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a) \(x\in\left\{34;51;68;85;102;119;136\right\}\)
b) \(x\in\left\{18;36;54;72\right\}\)
c) \(x\in\left\{12;18\right\}\)
d) \(x\in\left\{1;2;3;5;6\right\}\)
e) \(x\in\left\{3;4;6;10\right\}\)
k) \(x\in\left\{1;2;4;11\right\}\)
a, \(x-\left(36:18\right)=12\)
\(x-2=12\)
\(x=12+2\)
\(x=14\)
b, \(\left(x-36\right):18=12\)
\(x-36=12.18\)
\(x-36=216\)
\(x=216+36\)
\(x=252\)
a) x + 15 = 36 - 2x
x + 15 = 36 - (x + x )
15 =36 - ( x + x) - x
15 = 36 - x - x - x
15 = 36 - 3x
3x = 36 - 15
3x = 21
x = 21 : 3
=> x = 7
b) (x - 7) - (2x +5) = -14
x - 7 -( 2x + 5) = -14
x - (2x + 5) = -14 + 7 = -7
x - 2x - 5 = -7
x - 2x = -7 + 5 = -2
x - x + x = 2
x = 2 (-x + x cũng bằng chính nó)
=> x = 2
c) (x - 12) - 15 = (-7 + 20) - (18+x)
(x - 12) - 15 = 13 - (18 + x)
(x - 12) - 15 = 13 - 18 - x
(x - 12) - 15 = -5 - x
15 = (x - 12 ) - (-5 - x)
15 = x - 12 + 5 + x
15 = x + (-12) + 5 + x
15 = 2x + [(-12) + 5]
15 = 2x + -7
2x = -7 + 15
2x = 8
x = 8 : 2
=> x = 4
..................
a) 4x - 15 = - 75 - x
<=> 4x + x = -75 + 15
<=> 5x = -60
<=> x = -12
b) 72 - 3x = 5x + 8
<=> -3x -5x = 8 - 72
<=> -8x = -64
<=> x = 8
d) 3 . | x - 7 | = 21
<=> | x - 7 | = 7
\(\Leftrightarrow\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=0\end{cases}}\)
\(-7\left|x+3\right|=-49\)
\(\Leftrightarrow\left|x+3\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}\)
Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
a) 45 ⋮ x
Vì 45 ⋮ x nên x E Ư( 45 )
= { 1;3;5;9;15;45 }
mà x E Ư(45)
=> x E { 1;3;5;9;15;45 }
b) 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x và x lớn nhất
Vì 24 ⋮ x ; 36 ⋮ x ; 160 ⋮ x nên x E ƯC ( 24;36;160)
mà x lớn nhất
=> x E ƯCLN ( 24;36;160 )
Ta có
24 = 23 . 3
36 = 22.32
160 = 25 . 5
=> ƯCLN ( 24;36;160 ) = 22 = 4
a)\(\frac{1}{6}=\frac{x}{24}\Leftrightarrow6x=24\Leftrightarrow x=\frac{24}{6}\Leftrightarrow x=4\)
b) \(\frac{x}{8}=\frac{-1}{2}\Leftrightarrow2x=-8\Leftrightarrow x=-4\)
d) \(\frac{x}{-17}=\frac{-17}{x}\Leftrightarrow x^2=\left(-17\right)^2\Leftrightarrow x=\pm17\)
e) \(\frac{x+1}{20}=-\frac{1}{5}\Leftrightarrow5x+5=-20\Leftrightarrow5x=-25\Leftrightarrow x=-5\)
f) \(\frac{12}{7}=\frac{-36}{8-x}\Leftrightarrow96-12x=-252\Leftrightarrow12x=348\Leftrightarrow x=29\)
hok tốt!!
a) Có 1/6 = x/24
=>4/24=x/24
=>x=4
b) x/8= -1/2
mà -1/2= -4/8
Do x/8= -1/2
=>x= -4
Các câu khác bn làm tương tự
a, (x + 30) – 75 = 125
=> x + 30 = 125 + 75 = 200
=> x = 200 – 30
=> x = 170
Vậy x = 170
b, x – 72 : 36 = 18
=> x – 2 = 18
=> x = 18 + 2 = 20
Vậy x = 20
c, x – 17 = 54
=> x = 54 +17
=> x = 71.
Vậy x = 71
d, 36 – (x – 2) = 12
=> x – 2 = 36 – 12
=> x = 24 + 2 = 26
Vậy x = 26
e, 9x – 7 = 837
=>9x = 837 + 7 = 844
=> x = 844 9
Vậy x = 844 9
f, (x – 15) – 107 = 0
=> x – 15 = 107
=> x = 107 +15
=> x = 122.
Vậy x = 122
g, 134 + (116 – x) = 145
=> 116 – x = 145 – 134
=> x = 116 – 11
=> x = 5.
Vậy x = 5