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\(9^{2+9}=9^{11}\)
\(2^{2+4}=2^6\)
\(3^{4-2}=3^2\)
AI TÍCH MIK NHANH NHẤT
MIK SẼ TÍCH GẤP ĐÔI
\(9^2.9^9=9^{11}\)
\(2^2.2^4=2^6\)
\(3^4:3^2=3^2\)
k Ẻm đi nhá!
(x2-1)2=9
=> x2-1 = 3
x2 = 3+1
x2 = 4
=> x2 = 4 = 22 ( x2=22 )
<=> x = 2
12:{390:[5.102-(53+x.72)]} = 4
390:[5.102-(53+x.72)] = 12:4
390:[5.102-(53+x.72)] = 3
5.102-(53+x.72) = 390 : 3
5.102-(53+x.72) = 130
=> 500-(125+x+49)=130
125+x+49 = 500-130
125+x+49 = 370
125+x = 370-49
125+x = 321
x = 321-125
x = 106
53(3x+2):13=103:(135:134)
53(3x+2):13=103:13
53(3x+2):13= 1000/13
125(3x+2):13 = 1000/13
125(3x+2) = 1000/13 . 13
125(3x+2) = 1000
3x+2 = 1000:125
3x+2 = 8
3x = 8-2
3x = 6
x = 6:3
x = 2
b) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)
a,\(4^n.2^n=512\)
\(\Rightarrow2^{2n}.2^n=512\Rightarrow2^{3n}=2^9\Rightarrow3n=9\Rightarrow n=3\)
b,\(3^n+3^{n+3}=252\)( sửa đề )
\(\Rightarrow3^n.\left(1+3^3\right)=252\Rightarrow3^n.28=252\Rightarrow3^n=9\Rightarrow n=2\)
c,\(2.3^{2x+2}=18\)
\(\Rightarrow3^{2n+2}=9\Rightarrow2n+2=2\Rightarrow n=0\)
d,\(x^2=2^3+3^2+4^3\)
\(\Rightarrow x^2=8+9+64\Rightarrow x^2=81\Rightarrow x^2=9^2=\left(-9\right)^2\Rightarrow x=9\)hoặc \(x=-9\)
e,\(x^5=x^9\)
\(\Rightarrow x^9-x^5=0\Rightarrow x^5.\left(x^4-1\right)=0\Rightarrow\hept{\begin{cases}x^5=0\\x^4-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\\x=-1\end{cases}}}\)
f,\(\left(x-4\right)^3=\left(x-4\right)^{10}\)
\(\Rightarrow\left(x-4\right)^{10}-\left(x-4\right)^3=0\Rightarrow\left(x-3\right)^3.\left[\left(x-3\right)^7-1\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^3=0\\\left(x-3\right)^7=1\end{cases}\Rightarrow\hept{\begin{cases}x-3=0\\x-3=1\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\x=4\end{cases}}}\)
\(2^{2+10}=2^{12}\)
\(2^{10-1}=2^9\)
hazz , nãy giờ chưa được tích nào cả
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(9^{2+2+1}=9^5\)
\(3^5.2^5=\left(3.3.3.3.3\right).\left(2.2.2.2.2\right)=\left(3.2\right).\left(3.2\right).\left(3.2\right).\left(3.2\right).\left(3.2\right)=6.6.6.6.6=6^5\)
Tích nha mik tích lại
thanks nhiều