120 - (39+ 2x) = 11
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(120-\left(39+2x\right)=11\)
\(\Rightarrow39+2x=109\)
\(\Rightarrow2x=70\)
\(\Rightarrow x=35\)
Vậy x=35
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5x+x=39-3^{11}:3^9\)
\(\Leftrightarrow6x=39-3^2\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
b) \(2^x:2^5=16\)
\(\Leftrightarrow2^x:2^5=2^4\)
\(\Leftrightarrow2^x=2^4.2^5\)
\(\Leftrightarrow2^x=2^9\)
\(\Leftrightarrow x=9\)
c) \(7x-x=5^{21}:5^{19}+3.2^2-7^0\)
\(\Leftrightarrow6x=5^2+3.4-1\)
\(\Leftrightarrow6x=36\)
\(\Leftrightarrow x=6\)
d) \(7x-2x=6^{17}:6^{15}+44:11\)
\(\Leftrightarrow5x=6^2+4\)
\(\Leftrightarrow5x=40\)
\(\Leftrightarrow x=8\)
a)⇔6x=39-32
⇔6x=30
⇔ x=5
b)2x:25=16
⇔2x=24.25
⇔ 2x=29
⇔ x=9
c)⇔6x=52+3.22-1
⇔ 6x= 36
⇔ x=6
d)⇔5x=62+4
⇔ 5x=40
⇔ x=8
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2x + 15 = 45
2x = 45 - 15
2x = 30
x = 30 : 2
x = 15 (nhận)
Vậy x = 15
b) 120 - 2.(x + 3) = 22.52
120 - 2.(x + 3) = 1144
2.(x + 3) = 120 - 1144
2.(x + 3) = - 1024
x + 3 = -1024 : 2
x + 3 = -512
x = - 512 - 3
x = -515 (loại)
Vậy không tìm được x thỏa mãn x là số tự nhiên
c) 11 ⋮ (x - 2)
⇒ x - 2 ∈ Ư(11) = {-11; -1; 1; 11}
⇒ x ∈ {-9; 1; 3; 13}
Do x là số tự nhiên
⇒ x ∈ {1; 3; 13}
d) Do 12 ⋮ x và 18 ⋮ x nên x ∈ ƯC(12; 18)
12 = 2².3
18 = 2.3²
ƯCLN(12; 18) = 2.3 = 6
⇒ x ∈ ƯC(12; 18) = {1; ; 3; 6}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 72,5 - (2x + 1) = 630 : 9
⇒ 72,5 - (2x + 1) = 70
⇒ 2x + 1 = 2,5
⇒ 2x = 1,5
⇒ x = 0,75
b) (10 - 4x) + 120 : 23 = 17
⇒ (10 - 4x) + 120 : 8 = 17
⇒ (10 - 4x) + 15 = 17
⇒ 10 - 4x = 2
⇒ 4x = 8
⇒ x = 2
Vậy x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có : \(3^{39}=\left(3^{13}\right)^3=1594323^3\)
\(11^{21}=\left(11^7\right)^3=19487171^3\)
Vì \(1594323< 19487171\)
\(=>1594323^3< 19487171^3\)
\(=>3^{39}< 11^{21}\)
Vậy \(3^{39}< 11^{21}\)
b)Ta có : \(199^{20}=\left(199^4\right)^5=1568239201^5\)
\(2002^{15}=\left(2002^3\right)^5=8024024008^5\)
Vì \(1568239201< 8024024008\)
\(=>1568239201^5< 8024024008^5\)
\(=>199^{20}< 2002^{15}\)
Vậy \(199^{20}< 2002^{15}\)
c) Ta có:\(125^{90}=\left(125^3\right)^{30}=1953125^{30}\)
\(25^{120}=\left(25^4\right)^{30}=390625^{30}\)
Vì \(1953125>390625\)
\(=>1953125^{30}>390625^{30}\)
\(=>125^{90}>25^{120}\)
Vậy \(125^{90}>25^{120}\)
d)Ta có : \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=\left(7^3\right)^{100}=343^{100}\)
Vì \(243< 343\)
\(=>243^{100}< 343^{100}\)
\(=>3^{500}< 7^{300}\)
Vậy \(3^{500}< 7^{300}\)
Phù , cuối cùng cũng viết xong . Mỏi tay quá ! À , chúc bạn học tốt nhé !
