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H9
HT.Phong (9A5)
CTVHS
19 tháng 6 2023
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
PT
1
CM
8 tháng 9 2018
a) x + 4 = – 14 – 9
x + 4 = – 23
x = – 23 – 4
x = – 27
b) 3x = – 14 + 2
3x = – 12
x = – 12 : 3
x = – 4
c) 2| x | = 4 – (– 8)
2| x | = 12
| x | = 6
x = 6 hoặc x = – 6
d) |x – 2| = 7
x – 2 = 7 hoặc x – 2 = – 7
x = 9 hoặc x = – 5
3 tháng 2 2016
\(-\frac{8}{3}x-1=\frac{4}{-7}\)
\(-\frac{8}{3}x=\frac{4}{-7}+1\)
\(-\frac{8}{3}x=\frac{3}{7}\)
\(x=\frac{3}{7}:-\frac{8}{3}\)
\(x=-\frac{3}{7}.-\frac{3}{8}\)
\(x=\frac{9}{56}\)
\(\frac{x}{-3}=-\frac{3}{x}\)
\(\left(x\right)^2=-3.-3\)
\(\left(x\right)^2=9\)
\(x=\sqrt{9}\)
\(\Leftrightarrow x=3\)
DT
3
12 tháng 9 2018
a) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-4^3=32\)
12x = 96
x = 8
12 tháng 9 2018
b) \(\left(3x-2^4\right).7^3=2.7^4\)
3x - 24 = 14
3x = 30
x = 10
NA
1
SN
20 tháng 7 2021
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
15 tháng 4 2023
3:
\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)
4:
=>2/3:x=-2-1/3=-7/3
=>x=-2/3:7/3=-2/7
5:
AC=CB=12/2=6cm
IB=6/2=3cm
21 tháng 2 2020
A,-16+23+x=-16
<=>x=-16+16-23
<=>x=-23
B,10-2(4-3x)=-4
<=>10-8+6x=-4
<=>6x=-4-10+8
<=>6x=-6
<=>x=-1
C,-12+3(-x+7)=-18
<=>-12-3x+21=-18
<=>-3x=-18+12-21
<=>-3x=-27
<=>x=9
D,24:(3x-2)=-3
<=>24/(3x-2)=-3
<=>-3(3x-2)=24
<=>3x-2=-8
<=>3x=-6
<=>x=-2
E,|x+8|-7=8
<=>|x+8|=15
<=>x+8=+-15
Chia 2 TH:
TH1:x+8=15
<=>x=7
TH2:x+8=-15
<=>x=-23
F,-45:5(-3-2x)=3
<=>-9(-3-2x)=3
<=>27+18x=3
<=>18x=-24
<=>x=-4/3
G,|x-1|=0
<=>x-1=0
<=>x=1
H,-13|x|=-26
<=>|x|=2
<=>x=+-2
22 tháng 2 2020
a) -16 + 23 + x = -16
x = -16 + 16 - 23
x = -23
Vậy x = -23
b) 10 - 2. (4 - 3x) = -4
10 - 2 . 4 + 2. 3x = -4
10 - 8 + 6x = -4
6x = -4 - 10 + 8
6x = -6
x = (-6) : 6
x = -1
Vậy x = -1
c) -12 + 3. (-x + 7) = -18
-12 + 3. (-x) + 3. 7 = -18
-12 + (-3x) + 21 = -18
-3x = -18 + 12 - 21
-3x = -27
x = (-27) : (-3)
x = -9
Vậy x = -9
d) 24 : (3x - 2) = -3
3x - 2 = 24 : (-3)
3x - 2 = -8
3x = -8 + 2
3x = -6
x = (-6) : 3
x = -2
Vậy x = -2
e) lx + 8l - 7 = 8
lx + 8l = 8 + 7
lx + 8l = 15
\(\Rightarrow\hept{\begin{cases}x+8=15\\x=18-5\\x=13\end{cases}hay\hept{\begin{cases}x+8=-15\\x=-15-8\\x=-23\end{cases}}}\)
Vậy \(x\in\left\{13;-23\right\}\)
f) (-45) : 5. (-3 - 2x) = 3
(-9). (-3 - 2x) = 3
(-9). (-3) + 9. 2x = 3
27 + 18x = 3
18x = 3 - 27
18x = -24
x = (-24) : 18
x =\(\frac{-4}{3}\)
Vậy x =\(\frac{-4}{3}\)
g) lx - 1l = 0
\(\Rightarrow x-1=0\)
x = 0 + 1
x = 1
Vậy x = 1
h) (-13). lxl = -26
lxl = (-26) : (-13)
lxl = 2
\(\Rightarrow x=2\)
Vậy x = 2
Chúc bạn học tốt!!!
DT
0
3 tháng 6 2021
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
F
22 tháng 1 2020
Bài 1 : Bài giải
\(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)
M
22 tháng 1 2020
a, \(-2\left(-3-4x\right)-3\left(3x-7\right)=31\)
\(6+8x-9x-21=31\)
\(-x-15=31\)
\(-x=31+15\)
\(-x=46\)
\(x=-46\)
b, \(10\left(x-7\right)=8\left(x-4\right)+x\)
\(10x-70=8x-32+x\)
\(10x-70=9x-32\)
\(10x-9x=-32+70\)
\(x=38\)
c, \(2\left|x-1\right|=3\cdot\left(5-1\right)\)
\(2\left|x-1\right|=15-3\)
\(2\left|x-1\right|=12\)
\(\left|x-1\right|=12\text{ : }2\)
\(\left|x-1\right|=6\)
\(\Rightarrow\orbr{\begin{cases}x-1=-6\\x-1=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=7\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-5\text{ ; }7\right\}\)
d, \(2\left(x+3\right)-2\left(x-3\right)=x\)
\(2\left(x+3-x-3\right)=x\)
\(2\cdot0=x\)
\(x=0\)
e, \(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }4\right\}\)