Tim x,y la so nguyen
a) 2x-3y+5= xy
b) x mu 2 +2xy+2x+4y
B2 Tim x,y la so nguyen to
x mu 2 -2y mu 2 =1
B3 Tim so tu nhienx,y
7 .(x-2004) mu 2 = 23- y mu 2
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a: \(\left(x+5\right)^2>=0\forall x\)
\(\left(2y-8\right)^2>=0\forall y\)
Do đó: \(\left(x+5\right)^2+\left(2y-8\right)^2>=0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+5=0\\2y-8=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-5\\y=4\end{matrix}\right.\)
b: \(\left(x+3\right)\left(2y-1\right)=5\)
=>\(\left(x+3\right)\left(2y-1\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x+3;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;3\right);\left(2;1\right);\left(-4;-2\right);\left(-8;0\right)\right\}\)
theo mk nghĩ ;
(5x - 7)2 + (6-5y)2 + (5z)2 = 10
chi xay ra khi 4+4+2 (loại) vi k có so nao binh phuong =2
vay 9+1+0 =10 ta co;
x = 2
y=1
z=0
(hop ly va duy nhat)
a. ta có : 2x + 2x+3 = 144=
2x + 2xnhân 23=144
2x(1 + 23) =144
2x nhân 9 =144
2x = 144 : 9
2x = 16
2x = 24
suy ra : x = 4
( 2x + 1 )5 : ( 2x + 1 )2 = 1
\(\Rightarrow\)( 2x+ 1 )3 = 1 = 13
\(\Rightarrow\)2x + 1 = 1
\(\Rightarrow\)2x = 0
\(\Rightarrow\)x = 0
2x + 2 . 2x + ... + 10 . 2x= 10 . 11
\(\Rightarrow\)2x . ( 1 + 2 + 3 + ... + 9 + 10 ) = 110
\(\Rightarrow\)2x . 55 = 110
\(\Rightarrow\)2x = 110 : 55
\(\Rightarrow\)2x = 2
\(\Rightarrow\)x = 1
b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)