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a) \(\left(2x-1\right)+\frac{3}{15}=\frac{3}{2}\)
\(\Rightarrow2x-1=\frac{3}{2}-\frac{3}{15}=\frac{13}{10}\)
\(\Rightarrow2x=\frac{13}{10}+1=\frac{23}{10}\)
\(\Rightarrow x=\frac{23}{20}\)
b) \(x+\frac{46}{15}=1,5\)
\(\Rightarrow x+\frac{46}{15}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}-\frac{46}{15}\)
\(\Rightarrow x=\frac{-47}{30}\)
c) \(\left(-2x+1\right)+\frac{3}{15}=\frac{5}{3}\)
\(\Rightarrow-2x+1=\frac{5}{3}-\frac{3}{15}=\frac{22}{15}\)
\(\Rightarrow-2x=\frac{7}{15}\Rightarrow x=\frac{-7}{30}\)
1/ Ta có : ( -16 ) + 23 + x = ( -16 )
<=> 23 + x = ( -16 ) + 16
<=> 23 + x = 0
<=> x = 0 -23
<=> x = -23
Vậy x = -23
2/ Ta có : 2.x - 35 = 15
<=> 2.x = 15 + 35
<=> 2.x = 50
<=> x = 50 : 2
<=> x = 25
Vậy x = 25
1/ -16 + 23 + x = -16
[(-16) +23] +x = -16
7 + x = -16
x= -16 - 7
x= -23
2/ 2x - 35 = 15
2x= 15 + 35
2x = 50
x= 50:2
x= 25
Chúc bạn làm bài tốt
`a) 2^x div 4=16`
`<=> 2^x=64`
`<=> 2^x=2^6`
`<=> x=6`
`b) |x+1|=-2`
do `|x+1|>=0 AA x in ZZ`
$\to x\in\varnothing$
`c) 2x+15=-27`
`<=> 2x=-42`
`<=> x=-21`
a)x =-1
b)x = 7 phần 30
c)x = 1
d)x = 5/18
nếu đúng thì hãy cho mình nha
Bài1:
a. Đây là dãy số cách đều 3 đơn vị.
Có số số hạng là: ( 299 - 2) : 3 + 1 = 100 (số hạng)
Tổng của dãy số là : (299+1) x 100 : 2 = 15000
b. Đây là dãy số cách đều 5 đơn vị.
Có số số hạng là : (51 - 1) : 5 +1 = 11 ( Số hạng)
Tổng: ( 51 + 1) x 11 : 2 = 286
Bài 2:
(2x-15)^5 = (2x-15)^3
(2x-15)^2 = 1
(2x-15)^2 = 1^2
=> 2x-15 = 1
2x = 16
x = 8
\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)
Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16
7+x=-16
x=-16-7
x=-23
2) 2x – 35 = 15
2x=15+35
2x=50
x=50:2
x=25
3) 3x + 17 = 12
3x=12-17
3x=-5
x=-5/3
4) (2x – 5) + 17 = 6
2x-5=6-17
2x-5=-11
2x=-11+5
2x=-6
x=-6:2
x=-3
5) 10 – 2(4 – 3x) = -4
2(4-3x)=10-(-4)
2(4-3x)=14
4-3x=14:2
4-3x=7
3x=4-7
3x=-3
x=-3:3
x=-1
6) - 12 + 3(-x + 7) = -18
3(-x+7)=-18-(-12)
3(x+7)=-6
x+7=-6:3
x+7=-2
x=-2-7
x=-9
a, ( 1+x )^3 = (2x)^3
b, ( x-1 )^2=16
c, (x+1)^2=25
d, 4x^3+15=47
e,(2x-1)^5=x^5
Mn giải nhanh giúp mk vs
a,\(\left(1+x\right)^3=\left(2x\right)^3\)
=>\(1+x=2x\)
=>\(x-2x=-1\)
=>\(-x=-1\)
=>\(x=1\)
vậy \(x=1\)
b,\(\left(x-1\right)^2=16\)
=>\(\left(x-1\right)^2=4^2\)
=>\(x-1=4\)
=>\(x=4+1\)
=>\(x=5\)
Vậy\(x=5\)
c,\(\left(x+1\right)^2=25\)
=>\(\left(x+1\right)^2=5^2\)
=>\(x+1=5\)
=>\(x=5-1\)
=>\(x=4\)
Vậy \(x=4\)
d,\(4x^3+15=47\)
=>\(4x^3=47-15\)
=>\(4x^3=32\)
=>\(x^3=32:4\)
=>\(x^3=8\)
=>\(x^3=2^3\)
=>\(x=2\)
Vậy\(x=2\)
e,\(\left(2x-1\right)^5=x^5\)
=>\(2x-1=x\)
=>\(2x-x=1\)
=>\(x=1\)
Vậy\(x=1\)
ĐÚNG K MÌNH NHA
Sửa đề: \(1-\left(2x-\dfrac{1}{2}\right)^2=\dfrac{15}{16}\)
=>\(\left(2x-\dfrac{1}{2}\right)^2=1-\dfrac{15}{16}=\dfrac{1}{16}\)
=>\(\left[{}\begin{matrix}2x-\dfrac{1}{2}=\dfrac{1}{4}\\2x-\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\\2x=-\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{3}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)