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`#3107.101107`
a,
$2014^{15} . (2x - 60) = 2014^{16}$
$\Rightarrow 2x - 60 = 2014^{16} \div 2014^{15}$
$\Rightarrow 2x - 60 = 2014$
$\Rightarrow 2x = 2074$
$\Rightarrow x = 1037$
Vậy, `x = 1037`
b,
`x + 18 \div 3^2 = 5.4^2`
`\Rightarrow x + 18 \div 9 = 5.16`
`\Rightarrow x + 9 = 80`
`\Rightarrow x = 80 - 9`
`\Rightarrow x = 71`
Vậy, `x = 71`
c,
`7.(42 - x) = 5^3 + 134`
`\Rightarrow 7.(42 - x) = 125 + 134`
`\Rightarrow 7(42 - x) = 259`
`\Rightarrow 42 - x = 259 \div 7`
`\Rightarrow 42 - x = 37`
`\Rightarrow x = 42 - 37`
`\Rightarrow x = 5`
Vậy, `x = 5.`
`#3107.101107`
a,
\([(6.x - 39) \div3] .28 =5628\)
`\Rightarrow (6x - 39) \div 3 = 5628 \div 28`
`\Rightarrow (6x - 39) \div 3 = 201`
`\Rightarrow 6x - 39 = 201 . 3`
`\Rightarrow 6x - 39 = 603`
`\Rightarrow 6x = 42`
`\Rightarrow x = 42 \div 6`
`\Rightarrow x = 7`
Vậy, `x = 7`
b,
`5(7x - 45) = 2^3 . 5^2 - 3^2 . 20`
`\Rightarrow 5(7x - 45) = 8.5.5 - 9.4.5`
`\Rightarrow 5(7x - 45) = 5.(8.5 - 9.4)`
`\Rightarrow 5(7x - 45) = 5.(40 - 36)`
`\Rightarrow 7x - 45 = 5.(40 - 36) \div 5`
`\Rightarrow 7x - 45 = 4`
`\Rightarrow 7x = 49`
`\Rightarrow x = 7`
Vậy, `x = 7.`
\(2.\left(2x-\frac{4}{3}\right)^2+\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)^2=\frac{1}{2}-\frac{1}{4}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)^2=\frac{1}{4}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)=\sqrt{\frac{1}{4}}\)
\(\Rightarrow\left(2x-\frac{4}{3}\right)=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}+\frac{4}{3}\)
\(\Rightarrow2x=\frac{11}{6}\)
\(\Rightarrow x=\frac{11}{6}\div2\)
\(\Rightarrow x=\frac{11}{6}\times\frac{1}{2}\)
\(\Rightarrow x=\frac{11}{12}\)
a, \(x^2\) = \(x^3\)
\(x^3\) - \(x^2\) = 0
\(x^2\)( \(x\) -1) = 0
\(\left[{}\begin{matrix}x^2=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 0; 1}
e, 32\(x+1\) = 27
\(3^{2x}\)+1 = 33
2\(x\) + 1 = 3
2\(x\) = 2
\(x\) = 1
g, 62 = 6\(x-3\)
2 = \(x-3\)
\(x\) = 3 + 2
\(x\) = 5
\(a,x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(b,3^{2x+1}=27\\ \Rightarrow3^{2x+1}=3^3\\ \Rightarrow2x+1=3\\ \Rightarrow2x=3-1\\ \Rightarrow2x=2\\ \Rightarrow x=2:2\\ \Rightarrow x=1\)
\(c,6^2=6^{x-3}\\ \Rightarrow6^{x-3}=6^2\\ \Rightarrow x-3=2\\ \Rightarrow x=2+3\\ \Rightarrow x=5\)
A= \(\dfrac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)
a) \(6⋮\left(x-1\right)\left(đkxđ:x\ne1;x\inℕ\right)\)
\(\Rightarrow x-1\in U\left(6\right)=\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{2;3;4;7\right\}\)
b) \(14⋮\left(2x+3\right)\left(đkxđ:x\ne-\dfrac{3}{2};x\inℕ\right)\)
\(\Rightarrow2x+3\in U\left(14\right)=\left\{1;2;7;14\right\}\)
\(\Rightarrow x\in\left\{-1;-\dfrac{1}{2};2;\dfrac{9}{2}\right\}\)
\(\Rightarrow x\in\left\{-2\right\}\)
\(a,6⋮\left(x-1\right)\\ \Rightarrow\left(x-1\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\\ Ta.có:x-1=-6\Rightarrow x=-5\left(loại\right)\\ x-1=-3\Rightarrow x=-2\left(loại\right)\\ x-1=-2\Rightarrow x=-1\left(loại\right)\\ x-1=-1\Rightarrow x=0\left(nhận\right)\\ x-1=1\Rightarrow x=2\left(nhận\right)\\ x-1=2\Rightarrow x=3\left(nhận\right)\\ x-1=3\Rightarrow x=4\left(nhận\right)\\ x-1=6\Rightarrow x=7\left(nhận\right)\\ Vậy:x\in\left\{0;2;3;4;7\right\}\)
42 + (x - 6)2 : 3 = 45
(x - 6)2 : 3 = 45 - 42
(x - 6)2 : 3 = 3
(x - 6)2 = 3 x 3
(x - 6)2 = 32
x - 6 = 3
x = 3 + 6
x = 9
ủa lớp 1 học nhân chia r hẻ??????