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x^3 = 27
x^3 = 3^3
=> x = 3
(2x-1)^3 = 8
(2x-1)^3 = 2^3
2x-1 = 2
2x = 2+1
2x = 3
x = 3:2
=> ko có x phù hợp.
(x-2)^2 = 16
(x-2)^2 = 4^2
x-2 = 4
x = 4+2
x = 6
Chúc bn học tốt!
x^3 = 27
x^3 = 3^3
Vậy x = 3
Đề bài 2 hình như sai bạn ạ
( x - 2 )^2 = 16
( x- 2 )^2 = 4^2
x - 2 = 4
x = 4 + 2
x = 6
Vậy x = 6
3) Tìm số tự nhiên x, biết:
a) \(2^x=4\)
\(2^x=2^2\)
\(\Rightarrow x=2\)
___________
b) \(2^x=1\)
\(2^x=2^0\)
\(\Rightarrow x=0\)
___________
c) \(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
___________
d) \(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
___________
e) \(5^x=125\)
\(5^x=5^3\)
\(\Rightarrow x=3\)
___________
f) \(8^x=64\)
\(8^x=8^2\)
\(\Rightarrow x=2\)
___________
h) \(3^{x+1}=3^2\)
\(\rightarrow x+1=2\)
\(x=2-1\)
\(x=1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
a: 2^x=4
=>2^x=2^2
=>x=2
b: 2^x=1
=>2^x=2^0
=>x=0
c: 2^x=16
=>2^x=2^4
=>x=4
d; 3^x=9
=>3^x=3^2
=>x=2
e: 5^x=125
=>5^x=5^3
=>x=3
f: 8^x=64
=>8^x=8^2
=>x=2
f: 3^x+1=3^2
=>x+1=2
=>x=1
\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)
\(a)\) \(-\left(x+84\right)+213=-16\)
\(\Leftrightarrow\)\(-x-84+213=-16\)
\(\Leftrightarrow\)\(x=213-84+16\)
\(\Leftrightarrow\)\(x=145\)
Vậy \(x=145\)
\(b)\) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=\left|-1\right|\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)
Vậy \(x=0\) hoặc \(x=2\)
Chúc bạn học tốt ~
a) \(-\left(x+84\right)+213=-16\)
\(-\left(x+84\right)=-16-213\)
\(-\left(x+84\right)=-229\)
\(\Rightarrow x+84=229\)
\(\Rightarrow x=229-84=145\)
Vậy \(x=145\)
b) \(\left(x-1\right)^2=\left|\frac{1}{4}-\frac{1}{2}-\frac{3}{4}\right|\)
\(\left(x-1\right)^2=\left|\frac{-1}{4}-\frac{3}{4}\right|\)
\(\left(x-1\right)^2=\left|-1\right|\)
\(\left(x-1\right)^2=1\)
\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1+1=2\\x=-1+1=0\end{cases}}\)
Vậy \(x\in\left\{0;2\right\}\)
\(-3.\left(x-1\right)+2\left(x+2\right)=-16\)
\(-3x+3+2x+4=-16\)
\(-3x+2x=-16-3-4\)
\(-x=-23\)
\(x=23\)