=>2x.(1+23)=288
=>2x.9=288
=>2x=288:9
=>2x=32
=>x=5
2x+2x+3=288
2x.1+2x.23=288
2x(1+23)=288
2x.9=288
2x=288:9
2x=32
2x=25
=> x=5
Vậy x=5
x/y=5/4 và 2x^2-3y^2=288.tìm x và y
2^X + 2^X+3= 288 (2x-1)^3=27 (x+5)^3= -64 (2x -3)^2=9
cho 5x =2y ;3y=5z ;2x -3y+z=288 . Tìm xyz
Tìm x:
a)\(\left(\frac{1}{3}-\frac{1}{2}\right)^{x-1}=\frac{1}{36}\)
b)\(81^{-2x}.27^x=9^5\)
c)\(2^x+2^{x+3}=288\)
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}\)và 2x+y-z=81
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}\)và 5x-y+3z=124
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và x.y.z=810
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}\)và\(x^2.y^2.z^2=288^2\)
Tìm x, biết
a)(2x-1)^5-(2x-1)^8=0
b)(2x+1). (2x-3)<0
c)(x-1). (2x+3)>0
biết f(x)=x^2+2x+3/5 2x^3+3 ; g(x) = -2x^2-1/2x^3 -3x +6
tìm h(x) = 2f(x) - 1/2g(x)
\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)
\(\dfrac{x+1}{22}=\dfrac{6}{x}\)
\(\dfrac{2x-1}{2}=\dfrac{5}{x}\)
\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)
\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)
Tìm x
tìm x nguyên bt:
|5(2x+3)| + |2(2x+3)| + |2x+3| = 16
Tìm GTNN
D=|2x-4|+|2x+5|
E=|x+5|+|x+1|+4
Tìm GTLN
M=-|2x+3|+2x+4
N=-3×|x-4|+8-3x
