Tìm x,biết x:8=121(dư 5)
A.973 B.968 C.40 D.161
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NB
24 tháng 2 2020
\(\frac{x-971}{972}+\frac{x-973}{970}+\frac{x-975}{968}+\frac{x-977}{966}=\frac{x-972}{971}+\frac{x-970}{973}+\frac{x-968}{975}+\frac{x-966}{977}\)
\(\Rightarrow\frac{x-971}{972}-1+\frac{x-973}{970}-1+\frac{x-975}{968}-1+\frac{x-977}{966}-1=\frac{x-972}{971}-1+\frac{x-970}{973}-1+\frac{x-968}{975}-1+\frac{x-966}{977}-1\)\(\Rightarrow\frac{x-1943}{972}+\frac{x-1943}{970}+\frac{x-1943}{968}+\frac{x-1943}{966}=\frac{x-1943}{971}+\frac{x-1943}{973}+\frac{x-1943}{975}+\frac{x-966}{977}\)
\(\Leftrightarrow\frac{x-1943}{972}+\frac{x-1943}{970}+\frac{x-1943}{968}+\frac{x-1943}{966}-\frac{x-1943}{971}-\frac{x-1943}{973}-\frac{x-1943}{975}-\frac{x-966}{977}=0\)
\(\Leftrightarrow\left(x-1943\right).\left(\frac{1}{972}+\frac{1}{970}+\frac{1}{968}+\frac{1}{966}-\frac{1}{971}-\frac{1}{973}+\frac{1}{975}+\frac{1}{977}\right)=0\)
\(\Leftrightarrow\left(x-1943\right)=0\Leftrightarrow x=1943\)
Vậy...
TN
29 tháng 11 2018
a, x : 9 = 121 ( dư 8 )
=> x = 121 x 9 + 8
=> x = 1097
b, 925 ; x = 3 ( dư 1 )
=> x = ( 925 - 1 ) : 3
=> x = 308
c, x : 254 = 9 x ( 3 + 5 + 7 + 9 )
=> x : 254 = 216
=> x = 216 x 254
=> x = 54864
TN
1
24 tháng 8 2023
\(a.\dfrac{12+x}{42}=\dfrac{35}{42}\Leftrightarrow12+x=35\Leftrightarrow x=23\)
\(b.\dfrac{25-x}{40}=\dfrac{3}{8}\Leftrightarrow\dfrac{25-x}{40}=\dfrac{15}{40}\Leftrightarrow25-x=15\Leftrightarrow x=10\)
\(c.\dfrac{13+x}{20}=\dfrac{3}{4}\Leftrightarrow\dfrac{13+x}{20}=\dfrac{15}{20}\Leftrightarrow13+x=15\Leftrightarrow x=2\)
\(d.\dfrac{23-x}{25}=\dfrac{20}{25}\Leftrightarrow23-x=20\Leftrightarrow x=3\)
4 tháng 11 2021
a) x - 120: 30 = 40
x -40 =40
x =40+40
x =80
b) (x + 120) : 20 = 8
(x+ 120) = 8x20
x+120 =160
x = 160-120
x = 40
c) (x + 5). 3 = 300
x+5=300:3
x+5=100
x=100-5
x=95
d) x.2 + 21 : 3= 27
x.2 +7=27
x.2 = 27-7
x.2= 20
x=20:2
x=10
AH
Akai Haruma
Giáo viên
13 tháng 12 2022
Lời giải:
a. $121-3(x-5)=6$
$3(x-5)=121-6=115$
$x-5=115:3=\frac{115}{3}$
$x=\frac{115}{3}+5=\frac{130}{3}$
b.
$2x-138=2^3.3^2=72$
$2x=72+138=210$
$x=210:2=105$
c.
$x-3\vdots 7$
$\Rightarrow x-3\in\left\{0;7;14;21;28;35;42;49; 56;...\right\}$
Mà $10< x< 50$ nên $x\in\left\{14;21;28;35;42;49\right\}$
d.
$27\vdots x+1$
$\Rightarrow x+1\in\left\{\pm 1; \pm 3; \pm 9; \pm 27\right\}$
$\Rightarrow x\in\left\{0; -2; -4; 2; 8; -10; 26; -28\right\}$
13 tháng 12 2022
a ) 121-3.(x - 5 ) = 6
3.(x-5) = 121 -6
3. (x-5)=115
x-5 = 115:3
x-5=35
x=35+5
x = 40
b) 2x - 138 = 2'3. 3'2
2x -138=8.9
2x-138=72
2x=72+138
2x=210
x=210:2
x=105
c) theo bài ra : x-3 ∈ B(7)
ta có B(7)=(0,7,14,21,28,35,49,56,...)
=) x-3 ∈ ( 0,7,14,21,28,35,49,56,...)
=) x ∈( 3 , 10,17,24,31,38,42,58,..)
mà 10 <x<50 nên x ∈ ( 17 , 24 ,31,38,42 )
vậy x ∈(17,24,31,38,42)
20 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`(x-1954) \times 5=50`
`x-1954 = 50 \div 5`
`x-1954 = 10`
`x=10 + 1954`
`x= 1964`
`B.`
`48 - 40 \times x=8`
`40 \times x=48-8`
`40 \times x = 40`
` x=40 \div 40`
`x=1`
`C.`
`2+70 \div x = 3`
`70 \div x = 3-2`
`70 \div x = 1`
`x=70 \div 1`
`x=70`
`D. [3 \times (x+2) \div 7] \times 4=120`
`3 \times (x+2) \div 7 = 120 \div 4`
`3 \times (x+2) \div 7 =30`
`3 \times (x+2) = 30 \times 7`
`3 \times (x+2)=210`
`x+2=210 \div 3`
`x+2=70`
`x=70-2`
`x=68`
`E.`
`7,2 \div [ (0,6 \times x+8) \div 20 + 59] = 0,12?` Thiếu ngoặc bạn nhé._.
BD
20 tháng 6 2023
\(a,\left(x-1954\right)\times5=50\\ x-1954=50:5\\ x-1954=10\\ x=1954+10\\ x=1964\\b.48-40\times x=8\\40\times x=48-8\\40\times x=40\\ x=40:40\\ x=1 \\ c.2+70:X=3\\ 70:x=3-2\\ 70:x=1\\ x=70:1\\ x=70\\ d,\left[3\times\left(x+2\right):7\right]\times4=120\\ 3\times\left(x+2\right):7-120:4\\ 3\times\left(x+2\right):7=30\\ \left(x+2\right):7=30:3\\ \left(x+2\right):7=10\\ x+2=10\times7\\ x+2=70\\ x=70-2\\ x=68\)
x - 5 = 8 x 121
x - 5 = 968
x= 968 + 5
x= 973
chính là A