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a) 125 : x + ( 32 + 15 ) = 72
125 : x + 47 = 72
125 : x = 72 - 47
125 : x = 25
x = 125 : 25
x = 5
b) ( 35 + 135 ) - 2 . x = 86
170 - 2 . x = 86
2 . x = 170 - 86
2 . x = 84
x = 84 : 2
x = 42
c) 7 . x - 5 = 65
7 . x = 65 + 5
7 . x = 70
x = 70 : 7
x = 10
A) 125 / x + (32 + 15) = 72
125 / x + 47 = 72
125 / x = 72 - 47
125 / x = 25
x = 125 / 25
x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 125 : X + (32 + 15) = 72
125 : X + 47 = 72
X + 47 = 72 x 125
X + 47 = 62270
X = 62270 - 47
X = 62223
b) (35 + 15) - 2 x X = 86
50 - 2 x X = 86
50 - X = 86 + 2
50 - X = 88
X = 88+ 50
X = 138
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a) 2.(x+4)+5=65
2.(x+4)=65-5
2.(x+4)=60
x+4=60:2
x+4=30
x=30-4=26
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a) -9-25=(7-x)-(25+7)
<=> −9−25=7−x−25−7
<=> -9=7-x-7
<=> -9=-x
<=> x=9
b) (x-5)(x+8)=0
<=> x-5=0 hoặc x+8=0
<=> x=5 hoặc x=-8
d) Đề sai, x đâu???
...
\(c,0< \left|x-7\right|< 3\)
\(0< x-7< 3\)
\(7< x< 10\)
\(\Rightarrow x\in\left\{8;9\right\}\)
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12.(x-3):3=64+8
12.(x-3):3=72
12.(x-3) =72.3
12.(x-3) =216
x-3 =216:12
x-3 =18
x =18+3
x =21
100-7.(x-5)=65
7.(x-5)=100-65
7.(x-5)=35
x-5=35:7
x-5=5
x =5+5
x =10
Vậy x = 10
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\(\text{- ( 2789 _ 435 ) + ( 1789 _ 1435 )}\)
\(=-2789+435+1789-1435\)
\(=\left(-2789+1789\right)+\left(435-1435\right)\)
\(=-1000+-1000\)
\(=-2000\)
\(=-\left(-2010\right)+36.41-36.\left(-59\right)\)
\(=2010+36.\left(41+59\right)\)
\(=2010+36.100\)
\(=2010+3600\)
\(=5610\)
\(-75.\left(18-65\right)-65.\left(75-18\right)\)
\(=-75.18+75.65-65.75+65.18\)
\(=18.\left(-75+65\right)+75.\left(65-65\right)\)
\(=18.\left(-10\right)+75.0\)
\(=-180\)
\(-15:x=3\)
\(x=-15:3\)
\(x=-5\)
\(-3x+8=7\)
\(-3x=-1\)
\(x=\frac{1}{3}\)
\(\left(x-6\right).\left(7-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}}\)
\(\Rightarrow x\in\left\{6;7\right\}\)
\(2.\left(x-3\right)-3.\left(x-5\right)=4.\left(3-x\right)-18\)
\(2x-6-3x+15=12-4x-18\)
\(2x-3x+4x=12-18-15+6\)
\(3x=-15\)
\(\Rightarrow x=-5\)
\(-a.\left(c-d\right)-d.\left(a+c\right)=-c.\left(a+d\right)\)
\(-a.c+a.d-d.a+-d.c=-c.\left(a+d\right)\)
\(-c.\left(a+d\right)+a.\left(d-d\right)=-c.\left(a+d\right)\)
\(-c.\left(a+d\right)+a.0=-c.\left(a+d\right)\)
\(\Rightarrow-c.\left(a+d\right)=-c.\left(a+d\right)\)
(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
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
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a) pt <=> \(\frac{x\left(x+1\right)}{2}=500500\)
<=> \(x^2+x=1001000\)
<=> \(x^2-1000x+1001x-1001000=0\)
<=> \(\left(x-1000\right)\left(x+1001\right)=0\)
<=> \(\orbr{\begin{cases}x=1000\\x=-1001\end{cases}}\)
Do \(x>0\)=> \(x=1000\)
b)
<=> \(2x=210\)
<=> \(x=105\)
c)
<=> \(6x-81=3.7\)
<=> \(x=17\)
d)
<=> \(125-5\left(3x-1\right)=5^2\)
<=> \(5\left(3x-1\right)=100\)
<=> \(3x-1=20\)
<=> \(x=7\)
e)
<=> \(4^{x+1}+1=65\)
<=> \(4^{x+1}=64\)
<=> \(x+1=3\)
<=> \(x=2\)
j)
<=> \(2\left(2x-3\right)=14\)
<=> \(2x-3=7\)
<=> \(x=5\)
a) \(75:\left(x-2\right)+4=7\)
\(\Rightarrow75:\left(x-2\right)=7-4\)
\(\Rightarrow75:\left(x-2\right)=3\)
\(\Rightarrow x-2=75:3\)
\(\Rightarrow x-2=25\)
\(\Rightarrow x=27\)
b) \(125-3\left(x+3\right)=65\)
\(\Rightarrow3\left(x+3\right)=125-65\)
\(\Rightarrow3\left(x+3\right)=60\)
\(\Rightarrow x+3=60:3\)
\(\Rightarrow x+3=20\)
\(\Rightarrow x=17\)
a,\(75:\left(x-2\right)+4=7\\ \Rightarrow75:\left(x+2\right)=3\\ \Rightarrow x+2=25\\ \Rightarrow x=23.\)
b, \(125-3\left(x+3\right)=65\\ \Rightarrow3\left(x+3\right)=60\\ \Rightarrow x+3=20\\ \Rightarrow x=17.\)