=> x-3 = 25
<=> x= 25 +3
<=> x= 28
Vậy x= 28
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=> x-3 = 25
<=> x= 25 +3
<=> x= 28
Vậy x= 28
tim x,y,z biet \(\sqrt{\left(x-\sqrt{5}\right)^2}+\sqrt{\left(y+\sqrt{3}\right)^2}+\left|x-y-z\right|\)
Tim x ,y biet
a) x+\(\frac{1}{x}=1\)
b)x+\(\frac{2}{x}=5\)
c)x\(\sqrt{3}+3=x\sqrt{3}-x\)
d)\(\left(x-2\right)\sqrt{25n^2+5}+y-2=0\)
tim x biet\(\sqrt{x-7}\)=x-1
tim x biet
\(x=\sqrt{x}\)
Tim x biet 2.(x-3)-3.(x-1)-5
tim x biet
x-3/x+5=5/7
Bai 1
a, Tim x; y :\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
b, Tim x\(\in\)z de A\(\in z\) biet :
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
tim x biet
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y-\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
Tim x biet(4x-3)-(x+5)=(x+2)-2×(x-10)