD=BC
nên IA=IB
Xét ΔABC và ΔBAD có
AB chung
BC=AD
AC=BD
Do đó: ΔABC=ΔBAD
=>\(\widehat{CAB}=\widehat{DBA}\)
=>\(\widehat{OAB}=\widehat{OBA}\)
=>OA=OB
Ta có: OA+OC=AC
OB+OD=BD
mà OA=OB và AC=BD
nên OC=OD
ΔOCD cân tại O
mà ON là đường trung tuyến
nên ON\(\perp\)DC
ΔOAB cân tại O
mà OM là đường trung tuyến
nên OM\(\perp\)AB
mà AB//CD
nên OM\(\perp\)CD
Ta có: IA+AD=ID
IB+BC=IC
mà IA=IB và AD=BC
nên ID=IC
=>ΔIDC cân tại I
mà IN là đường trung tuyến
nên IN\(\perp\)DC
Ta có: OM\(\perp\)CD
ON\(\perp\)CD
mà OM,ON có điểm chung là O
nên O,M,N thẳng hàng(1)
Ta có: IN\(\perp\)DC
ON\(\perp\)CD
mà IN,ON có điểm chung là N
nên I,N,O thẳng hàng(2)
Từ (1),(2) suy ra I,M,O,N thẳng hàng
DT
Dang Tung
CTVHS
30 tháng 6
\(P=4x^2+2y^2-4xy-4x-8y+2050\\ =\left(4x^2-4xy+y^2\right)+y^2-4x-8y+2050\\ =\left(2x-y\right)^2-2.\left(2x-y\right).1+1^2+y^2-10y+2049\\ =\left(2x-y-1\right)^2+\left(y^2-10y+25\right)+2024\\ =\left(2x-y-1\right)^2+\left(y-5\right)^2+2024\ge2024\forall x,y\)
Dấu = xảy ra khi: \(\left(2x-y-1\right)^2=\left(y-5\right)^2=0\\ \Leftrightarrow\left(x;y\right)=\left(3;5\right)\)
Vậy min P = 2024 tại (x;y)=(3;5)
NQ
1
TT
1
30 tháng 6
a: \(12,7-x=25,7+23,4\)
=>\(12,7-x=49,1\)
=>x=12,7-49,1=-36,4
b: \(\dfrac{13}{4}:x=\dfrac{12}{9}-\dfrac{4}{3}\)
=>\(\dfrac{13}{4}:x=\dfrac{4}{3}-\dfrac{4}{3}=0\)
=>\(x\in\varnothing\)
30 tháng 6
a: \(7\dfrac{13}{4}+6\dfrac{5}{3}\)
\(=\dfrac{41}{4}+\dfrac{23}{3}\)
\(=\dfrac{41\cdot3+23\cdot4}{12}=\dfrac{215}{12}\)
b: \(\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\dfrac{15}{60}+\dfrac{12}{60}-\dfrac{10}{60}\)
\(=\dfrac{17}{60}\)
c: \(\dfrac{14}{7}+\dfrac{2}{9}-\dfrac{4}{5}\)
\(=2+\dfrac{2}{9}-\dfrac{4}{5}\)
\(=\dfrac{90}{45}+\dfrac{10}{45}-\dfrac{36}{45}=\dfrac{64}{45}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0;y>=0\\x^2+y^2\ne1^2+1^2=2\end{matrix}\right.\)
\(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}+1\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\left(\sqrt{xy}+1\right)\left(\sqrt{xy}+\sqrt{x}\right)+xy-1}{xy-1}\)
\(=\dfrac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}+xy-1}{xy-1}\)
\(=\dfrac{-2\sqrt{x}-2}{xy-1}\)
\(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt[]{xy}+1}\)
\(=\dfrac{xy-1-\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}\)
\(=\dfrac{xy-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{xy-1}\)
\(=\dfrac{-2\sqrt{xy}-2x\sqrt{y}}{xy-1}\)
\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\dfrac{\sqrt{x}+1}{\sqrt[]{xy}+1}\right)\)
\(=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}:\dfrac{-2\sqrt{xy}\left(\sqrt{x}+1\right)}{xy-1}\)
\(=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}\cdot\dfrac{xy-1}{-2\sqrt{xy}\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{xy}}\)