2 mũ x + 26 = 3 mũ y
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\(\frac{x}{y}=\frac{2}{3}\)và\(x^2+y^2=26\)
\(\Leftrightarrow\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x^2}{2^2}=\frac{y^2}{3^2}\Leftrightarrow\frac{x^2}{4}=\frac{y^2}{9}\)
+)APTC của dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{x^2+y^2}{4+9}=\frac{26}{13}=2\left(1\right)\)
+)Từ 1 suy ra:
\(\frac{x^2}{4}=2\Leftrightarrow x^2=8\Leftrightarrow x\in\left\{\sqrt{8};-\sqrt{8}\right\}\)
\(\frac{y^2}{9}=2\Leftrightarrow y^2=18\Leftrightarrow y\in\left\{\sqrt{18};-\sqrt{18}\right\}\)
Vậy \(x\in\left\{\sqrt{8};-\sqrt{8}\right\}\)
\(y\in\left\{\sqrt{18};-\sqrt{18}\right\}\)
Chúc bạn học tốt
a, (-0,2)2 \(\times\) 5 - \(\dfrac{2^{13}\times27^3}{4^6\times9^5}\)
= 0,04 \(\times\) 5 - \(\dfrac{2^{13}\times3^9}{2^{12}\times3^{10}}\)
= 0,2 - \(\dfrac{2}{3}\)
= \(\dfrac{2}{10}\) - \(\dfrac{2}{3}\)
= - \(\dfrac{7}{15}\)
b, \(\dfrac{5^6+2^2.25^3+2^3.125^2}{26.5^6}\)
= \(\dfrac{5^6+4.5^6+8.5^6}{26.5^6}\)
= \(\dfrac{5^6.\left(1+4+8\right)}{26.5^6}\)
= \(\dfrac{1}{2}\)
a, (-0,2)2 ×× 5 - 213×27346×9546×95213×273
= 0,04 ×× 5 - 213×39212×310212×310213×39
= 0,2 - 2332
= 210102 - 2332
= - 715157
b, 56+22.253+23.125226.5626.5656+22.253+23.1252
= 56+4.56+8.5626.5626.5656+4.56+8.56
= 56.(1+4+8)26.5626.5656.(1+4+8)
= 1221
=9x2^25-2^2x2^26/2^24x5^2-2^27x3
=9x2^25-2^28/2^24x5^2-2^27x3
=2^25x(9-2^3)/2^24x(5^2-2^3x3)
=2^25/2^24
=2^1=2
Ta có: 72x-62x = 13.23-26 => x(72-62) = 13.8-26 => 13x = 78 => x = 78:13 => x = 6 Vậy, x = 6.
a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
mày ko chả lời thì thôi
\(2^x+26=3^y\)
\(\Leftrightarrow2^x=3^y-26\)
Để phương trình có nghiệm thì \(3^y>26\)
hay y>3