giúp mình với nhanh nhanh nha mình đag cần gấp á
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=27.\left(-53\right)+27.\left(-47\right)\\ =27.\left[\left(-53\right)+\left(-47\right)\right]\\ =27.-100\\ =-2700\)
b: \(BC\cdot\sin B\cdot\sin C\)
\(=BC\cdot\dfrac{AC}{BC}\cdot\dfrac{AB}{BC}\)
\(=\dfrac{BC\cdot AH\cdot BC}{BC^2}=AH\)
Ta có: \(\left|y+3\right|\ge0\forall y\)
\(\left|2x+y\right|\ge0\forall x,y\)
Do đó: \(\left|y+3\right|+\left|2x+y\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}y+3=0\\2x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-3\end{matrix}\right.\)
\(\frac{3}{x}+\frac{y}{2}=\frac{1}{16}\)
=>\(\frac{6+xy}{2x}=\frac{1}{16}\)
=>16(xy+6)=2x
=>8(xy+6)=x
=>8xy-x=-48
=>x(8y-1)=-48
=>(x;8y-1)∈{(1;-48);(-48;1);(-1;48);(48;-1);(2;-24);(-24;2);(-2;24);(24;-2);(3;-16);(-16;3);(-3;16);(16;-3);(4;-12);(-12;4);(-4;12);(12;-4);(6;-8);(-8;6);(-6;8);(8;-6)}
=>(x;8y)∈{(1;-47);(-48;2);(-1;49);(48;0);(2;-23);(-24;3);(-2;25);(24;-1);(3;-15);(-16;4);(-3;17);(16;-2);(4;-11);(-12;5);(-4;13);(12;-3);(6;-7);(-8;7);(-6;9);(8;-5)}
=>(x;y)∈{\(\left(1;-\frac{47}{8}\right);\left(-48;\frac14\right);\left(-1;\frac{49}{8}\right);\left(48;0\right);\left(2;-\frac{23}{8}\right);\left(-24;\frac38\right);\left(-2;\frac{25}{8}\right);\left(24;-\frac18\right);\left(3;-\frac{15}{8}\right);\left(-16;\frac12\right)\) ; \(\left(4;-\frac{11}{8}\right);\left(-12;\frac58\right);\left(-4;\frac{13}{8}\right);\left(12;-\frac38\right);\left(6;-\frac78\right);\left(-8;\frac78\right)\) ; \(\left(-6;\frac98\right);\left(8;-\frac58\right)\) }






Bài 2:
1.
$=(\frac{1}{7}+\frac{2}{7}+\frac{4}{7})+(\frac{3}{11}+\frac{8}{11})+2010$
$=\frac{7}{7}+\frac{11}{11}+2010=1+1+2010=2012$
2.
$=(\frac{2}{3}-\frac{2}{3})+(\frac{-3}{4}+\frac{3}{4})+(\frac{4}{5}-\frac{4}{5})+(\frac{-5}{6}+\frac{5}{6})+(\frac{6}{7}-\frac{6}{7})+(\frac{-7}{8}+\frac{7}{8})+\frac{1}{2012}$
$=0+0+0+0+0+0+\frac{1}{2012}=\frac{1}{2012}$
3.
$=\frac{1}{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{1}{20-19}{19.20}$
$=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+..+\frac{1}{19}-\frac{1}{20}$
$=\frac{1}{4}-\frac{1}{20}=\frac{1}{5}$
4.
\(=182\left[\frac{1+\frac{1}{5}-\frac{2}{7}+\frac{5}{13}}{2(1+\frac{1}{5}-\frac{2}{7}+\frac{5}{13})}: \frac{4(1-\frac{1}{33}-\frac{1}{57}+\frac{1}{7000})}{1-\frac{1}{33}-\frac{1}{57}+\frac{1}{7000}}\right]:\frac{10101\times 91}{10101\times 80}\)
$=182(\frac{1}{2}:4):\frac{91}{80}=182\times \frac{1}{8}\times \frac{80}{91}$
$=\frac{91\times 2\times 80}{8\times 91}=\frac{160}{8}=20$