55 ⋮ (x-2)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$(-55).(-17)-55.(-2+17)=(-55)(-17)-55.15=(-55)(-17)+(-55).15$
$=(-55)(-17+15)=(-55).(-2)=55.2=110$
![](https://rs.olm.vn/images/avt/0.png?1311)
d,155-10(x+1)=55
10(x+1) = 155 - 55
10(x+1) = 100
x+1 = 100 : 10
x+1 = 10
x = 10 - 1 =9
e,6(x+\(2^3\))+40=100
6(x+\(2^3\)) = 100 - 40
6(x+\(2^3\)) = 60
(x+\(2^3\)) = 60 : 6
(x+\(2^3\)) = 10
x = 10 - \(2^3\)
x = 10 - 8 = 2
f,\(2^2\)(x+\(3^2\))-5=55
\(2^2\)(x+\(3^2\)) = 55 + 5
\(2^2\)(x+\(3^2\)) = 60
(x+\(3^2\)) = 60 : \(2^2\)
(x+\(3^2\)) = 60 : 4
(x+\(3^2\)) = 15
x + 9 = 15
x = 15-9=6.
Hok tốt !
d) \(155-10\left(x+1\right)=55\)
\(\Rightarrow10\left(x+1\right)=155-55\)
\(\Rightarrow10\left(x+1\right)=100\)
\(\Rightarrow x+1=100:10\)
\(\Rightarrow x+1=10\)
\(\Rightarrow x=10-1\)
\(\Rightarrow x=9\)
Vậy x = 9
e) \(6\left(x+2^3\right)+40=100\)
\(\Rightarrow6\left(x+2^3\right)=100-40\)
\(\Rightarrow6\left(x+2^3\right)=60\)
\(\Rightarrow x+2^3=60:6\)
\(\Rightarrow x+2^3=10\)
\(\Rightarrow x+8=10\)
\(\Rightarrow x=10-8\)
\(\Rightarrow x=2\)
Vậy x = 2
f) \(2^2\left(x+3^2\right)-5=55\)
\(\Rightarrow2^2\left(x+3^2\right)=55+5\)
\(\Rightarrow2^2\left(x+3^2\right)=60\)
\(\Rightarrow4\left(x+9\right)=60\)
\(\Rightarrow x+9=60:4\)
\(\Rightarrow x+9=15\)
\(\Rightarrow x=15-9\)
\(\Rightarrow x=6\)
Vậy x = 6
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ 22(x + 32) - 5 = 55 b/ 155 - 10(x+1) = 55
4(x + 32) = 60 10(x + 1) = 100
x + 32 = 60 : 4 x + 1 = 10
x + 32 = 15 x = 10 - 1
x = 15 - 32 x = 9
x = - 17
c/ 5. 22 + (x+3) = 52 53 - 5 (4 + x) = 15
20 + (x+3) = 25 125 - 5(4 +x) = 15
x+ 3 = 5 5(4 + x) = 110
x = 5-3 4 + x= 110 : 2
x = 2 4 + x = 55
x = 51
ủng hộ mk nha !!!! ^_^
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{x}{14}-\dfrac{1}{7}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x}{14}=\dfrac{-3}{4}+\dfrac{1}{7}=\dfrac{-21}{28}+\dfrac{4}{28}=\dfrac{-17}{28}\)
hay \(x=\dfrac{-17\cdot14}{28}=\dfrac{-17}{2}\)
Vậy: \(x=-\dfrac{17}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=55\)
\(\Leftrightarrow x^4-10x^2-4x^2+40=55\)
\(\Leftrightarrow x^4-14x^2-15=0\)
Đặt \(t=x^2\left(t\ge0\right)\), ta có: \(t^2-14t-15=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=15\\t=-1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{15}\\x=-\sqrt{15}\end{matrix}\right.\)
<=>\(\left(x^2-4\right)\left(x^2-10\right)=55\)
<=>\(x^4-14x^2+40=55\)
<=>\(x^4-14x^2-15=0\)
<=>\(\left(x^2-15\right)\left(x^2+1\right)=0\)
<=>\(x^2-15=0\)(cái kia lun lớn hơn 0)
<=>\(x^2=15\)
<=>\(x=\pm\sqrt{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x:2+250.4=55.61+69.55-45.37-93.45\)
\(\Leftrightarrow x:2+1000=55\left(61+69\right)-45\left(37+93\right)\)
\(\Leftrightarrow x:2+1000=55.130-45.130\)
\(\Leftrightarrow x:2+1000=130\left(55-45\right)\)
\(\Leftrightarrow x:2+1000=130.10\)
\(\Leftrightarrow x:2+1000=1300\)
\(\Leftrightarrow x:2=1300-1000=300\)
\(\Leftrightarrow x=300.2=600\)
Vậy \(x=600\)
X:2+250*4=55*61+69*55-45*37-93*45
X:2+250*4=(55*61+69*55)-(45*37+93*45)
X:2+1000=55*(61+69)-45*(37+93)
X:2+1000=55*130-45*130
X:2+1000=(55-45)*130
X:2+1000=10*130
X:2+1000=1300
X:2=1300-1000
X:2=300
X=300*2
X=600
Vậy X = 600
![](https://rs.olm.vn/images/avt/0.png?1311)
= 11/55+ 10/55<x/55<2/5+1/55
=2/5<x/55<23/55
=22/55<x/55<23/55
=> x thuộc rỗng
x/21=4/7+(-7)/3
= x/21=12/21+(-49)/21
= x/21=-37/21
=> x = -37
Đề là gì vậy em