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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
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!!!
\(\text{- ( 2789 _ 435 ) + ( 1789 _ 1435 )}\)
\(=-2789+435+1789-1435\)
\(=\left(-2789+1789\right)+\left(435-1435\right)\)
\(=-1000+-1000\)
\(=-2000\)
\(=-\left(-2010\right)+36.41-36.\left(-59\right)\)
\(=2010+36.\left(41+59\right)\)
\(=2010+36.100\)
\(=2010+3600\)
\(=5610\)
\(-75.\left(18-65\right)-65.\left(75-18\right)\)
\(=-75.18+75.65-65.75+65.18\)
\(=18.\left(-75+65\right)+75.\left(65-65\right)\)
\(=18.\left(-10\right)+75.0\)
\(=-180\)
\(-15:x=3\)
\(x=-15:3\)
\(x=-5\)
\(-3x+8=7\)
\(-3x=-1\)
\(x=\frac{1}{3}\)
\(\left(x-6\right).\left(7-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}}\)
\(\Rightarrow x\in\left\{6;7\right\}\)
\(2.\left(x-3\right)-3.\left(x-5\right)=4.\left(3-x\right)-18\)
\(2x-6-3x+15=12-4x-18\)
\(2x-3x+4x=12-18-15+6\)
\(3x=-15\)
\(\Rightarrow x=-5\)
\(-a.\left(c-d\right)-d.\left(a+c\right)=-c.\left(a+d\right)\)
\(-a.c+a.d-d.a+-d.c=-c.\left(a+d\right)\)
\(-c.\left(a+d\right)+a.\left(d-d\right)=-c.\left(a+d\right)\)
\(-c.\left(a+d\right)+a.0=-c.\left(a+d\right)\)
\(\Rightarrow-c.\left(a+d\right)=-c.\left(a+d\right)\)
(3a+2).(2a–1)+(3–a).(6a+2)–17.(a–1)
=6a²−3a+4a−2+18a+6−6a²−2a−17a+17
=(6a²−6a²)+(−3a+4a+18a−2a−17a)+(17−2+6)
=0+0+21
=21
học tốt
a, 440 + 2 . (125 - x) = 546
2 . (125 - x) = 546 - 440
2 . (125 - x) = 106
125 - x = 106 : 2
125 - x = 53
x = 125 - 53
x = 72
Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
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