Question

Решить систему: log(2)(x+y)+2log(4)(x-y)=3
                           3^(2+log(3)(2x-y)=45

Решить не равенство: log(1/4)(2x-5>-1

Answer 1

$left { {{log_2(x+y)+2log_4(x-y)=3} atop {3^{2+log_3(2x-y)}=45}} right. left { {{log_2(x+y)+2log_{2^2}(x-y)=3} atop {3^2*3^{log_3(2x-y)}=45}} right. \ left { {{log_2(x+y)+2*frac{1}{2}log_{2}(x-y)=3} atop {9*3^{log_3(2x-y)}=45}} right. left { {{log_2(x+y)+log_{2}(x-y)=3} atop {2x-y}=5}} right. \ left { {{log_2((x+y)(x-y))=3} atop {2x-y}=5}} right. left { {{x^2-y^2=8} atop {2x-y}=5}} right.$
$left { {{x^2-y^2=8} atop {y=2x-5}} right. left { {{x^2-(2x-5)^2=8} atop {y=2x-5}} right.\ x^2-4x^2+20x-25-8=0 \ -3x^2+20x-33=0 \ 3x^2-20x+33=0 \ D=400-396=4 \ x_1=frac{20+2}{6}= 3frac{2}{3} x_2=frac{20-2}{6}=3 \ y_1=2*3frac{2}{3}-5=2frac{1}{3} y_2=2*3-5=1 \ \ (3frac{2}{3};2frac{1}{3}) (3;1)$


$log_frac{1}{4}(2x-5)>-1 \ left { {{2x-5>0} atop {2x-5<(frac{1}{4})^{-1}}} right. left { {{2x>5} atop {2x-5<4} right. left { {{x>2,5} atop {2x<9} right. left { {{x>2,5} atop {x<4,5} right.$
x∈(2,5; 4,5)