Cần gấp các cậu ơi, giúp mình câu này với
Tìm cặp số x, y biết:
1. (3x+5)2016 + (5y-4)2018 =0
2. \(\left(\frac{3x-5}{9}\right)^{2k}+\left(\frac{3y+1,4}{5}\right)^{2k}\le0\left(k\in N\right)\)
Bài 1 :
a) Tìm giá trị nhỏ nhất của biểu thức B= |x-2013| .2 + |2x-2014|
b) Tìm x,y,z biết : \(\left|3x-5\right|+\left(5y+7\right)^{2018}+\left(2z-3\right)^{2020}\le0\)
Bài 2 :
a) Tìm a,b biết \(\frac{a+b}{10}=\frac{a-2b}{7}\)và ab=9
b) Tìm GTLN của A : \(A=\frac{15\left|x+2018\right|+32}{6\left|x+2018\right|+8}\)
GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
hệ phương trình
1, \(\left\{{}\begin{matrix}3x=6\\x-3y=2\end{matrix}\right.\)
2,\(\left\{{}\begin{matrix}3x+5y=15\\2y=-7\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}7x-2y=1\\3x+y=6\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}3\left(x+y\right)+9=2\left(x-y\right)\\2\left(x+y\right)=3\left(x-y\right)+11\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}3\left(x+y\right)+5\left(x-y\right)=12\\-5\left(x+y\right)+2\left(x-y\right)=11\end{matrix}\right.\)
6 , \(\left\{{}\begin{matrix}2\left(3x-2\right)-4=5\left(3y+2\right)\\4\left(3x-2\right)+7\left(3y+2\right)=-2\end{matrix}\right.\)
7, \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{4}{5}\\\frac{1}{x}-\frac{1}{y}=\frac{1}{5}\end{matrix}\right.\)
8 , \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
có ái đó giúp mình với mình đang cần gấp
Tìm cặp số x,y thoả mãn điều kiện sau:
\(\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}=0\)
\(\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}=0\)
Ta có : \(\hept{\begin{cases}\left(\frac{3x-5}{9}\right)^{2018}\ge0\forall x\\\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall y\end{cases}}\Rightarrow\left(\frac{3x-5}{9}\right)^{2018}+\left(\frac{3y+0,4}{3}\right)^{2020}\ge0\forall x,y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{3x-5}{9}=0\\\frac{3y+0,4}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x-5=0\\3y+0,4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{2}{15}\end{cases}}\)
\(\left(3x-5\right).\left(-2x-7\right)=0\)
\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)
\(x^2-9=\left(1-4x\right)\left(x+3\right)\)
\(x^3-2x=-x^2+2\)
\(9x^2-16-x\left(3x+16\right)=0\)
\(\frac{2+4+...+2016+2018}{1019090}=-3x^2-4x\)
mình cần gấp mong các bạn giúp đỡ
\(\left(3x-5\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-5=0\\-2x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=5\\-2x=7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{-7}{2}\end{cases}}}\)
\(9x^2-1=\left(1+3x\right)\left(2x-3\right)\)
\(\Leftrightarrow9x^2-1=2x-3+6x^2-9x\)
\(\Leftrightarrow9x^2-1=-7x-3+6x^2\)
\(\Leftrightarrow9x^2-1+7x+3-6x^2=0\)
\(\Leftrightarrow3x^2+2+7x=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{3}\end{cases}}\)
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
Câu hỏi: Rút gọn biểu thức A = \(\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(\left(2k\right)^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(\left(2k-1\right)^4+\frac{1}{4}\right)}\) (k thuộc N*)
Giải hệ phương trình sau:
a)\(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\) b) \(\left\{{}\begin{matrix}2x-5y=11\\3x+4y=5\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
Giúp em với ạ, em cần gấp lắm
a, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3.3-5=4\\x=\frac{33}{11}=3\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 3;4 )
b, Làm tương tự a
c, Ta có : \(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{15}{x-y+2}+\frac{10}{x+y-1}=20\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{29}{x-y+2}=29\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-y+2=1\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{3}{y-1-y+2}+\frac{2}{y-1+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\3+\frac{2}{2y-2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{2}{2y-2}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\2y-2=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-1=1\\y=2\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 1;2 )
Tim k thuoc N biet, \(x^3y^5+3x^3y^5+.....+\left(2k-1\right)x^3y^5=3249x^3y^5\)
\(x^3y^5+3x^3y^5+...+\left(2k-1\right)x^3y^5=3249x^3y^5\)
\(\Leftrightarrow x^3y^5\left[1+2+3+...+\left(2k-1\right)\right]=3249x^3y^5\)
\(\Leftrightarrow1+3+5+...+\left(2k-1\right)=3249\)
\(\Leftrightarrow\frac{\left[\left(2k-1\right)+1\right].\left(\frac{\left(2k-1\right)-1}{2}+1\right)}{2}=3249\)
\(\Leftrightarrow\frac{2k.\left(k-1+1\right)}{2}=3249\)
\(\Leftrightarrow\frac{2k^2}{2}=3249\)
\(\Leftrightarrow k^2=3249=57^2\) ( ko xét k = - 57 vì theo quy luật thi k luôn dương )
\(\Rightarrow k=57\)
\(b.\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
=>\(3x-\frac{1}{2}=0;\frac{1}{2}y+\frac{3}{5}=0\left(\left|3x-\frac{1}{2}\right|;\left|\frac{1}{2}y+\frac{3}{5}\right|\ge0\right)\)
=>\(x=\frac{1}{6};y=\frac{-6}{5}\)
b)\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
Ta lại có:
\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\ge0\)
=>\(\frac{3}{2}x+\frac{1}{9}=0;\frac{1}{5}y-\frac{1}{2}=0\Rightarrow x=-\frac{2}{27};y=\frac{5}{2}\)