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Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
a) (x - 12) - (2x + 31) = -6 - 5
=> x - 15 - 2x - 31 = -11
=> -x - 46 = -11
=> -x = -11 + 46
=> -x = 35
=> x = -35
b) 2(x - 4)2 - 48 = -16
=> 2(x - 4)2 = -16 + 48
=> 2(x - 4)2 = 32
=> (x - 4)2 = 32 : 2
=> (x - 4)2 = 16
=> (x - 4)2 = 42
=> \(\orbr{\begin{cases}x-4=4\\x-4=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=8\\x=0\end{cases}}\)
Vậy ...
c) (3x + 9)(11 - x) = 0
=> \(\orbr{\begin{cases}3x+9=0\\11-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}3x=-9\\x=11\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=11\end{cases}}\)
Vậy ...
d) 2 - |x + 5| = 7
=> |x + 5| = 2 - 7
=> |x + 5| = -5
=> ko có giá trị x thõa mãn
vì |x + 5| \(\ge\)0 mà |x + 5| = -5
d) 10 - 2|x + 5| = 2
=> 2|x + 5| = 10 - 2
=> 2|x + 5| = 8
=> |x + 5| = 8 : 2
=> |x + 5| = 4
=> \(\orbr{\begin{cases}x+5=4\\x+5=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=-9\end{cases}}\)
Vậy ...
Tự ghi đề nhé!
a. \(\frac{-3}{4}:x=\frac{-11}{36}+\frac{1}{2}\)
= 7/36
x = 7/36 : -3/4 = -7/27
a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha
Dạng 3 :
a) 3x - 10 = 2x + 13
=> 3x - 2x = 13 - 10
=> x = 3
b) x + 12 = -5 - x
=> x + x = -5 - 12
=> 2x = -17
=> x = -8,5
c) x + 5 = 10 - x
=> x + x = 10 - 5
=> 2x = 5
=> x = 2,5
d) 6x + 23 = 2x - 12
=> 2x - 6x = 23 + 12
=> -4x = 35
=> x = -8,75
e) 12 - x = x + 1
=> x + x = 12 - 1
=> 2x = 11
=> x = 5,5
f) 14 + 4x = 3x + 20
=> 4x - 3x = 20 - 14
=> x = 6
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
1)
a)-24+3(x-4)=111
3(x-4)=111-(-24)
3(x-4)=111+24
3(x-4)=135
x-4=135:3
x-4=45
x =45+4
x =49
b)(2x-4)(3x+63)=0
\(\Rightarrow\)\(\orbr{\begin{cases}2x-4=0\\3x+63=0\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=2\\x=-21\end{cases}}\)
Vậy x\(\in\){2;-21}
c)|x-7|-4=(-2)4
|x-7| =(-2)4+4
|x-7| =16+4
|x-7| =20
\(\Rightarrow\)\(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=14\\x=0\end{cases}}\)
Vậy x\(\in\){14;0}
d)(x-1)2=144
(x-1)2=122
\(\Rightarrow\)x-1=12
x =12+1
x =13
e)(x+7)3=-8
(x+7)3=(-2)3
\(\Rightarrow\)x+7=-2
x =-2-7
x =-9
2)
a)Ta có:
\(3n+12⋮n-3\)
\(\Rightarrow3n-9+21⋮n-3\)
\(\Rightarrow3\left(n-3\right)+21⋮n-3\)
\(\Rightarrow21⋮n-3\)
\(\Rightarrow n-3\inƯ\left(21\right)\)
\(\Rightarrow n-3\in\left\{1;3;7;21\right\}\)
Ta có bảng sau:
n-3 | 1 | 3 | 7 | 21 |
n | 4 | 6 | 10 | 24 |
Vậy\(n\in\left\{4;6;10;24\right\}\)
b)Ta có:
\(n+9⋮n-1\)
\(\Rightarrow n-1+10⋮n-1\)
\(\Rightarrow10⋮n-1\)
\(\Rightarrow\)\(n-1\inƯ\left(10\right)\)
\(\Rightarrow n-1\in\left\{1;2;5;10\right\}\)
Ta có bảng sau:
n-1 | 1 | 2 | 5 | 10 |
n | 2 | 3 | 6 | 11 |
Vậy \(n\in\left\{2;3;6;11\right\}\)
a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
52 - 2x = - 11
2x = 52 - ( -11)
2x = 63
x = 63/2
b, (-3) .x - 61 = 5
-3 x = 5 + 61
- 3x = 66
x = 66 : ( -3)
x = - 22
c, 12 - 2. ( x-4) = 13. 22
12 - 2x + 8 = 52
20 - 2x = 52
2x = 20 - 52
2x = - 32
x = -32 : 2
x = -16
d, 12 - (3x + 4) = - 7
12 - 3x - 4 = - 7
8 - 3x = -7
3x = 8 - (-7)
3x = 15
x = 15:3
x = 5
a. -2x= 63
<=> x= -31,5
b.(-3)x=66
<=>x=-22
c. -2(x-4)=12-13x2^2
<=> -2(x-4)=-40
<=> x-4 = 20
<=> x=24
d.3x+4=12-(-7)
<=>3x=15
<=> x= 5