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Tìm x, biết:
a.
75883-(31200+x)=999
31200+x =75883-999
31200+x= 74884
x = 74884-31200
x = 43684
b.
24 : ( x +1 ) = 8
x+1=24:8
x+1=3
x=3-1
x=2
c.
1+2+3+...+x=55
Dãy trên có số số hạng là:
(x−1):1+1=x−1+1=x
Ta có:
(x+1).x:2=55
(x+1).x=55.2
(x+1).x=110
(x+1).x=11.10
⇒x=10
a, ---> 31200 + x = 75883 - 999
---> 31200 + x = 74884
---> x = 74884 - 31200 = 43684
b, ---> x + 1 = 3
---> x = 2
c, Số số hạng là : x
Tổng = ( x + 1 ) * x : 2 = 55
---> x ( x + 1 ) = 110
---> 10 ( 10 + 1 ) = 110
---> x = 10
xin tiick
Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$
$x-x\times \frac{1}{2}=\frac{3}{2}$
$x\times (1-\frac{1}{2})=\frac{3}{2}$
$x\times \frac{1}{2}=\frac{3}{2}$
$x=\frac{3}{2}: \frac{1}{2}=3$
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
675,4 - X x 2,5 = 0,4
X x 2,5 = 675,4 - 0,4
X x 2,5 = 675
X = 675 : 2,5
X = 27
chúc bạn HT
x = 6,51 ; 6,52 ; 6,53 ; 6,54 ; 6,55 ; 6,56 ; 6,57 ; 6,58 ; 6;59
`1,8 xx 2 xx 562 + 0,36 xx 4830`
`=3,6 xx 562 + 1738,8`
`=2023,2 + 1738,8`
`=3762`
Sao lại = ... hả bạn? Nếu $35\times x+45$ không có giá trị nào thì cũng không tìm được $x$ nhé.
\(a,\left(x-15\right)\times7-270=169\times45\\ \left(x-15\right)\times7-270=7605\\ \left(x-15\right)\times7=7605+270\\ \left(x-15\right)\times7=7875\\ x-15=7875:7\\ x-15=1125\\ x=1125+15\\ x=1140\\ b,64-\left(34\times x-8\right)=4\\ 34\times x-8=64-4\\ 34\times x-8=60\\ 34\times x=60+8\\ 34\times x=68\\ x=68:34\\ x=2\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`13/50 + 9% + 41/100 + 0,24`
`= 0,26 + 0,09 + 0,41 + 0,24`
`= (0,26 + 0,24) + (0,09 + 0,41)`
`= 0,5 + 0,5`
`= 1`
`b)`
`2018 \times 2020 - 1/2017 + 2018 \times 2019`
`= 2018 \times (2020 + 2019) - 1/2017`
`= 2018 \times 4039 - 1/2017`
`= 8150702`
`c)`
`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)
`=`\(1-\dfrac{1}{7}\)
`= 6/7`
\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)
câu a bạn ghi thế thì mình chịu nha
b, \(24:\left(x+1\right)=8\)
\(x+1=24:8\)
\(x+1=3\)
\(x=3-1=2\)
c, \(1+2+3+...+x=55\)
SSH: \(\left(x-1\right):1+1=x\)
tổng: \(\left(x+1\right).x:2=55\)
\(\left(x+1\right).x=55.2\)
\(\left(x+1\right).x=110\)
\(\Rightarrow\left(x+1\right).x=11.10\)
vậy x=10