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\(\left(x+3\right)^4=64\)
\(\Rightarrow\left(x+3\right)^4=\pm4^4\)
\(\Rightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
\(3^x+3^{x+2}=90\)
\(\Rightarrow3^x+3^x.3^2=90\)
\(\Rightarrow3^x.1+3^x.9=90\)
\(\Rightarrow3^x\left(1+9\right)=90\)
\(\Rightarrow3^x.10=90\)
\(\Rightarrow3^x=9\Rightarrow x=2\)
a.
\(\frac{x}{4}=\frac{3}{2}\)
\(x=\frac{3}{2}\times4\)
\(x=6\)
b.
\(\frac{x}{16}=\frac{9}{x}\)
\(x\times x=16\times9\)
\(x^2=144\)
\(x^2=\left(\pm12\right)^2\)
\(x=\pm12\)
Vậy \(x=12\) hoặc \(x=-12\)
c.
\(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2=\frac{24}{25}\times6\)
\(x^2=\frac{144}{25}\)
\(x^2=\left(\pm\frac{12}{5}\right)^2\)
\(x=\pm\frac{12}{5}\)
Vậy \(x=\frac{12}{5}\) hoặc \(x=-\frac{12}{5}\)
d.
\(\frac{72-9}{7}=\frac{x-40}{9}\)
\(\frac{x-40}{9}=\frac{63}{7}\)
\(x-40=\frac{63}{7}\times9\)
\(x-40=81\)
\(x=81+40\)
\(x=121\)
a) Ta có:
128 = (122)4 = 1444
812 = (83)4 = 5124
Vì 1444 < 5124
=> 128 < 812
b) (-5)39 = -539 =-(53)13 = -12513
(-2)91 = -291 = -(27)13 = -12813
Vì -12513 > -12813
=> (-5)39 > (-2)91
Mấy dấu chấm là dấu nhân đó
a,Ta có \(\left(3^3\right)^n:3^n=9\Leftrightarrow3^{3n}:3^n=3^2\Leftrightarrow3n-n=2\Leftrightarrow n=1\)
b,TA có \(\dfrac{5^2}{5^n}=5^1\Leftrightarrow2-n=1\Leftrightarrow n=1\)
Các câu sau để bn tự làm
a) 27n : 3n = 9
\(\Leftrightarrow\) (27 : 3)n = 9
\(\Leftrightarrow\) 9n = 9
\(\Leftrightarrow\) n = 1
b) \(\dfrac{25}{5^n}=5\)
\(\Leftrightarrow\dfrac{5^2}{5^n}=5\)
\(\Leftrightarrow5^n.5=5^2\)
\(\Leftrightarrow5^{n+1}=5^2\)
\(\Leftrightarrow n+1=2\)
n = 2 - 1
n = 1
c) \(\dfrac{81}{\left(-3\right)^n}=-243\)
\(\Leftrightarrow\dfrac{\left(-3\right)^4}{\left(-3\right)^n}=\left(-3\right)^5\)
\(\Leftrightarrow\left(-3\right)^n.\left(-3\right)^5=\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^{n+5}=\left(-3\right)^4\)
\(\Leftrightarrow n+5=4\)
n = 4 - 5
n = -1
\(\dfrac{4,5:\left[47,375-\left(26\dfrac{1}{3}-18.0,75\right).2,4:0,88\right]}{17,81:1,37-23\dfrac{2}{3}:1\dfrac{5}{6}}\)
\(=\dfrac{\dfrac{9}{2}:\left[\dfrac{379}{8}-\left(\dfrac{79}{3}-\dfrac{27}{2}\right).\dfrac{30}{11}\right]}{13-\dfrac{71}{3}.\dfrac{6}{11}}\)
\(=\dfrac{\dfrac{9}{2}:\left[\dfrac{379}{8}-\dfrac{77}{6}.\dfrac{30}{11}\right]}{13-\dfrac{142}{11}}\)
\(=\dfrac{\dfrac{9}{2}:\left[\dfrac{397}{8}-35\right]}{\dfrac{1}{11}}\)
\(=\left(\dfrac{9}{2}:\dfrac{99}{8}\right):\dfrac{1}{11}\)
\(=\dfrac{4}{11}.11=4\)
Vì |a - b| + (2a + b + 6)2 = 0
Ta có:
\(\begin{cases}a-b\ge0\\\left(2a+b+6\right)^2\ge0\end{cases}\)\(\Rightarrow\left|a+b\right|+\left(2a+b+6\right)^2\ge0\)
Dấu "=" xảy ra \(\begin{cases}a-b=0\\\left(2a+b+6\right)^2=0\end{cases}\)
\(\Rightarrow\begin{cases}a=b\\2a+b+6=0^{\left(1\right)}\end{cases}\)
Thay b = a vào (1), ta được:
\(2a+a+6=0\Rightarrow3a=-6\Rightarrow a=-2\)
Vậy a = b = -2
làm gì có x