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![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(2x+\dfrac{11}{6}+\dfrac{23}{12}+\dfrac{39}{20}+\dfrac{59}{30}+\dfrac{83}{42}+\dfrac{111}{56}=12\)
\(\Leftrightarrow2x+\left(2+2+2+2+2+2\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\right)=12\)
\(\Leftrightarrow2x-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
\(\Leftrightarrow2x-\dfrac{3}{8}=0\)
hay x=3/16
a) 7^2.5-(2x+1)=630:9
b) (10-4x)+120:2^3=17
c) [124-(20-4X)]:30+7=11
d) |2x-5|=1
nhớ ghi đầy đủ bước nhá
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\)\(.\)\(7^2.5-\left(2x+1\right)=630:9\)
\(49.5-\left(2x+1\right)=70\)
\(245-\left(2x+1\right)=70\)
\(2x+1=245-70\)
\(2x+1=175\)
\(2x=175-1\)
\(2x=174\)
\(x=174:2\)
\(x=87\)
Vậy \(x=87\)
\(b.\)\(\left(10-4x\right)+120:2^3=17\)
\(\left(10-4x\right)+120:8=17\)
\(\left(10-4x\right)+15=17\)
\(10-4x=17-15\)
\(10-4x=2\)
\(4x=10-2\)
\(4x=8\)
\(x=8:4\)
\(x=2\)
Vậy \(x=2\)
\(c.\)\(\left[124-\left(20-4x\right)\right]:30+7=11\)
\(\left[124-\left(20-4x\right)\right]:30=11-7\)
\(\left[124-\left(20-4x\right)\right]:30=4\)
\(\left[124-\left(20-4x\right)\right]=4.30\)
\(\left[124-\left(20-4x\right)\right]=120\)
\(124-20+4x=120\)
\(104+4x=120\)
\(4x=120-104\)
\(4x=16\)
\(x=16:4\)
\(x=4\)
Vậy \(x=4\)
\(d.\)\(\left|2x+5\right|=1\)
\(\Rightarrow2x+5\in\left\{-1;1\right\}\)
\(\Rightarrow2x\in\left\{-6;-4\right\}\)
\(\Rightarrow x\in\left\{-3;-2\right\}\)
Vậy \(x\in\left\{-3;-2\right\}\)
học tốt
a) \(7^2\cdot5-\left(2x+1\right)=630:9\)
\(\Leftrightarrow49\cdot5-2x-1=70\)
\(\Leftrightarrow244-2x=70\)
\(\Leftrightarrow2x=244-70\)
\(\Leftrightarrow2x=174\)
\(\Leftrightarrow x=87\)
Vậy x=87.
b) \(\left(10-4x\right)+120:2^3=17\)
\(\Leftrightarrow10-4x+120:8=17\)
\(\Leftrightarrow10-4x+15=17\)
\(\Leftrightarrow25-4x=17\)
\(\Leftrightarrow4x=25-17=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
c)\([124-\left(20-4x\right)]:30+7=11\)
\(\Leftrightarrow\left(124-20+4x\right):30=4\)
\(\Leftrightarrow104+4x=120\)
\(\Leftrightarrow4x=120-104=16\)
\(\Leftrightarrow x=4\)
Vậy x=4
d) \(|2x-5|=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
Vậy \(x\in\text{ }\left\{3;2\right\}\)
120 - (39+2x) = 11
(39+2x) = 120 - 11
39 + 2x = 109
2x = 109 - 39
2x = 70
x = 70 : 2
x = 35
HOK TỐT
Trả lời :
x = 35
Hok tốt Phương
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