(x - 8)9 = 8
<=> \(\left[\left(x-8\right)^3\right]^3=2^3\)
<=> (x - 8)3 = 2
<=> x - 8 = \(\sqrt[3]{2}\)
<=> x = \(\sqrt[3]{2}+8\)
(x - 8)9 = 8
<=> \(\left[\left(x-8\right)^3\right]^3=2^3\)
<=> (x - 8)3 = 2
<=> x - 8 = \(\sqrt[3]{2}\)
<=> x = \(\sqrt[3]{2}+8\)
tìm nghiệm
G(x)=8(x+1^3+1
H(x)=8/9-2(x-1)^2
K(x)=(x+8)(x^2-9/25)
Tìm x, biết:
a) \(\left(-\dfrac{5}{9}\right)^{10}\) : x = \(\left(\dfrac{-5}{9}\right)^8\)
b) x ; \(\left(\dfrac{-5}{9}\right)^8\) = \(\left(\dfrac{-9}{5}\right)^8\)
c) x3 = -8
Tìm x, biết
8/3 : x = 16/9 : 8/3
tìm x :
\(\frac{5}{8}.8^{x+2}-\frac{3}{5}.8^x=\frac{5}{3}.8^{11}-\frac{3}{5}.8^9\)
Tìm x
x . 6^12 . 8^6 = 2^30 . 9^8
Tìm x biết:
a) 6* 8^x-1 + 8^x+1 = 6* 8^19 + 8^21
b) 4* 3^x-1 + 2* 3^x+2 = 4* 3^6 +2* 3^9
Tìm x
\(\frac{x+5}{7}-\frac{x+18}{8}+\frac{x+8}{9}=0\)
Cho \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\). Tìm x.
tìm x
x-6/7+x-7/8+x-8/9=x-9/10+x-10/11+x-11/12
mình đang rất cần
cho bạn giải dc 1 like
cảm ơn nha!