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1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`a)`
`5(x + 35) = 515`
`=> x + 35 = 515 \div 5`
`=> x+ 35 = 103`
`=> x = 103 - 35`
`=> x = 68`
Vậy, `x = 68`
`b)`
`12x - 33 = 3^2 * 3^3`
`=> 12x - 33 = 3^5`
`=> 12x = 3^5 + 33`
`=> 12x = 276`
`=> x = 276 \div 12`
`=> x = 23`
Vậy, `x = 23`
`c)`
`6x - 5 = 19`
`=> 6x = 19 + 5`
`=> 6x = 24`
`=> x = 24 \div 6`
`=> x = 4`
Vậy, `x = 4`
`d)`
`4(x - 12) + 9 = 17`
`=> 4(x - 12) = 17 - 9`
`=> 4(x - 12) = 8`
`=> x - 12 = 8 \div 4`
`=> x - 12 = 2`
`=> x = 14`
Vậy, `x = 14`
`e)`
123 - 5(x + 4) = 38`
`=> 5(x + 4)= 123 - 38`
`=> 5(x + 4) =85`
`=> x + 4 = 85 \div 5`
`=> x + 4 = 17`
`=> x = 13`
Vậy, `x = 13`
`f)`
`(3x - 2^4) * 7^3 = 2 * 7^4`
`=> 3x - 2^4 = 2 * 7^4 \div 7^3`
`=> 3x - 2^4 = 2* 7`
`=> 3x - 2^4 = 14`
`=> 3x = 14 + 2^4`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x = 10`
Vậy, `x = 10.`
![](https://rs.olm.vn/images/avt/0.png?1311)
1, \(x=-\dfrac{7}{3}-\dfrac{1}{3}=-\dfrac{8}{3}\)
2, \(x=\dfrac{1}{8}-\dfrac{3}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)
3, \(x=\dfrac{6}{5}+\dfrac{4}{5}=\dfrac{10}{5}=2\)
4, \(x=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)
5, \(x+\dfrac{7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{5}{3}-\dfrac{7}{3}=-\dfrac{2}{3}\)
\(=x+\dfrac{1}{3}=\dfrac{-7}{3}\Leftrightarrow x=\dfrac{-8}{3}\)
\(=\dfrac{1}{8}-x=\dfrac{3}{8}\Leftrightarrow x=\dfrac{-1}{4}\)
\(=x-\dfrac{4}{5}=\dfrac{6}{5}\Leftrightarrow x=2\)
\(=\dfrac{2}{3}+x=\dfrac{7}{3}\Leftrightarrow x=\dfrac{5}{3}\)
\(=x-\dfrac{-7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{-2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)
=> \(x=56\)
2) => 18x = 18
=> x = 1
3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)
=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)
=> \(x=\dfrac{-1}{2}\)
4) 45%.x =\(\dfrac{3}{5}\)
=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)
=> \(x=\dfrac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.219 - 7(x+1) = 100
7(x+1) = 219 - 100
7(x+1) = 119
x + 1 = 119 : 7
x + 1 = 17
x = 17 - 1
x = 16
b. (3x - 6 ) . 3 = 36
3x - 6 = 36 : 3
3x - 6 = 12
3x = 12 + 6
3x = 18
x = 18 : 3
x = 6
c.716 - ( x-143) = 659
x-143 = 716 - 659
x-143 = 57
x = 57 + 143
x = 200
b. 30 - [4(x-2)+15] = 3
4(x-2) + 15 = 30 - 3
4(x-2)+15 = 27
4(x-2) = 27 - 15
4(x-2) = 12
x-2 = 12 : 4
x-2 = 3
x = 2 + 3 = 5
e.[(8x - 12) : 4] .33 = 36
[(8x - 12) : 4] . 27 = 729
(8x - 12) : 4 = 729 : 27 = 27
8x - 12 = 27 . 4 = 108
8x = 108 + 12 = 120
x = 120 : 8 = 15
a) \(\Leftrightarrow7\left(x+1\right)=119\\ \Leftrightarrow x+1=17\\ \Leftrightarrow x=16\)
b) \(\Leftrightarrow9\left(x-2\right)=36\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\)
c) \(\Leftrightarrow x-143=57\\ \Leftrightarrow x=200\)
d) \(\Leftrightarrow4\left(x-2\right)+15=27\\ \Leftrightarrow4\left(x-2\right)=12\\ \Leftrightarrow x-2=3\\ \Leftrightarrow x=5\)
e) \(\Leftrightarrow\left(2x-3\right).4:4=3^3\\ \Leftrightarrow2x-3=27\\ \Leftrightarrow2x=24\\ \Leftrightarrow x=12\)
`x+(-1/2)=3`
`=>x= 3-(-1/2)`
`=>x= 3+1/2`
`=>x= 6/2+1/2`
`=>x=7/2`
`---------`
`x-3/4 =4/3`
`=>x= 4/3 + 3/4`
`=>x= 16/12+9/12`
`=>x= 25/12`
`------------`
`3/4+(x-1/2)=5/6`
`=>x-1/2=5/6-3/4`
`=>x-1/2=20/24 - 18/24`
`=>x-1/2=2/24`
`=>x-1/2=1/12`
`=>x= 1/12 +1/2`
`=>x= 1/12+ 6/12`
`=>x=7/12`
ây