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a,
128-3x-12=23
3x=128-12-23
3x=93
x=93:3
= 31
b,
(12x+84+55):5=35
12x+84+55=35.5
12x+84+55=175
12x=175-55-84
12x=36
x=36:12
x=3
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
\(a,128-3.\left(x+4\right)=23\\ \Rightarrow3.\left(x+4\right)=105\\ \Rightarrow x+4=35\\ \Rightarrow x=31\\ b,\left[\left(4x+28\right).3+55\right]:5=35\\ \Rightarrow\left(4x+28\right).3+55=175\\ \Rightarrow4x+28.3=120\\ \Rightarrow4x+28=60\\ \Rightarrow4x=32\\ \Rightarrow x=8.\)
c) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-64=4.8^4:8^3\)
\(12x-64=32\)
\(12x=32+64\)
\(12x=96\)
\(x=\dfrac{96}{12}\)
\(x=8\)
d) \(720:\left[41-\left(2x-5\right)\right]:5=35\)
\(720:\left(41-2x+5\right):5=35\)
\(720:\left(46-2x\right)=35.5\)
\(720:\left(46-2x\right)=175\)
\(46-2x=720:175\)
\(46-2x=\dfrac{144}{35}\)
\(2x=46-\dfrac{144}{35}\)
\(2x=\dfrac{1466}{35}\)
\(x=\dfrac{1466}{35}:2\)
\(x=\dfrac{733}{35}\)
a, 128 – 3(x+4) = 23
b, 12 x - 4 3 . 8 3 = 4 . 8 4
c, [(4x+28).3+55]:5 = 35
d, 720:[41 – (2x – 5)] = 2 3 . 5
Bài toán 1 : Tìm x nguyên biết.
a. 3 ≤ x – 2 < 5
=>3-2 < x-2 < 5-2
=>1 < x < 3
=>x=1
Vậy x=1
b. 0 ≤ x – 5 ≤ 2
=>0-5 < x-5 < 2-5
=>-5 < x < -3
=>x=-4
Vậy x=-4
Bài toán 2 : Tính hợp lý.
a. 4567 + (1234 – 4567) -4
=4567+1234-4567-4
=(4567-4567)+(1234-4)
=0+1230
=1230
b. 2001 – (53 + 1579) – (-53)
=2001-53-1579+53
=(2001-1579)+(-53+53)
=422+0
=422
c. 35 – 17 + 2017 – 35 + (-2017)
=(35-35)+[2017+(-2017)]-17-35
=0+0-17-35
=0-17-35
=-52
d. 37 + (-17) – 37 + 77
=(37-37)+[-17+77]
=0+60
=60
e. –(-219) + (-219) – 401 + 12
=0-401+12
=-389
f. |-85| – (-3).15
=85+3.15
=85+45
=130
g. 11.107 + 11.18 – 25.11
=11.(107+18-25)
=11.100
=1100
h. 115 – (-85) + 53 – (-500 + 53)
=115+85+53+500-53
=(53-53)+(115+85+500)
=0+700
=700
k. (-18) + (-31) + 98 + |-18| + (-69)
=-18-31+98+18-69
=(-18+18)+(-31-69)+98
=0+(-100)+98
=-100+98
=-2
Bài toán 4 : Tìm x, biêt.
a. 5x – 16 = 40 + x
5x - x = 40 + 16
4x = 56
x = 56 : 4
x=14
b. 4x – 10 = 15 – x
4x + x = 15 + 10
5x = 25
x = 25 : 5
x = 5
c. -12 + x = 5x – 20
-12 + 20 = 5x - x
8 = 4x
x = 8 : 4
x = 2
d. 7x – 4 = 20 + 3x
7x - 3x = 20 + 4
4x=24
x=24:4
x=6
e. 5x – 7 = – 21 – 2x
5x + 2x = -21+7
7x = -14
x = -14 : 7
x = -2
f. x + 15 = 7 – 6x
x + 6x = 7 - 15
7x = -8
x = -8/7
g. 17 – x = 7 – 6x
17 - 7 = -6x + x
10 = -7x
x=10/-7
h. 3x + (-21) = 12 – 8x
3x + 8x = 12 + 21
11x = 33
x = 33:11
x=3
k. 125 : (3x – 13) = 25
3x-13=125:25
3x-13=5
3x=5+13
3x=18
x=18:3
x=6
l. 541 + (218 – x) = 735
218-x=735-541
218-x=693
x=218-693
x=-475
\(a.5\cdot\left(12-x\right)-20=30\\ 60-5x-20=30\\ 5x=10\\ x=2\\ b.\left(50-6x\right)\cdot18=8\cdot9\cdot5\\ 900-108x=360\\ 180x=540\\ x=3\\ c.128-3\cdot\left(x+4\right)=23\\ 128-3x-12=23\\ 3x=93\\ x=31\\ d.\left[\left(4x+28\right):3+55\right]:5=35\\ \left(4x+28\right):3=120\\ 4x+28=360\\ 4x=332\\ x=83\\ e.6x+4x=2010\\ x\cdot\left(6+4\right)=2010\\ 10x=2010\\ x=201\\ f.200-\left(2x+6\right)=64\\ 2x+6=136\\ 2x=130\\ x=65\)
\(g.135-5\cdot\left(x+4\right)=35\\ 5\cdot\left(x+4\right)=100\\x+4=20\\x=16\)
\(h.12x+13x=2000\\ x\cdot\left(12+13\right)=2000\\ x\cdot25=2000\\ x=80\\ i.\left(x-4\right)\cdot\left(x-3\right)=0\\ \Rightarrow x-4=0\Rightarrow x=4\\ x-3=0\Rightarrow x=3\\ \text{ vậy x = 4 hoặc x = 3 thì }\left(x-4\right)\cdot\left(x-3\right)=0\\ \)