Tìm x,y ∈ N biết
\(\dfrac{x}{9}\)=\(\dfrac{3}{y}\)+\(\dfrac{1}{18}\) Help me
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\(3^{8x+4}=81^{x+3}\)
\(3^{8x+4}=\left(3^4\right)^{x+3}\)
\(3^{8x+4}=3^{4x+12}\)
\(\Rightarrow8x+4=4x+12\)
\(\Rightarrow8x-4x=12-4\)
\(\Rightarrow4x=8\Rightarrow x=2\)
38.x + 4 = 81x + 3
38.x + 4 = (34)x + 3
38.x + 4 = 34.x + 12
8.x + 4 = 4.x + 12
8.x - 4.x = 12 - 4
4.x = 8
x = 8 : 4
x = 2
\(2^x=2+2+2^2+...+2^{30}\)
\(\Rightarrow2^x=1+\left(1+2+2^2+...+2^{30}\right)\)
\(\Rightarrow2^x=1+\dfrac{2^{30+1}-1}{2-1}\)
\(\Rightarrow2^x=1+2^{31}-1\)
\(\Rightarrow2^x=2^{31}\)
\(\Rightarrow x=31\)
2x + 2x + 1 = 96
2x + 2x . 2 = 96
2x . (1 + 2) = 96
2x . 3 = 96
2x = 96 : 3
2x = 32
2x = 25
x = 5
`#040911`
\(2^{10}\div8^5\\ =2^{10} \div\left(2^3\right)^5\\=2^{10}\div2^{15}=2^{-5} \)
Đề có nhầm gì k bạn?
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\(920+3^2.\left(2^2.5^3-6.5^2\right)\\ \\ \\ =920+9.\left(4.125-6.25\right)\\ \\ \\ =920+9.\left(500-150\right)\\ \\ \\ =920+9.350=920+3150=4070\)
920 + 32 . (22 . 53 - 6 . 52)
= 920 + 9 . (4 . 125 - 6 . 25)
= 920 + 9 . (500 - 150)
= 920 + 9 . 350
= 920 + 3150
= 4070
\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)
\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)
\(\Rightarrow2xy=54+y\)
\(\Rightarrow2xy-y=54\)
\(\Rightarrow xy-\dfrac{y}{2}=27\)
\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)
\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)