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a, \(15:\left(x+2\right)=3\Leftrightarrow x+2=3\Leftrightarrow x=1\)
b, \(20:\left(x+1\right)=2\Leftrightarrow x+1=10\Leftrightarrow x=9\)
c, \(240:\left(x-5\right)=2^2.5^2-20=80\Leftrightarrow x-5=3\Leftrightarrow x=8\)
d, \(96-3\left(x+1\right)=42\Leftrightarrow3\left(x+1\right)=54\Leftrightarrow x+1=18\Leftrightarrow x=17\)
a) 15 : (x + 2) = 3
x + 2 = 15 : 3
x + 2 = 5
x = 5 – 2 = 3
b) 20 : (1 + x) = 2
1 + x = 20 : 2
1 + x = 10
x = 10 – 1 = 9
c) 240 : (x – 5) = 22.52 – 20
240 : (x – 5) = 4.25 – 20
240 : (x - 5) = 100 – 20
240 : (x - 5) = 80
x – 5 = 240 : 80
x – 5 = 3
x = 3 + 5 = 8
d) 96 - 3(x + 1) = 42
3(x + 1) = 96 – 42
3(x + 1) = 54
x + 1 = 54 : 3
x + 1 = 18
x = 18 – 1
x = 17
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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)
a ) \(15:\left(x+2\right)=3\)
\(\left(x+2\right)=15:3\)
\(x+2=5\)
\(x=5-2\)
\(x=3\)
Vậy \(x=3\)
b) \(20:\left(1+x\right)=2\)
\(\left(1+x\right)=20:2\)
\(1+x=10\)
\(x=10-1\)
\(x=9\)
Vậy \(x=9\)
c) \(240:\left(x-5\right)=2^2.5^2=20\)
\(240:\left(x-5\right)=4.25=20\)
\(240:\left(x-5\right)=100=20\)
\(\Rightarrow\orbr{\begin{cases}240:\left(x-5\right)=100\\240:\left(x-5\right)=20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)=240:100\\\left(x-5\right)=240:20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-5=2,4\\x-5=12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2,4+5\\x=12+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7,4\\x=17\end{cases}}\)
Vậy \(x=7,4\)hoặc \(x=17\)
d) \(96-3\left(x+1\right)=12\)
\(3\left(x+1\right)=96-12\)
\(3\left(x+1\right)=84\)
\(\left(x+1\right)=84:3\)
\(x+1=28\)
\(x=28-1\)
\(x=27\)
Vậy \(x=27\)
Chúc bạn học tốt !!!
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Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}\frac{2}{3}.\frac{3}{4}.......\frac{99}{100}\)
\(=\frac{1}{100}\)
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Ta có: \(240:\left(x-5\right)=2^2\cdot5^2-20\)
\(\Leftrightarrow240:\left(x-5\right)=80\)
\(\Leftrightarrow x-5=3\)
hay x=8
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a, Do UCLN là 5 nên a, b chia hết cho 5 => tận cùng là 0 hoặc 5
Ta có 20 = 15 + 5 = 18 + 2=19+1=17+3=16+4=14+6=13+7=12+8=11+9
=> 2 số a và b là 15 và 5 hoặc 5 và 15
Bài sau làm tương tự em nhé :)
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ko chép lại đầu bài nha
=(53x+800):7020=(49x+147):6000
=(53x+800).6000=70
x=145,0369515
lưu ý:mik làm theo cách suy luận nhé @@@
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\(240-\left(x+5\right)=137\)
\(x+5=240-137\)
\(x+5=103\)
\(x=103-5=98\)
b) \(\left(2x-10\right)-20=110\)
\(2x-10=110+20\)
\(2x-10=130\)
\(2x=130+10\)
\(2x=140\)
\(x=140:2=70\)
\(a,240-\left(x+5\right)=137\\ x+5=103\\ x=98\\ b,\left(2x-10\right)-20=110\\ 2x-10=130\\ 2x=140\\ x=70\)
tìm x giúp m nha đúng mình tíck
\(\frac{x+10}{270}+\frac{x+20}{260}=\frac{x+3}{250}+\frac{x+3}{240}\)
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\(\Leftrightarrow\frac{x+10}{270}+1+\frac{x+20}{260}+1=\frac{x+30}{250}+1+\frac{x+40}{240}+1\)
\(\Leftrightarrow\frac{x+280}{270}+\frac{x+280}{260}=\frac{x+280}{250}+\frac{x+280}{240}\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}\right)=\left(x+280\right)\left(\frac{1}{250}+\frac{1}{240}\right)\)
\(\Leftrightarrow\left(x+280\right)\left(\frac{1}{270}+\frac{1}{260}-\frac{1}{250}-\frac{1}{240}\right)=0\)
=>x=-280
\(\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)\cdot...\cdot\left(x+20\right)=240\)
\(\left(x+x+x+...+x\right)\cdot\left(1+2+3+...+20\right)=240\)
\(20x\cdot210=240\)
\(20x=\frac{240}{210}=\frac{8}{7}\)
\(\Rightarrow x=\frac{8}{7}\div20=\frac{2}{35}\)
(x+1)(x+2)(x+3)....(x+20)=240
=> x(1+2+3+...+20) = 240
=> x.210=240
x = 240:210
x=8/7