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a)\(x^{15}=x\Rightarrow x\in\left\{0;1\right\}\) d)Là ý b
b)\(\left(2x+1\right)^3=125\\ \left(2x+1\right)^3=5^3\Rightarrow2x+1=5\\ 2x=4\\ x=2\) e)\(\left(x-3\right)^2=25\\ \Rightarrow\left(x-3\right)^2=5^2\\ \Rightarrow\hept{\begin{cases}x-3=5\Rightarrow x=8\\x-3=-5\Rightarrow x=-2\end{cases}}\)
c)\(\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow x-5\in\left\{0;1\right\}\Rightarrow x\in\left\{5;6\right\}\)
a)\(3^2.9^3=9.9^3=9^{1+3}=9^4\)
b)\(2^2.5^2=4.25=100=10^2\)
c)\(8^5.2^3=8^5.8=8^{5+1}=8^6\)
d)\(9^8:3^2=9^8:9=9^{8-1}=9^7\)
2x x 16 = 128
2x = 128 : 16
2 x = 8
2x = 23
3x : 9 = 27
3x = 27 x 9
3x =243
3x = 35
[ 2x + 1 ]3 = 27
2x3 + 13 = 27
2x3 +1 = 27
2x3 = 27 - 1
2x3 = 26
\(23\left(x-1\right)+19=65\)
\(23\left(x-1\right)=65-19\)
\(23\left(x-1\right)=46\)
\(x-1=46:23\)
\(x-1=2\)
\(x=2+1\)
\(x=3\)
\(5x+3x=88\)
\(x\left(5+3\right)=88\)
\(x.8=88\)
\(x=88:8\)
\(x=11\)
\(x^3=64\)
\(x^3=4^3\)
\(\Rightarrow x=4\)
\(\left(5x-4\right):7-2=6\)
\(\left(5x-4\right):7=6+2\)
\(\left(5x-4\right):7=8\)
\(5x-4=8.7\)
\(5x-4=56\)
\(5x=56+4\)
\(5x=60\)
\(x=60:5\)
\(x=12\)
\(x^{50}=x\)
\(\Rightarrow x=1\)
\(4.2^x-3=125\)
\(4.2^x=125+3\)
\(4.2^x=128\)
\(2^x=128:4\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
k mk nha
a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
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