1.(x+1/3)^2=16
2.x+3/10=7/5
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MH
3 tháng 3 2023
\(\dfrac{45}{135}=\dfrac{45:45}{135:45}=\dfrac{1}{3}\)
\(\dfrac{117}{234}=\dfrac{117:117}{234:117}=\dfrac{1}{2}\)
\(\dfrac{1515}{2727}=\dfrac{1515:303}{2727:303}=\dfrac{5}{9}\)
\(\dfrac{232323}{494949}=\dfrac{232323:10101}{494949:10101}=\dfrac{23}{49}\)
\(\left(x-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{1}{2}\)
\(x-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{1}{4}\)
\(x-\dfrac{1}{3}=\dfrac{2}{4}-\dfrac{1}{4}\)
\(x-\dfrac{1}{3}=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}+\dfrac{1}{3}\)
\(x=\dfrac{3}{12}+\dfrac{4}{12}\)
\(x=\dfrac{7}{12}\)
\(x-\left(\dfrac{1}{7}+\dfrac{2}{5}\right)=\dfrac{16}{35}\)
\(x-\left(\dfrac{5}{35}+\dfrac{14}{35}\right)=\dfrac{16}{35}\)
\(x-\dfrac{19}{35}=\dfrac{16}{35}\)
\(x=\dfrac{16}{35}+\dfrac{19}{35}\)
\(x=\dfrac{35}{35}\)
\(x=1\)
\(x+\dfrac{3}{7}=\dfrac{2}{5}+\dfrac{3}{10}\)
\(x+\dfrac{3}{7}=\dfrac{4}{10}+\dfrac{3}{10}\)
\(x+\dfrac{3}{7}=\dfrac{7}{10}\)
\(x=\dfrac{7}{10}-\dfrac{3}{7}\)
\(x=\dfrac{49}{70}-\dfrac{30}{70}\)
\(x=\dfrac{19}{30}\)
3 tháng 3 2022
a: =>x/16=3/8+7/2=3/8+28/8=31/8=62/16
=>x=62
b: =>3/5:x=17/10-2/5=17/10-4/10=13/10
=>x=3/5:13/10=3/5x10/13=30/65=6/13
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 12 2023
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 12 2023
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
9 tháng 1 2020
1) <=> 8x-16-21-7x=-4
<=> x-37=-4
<=> x=33
2) <=> 12x-48+6x-12-16x-48=-28
<=> 2x-108=-28
<=> 2x = 80
<=> x = 40
3)<=> 4x-20-35+7x+50-10x=-3
<=> x-5=-3
<=> x=2
4) <=> -3x-15=-45-21
<=> 51=3x
<=> 17=x
5) 4x-28+15=-2x+14-10
<=> 6x-13=4
<=> 6x=17
<=> x = \(\frac{17}{6}\)
Chúc bạn học tốt
12 tháng 5 2022
Bài 1:
1: =-5/24+16/27+3/4
=-5/24+18/24+16/27
=13/24+16/27
=117/216+128/216=245/216
2: =-1/3+1/3+6/7=6/7
3: \(=\dfrac{1}{2}-\dfrac{7}{12}+\dfrac{1}{2}=1-\dfrac{7}{12}=\dfrac{5}{12}\)
4: \(=-\dfrac{5}{8}+\dfrac{14}{25}-\dfrac{6}{10}=\dfrac{-125+112-120}{200}=\dfrac{-133}{200}\)
NM
21 tháng 8 2017
a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0
KL
22 tháng 8 2017
a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran
30 tháng 3 2020
1/ 10(X-7)-8(X+5)=6(-5)+24
10x - 70 - 8x - 40 = -30 +24
2x - 110 = -6
2x = 104
x=52
2/ 8(X-|-7|)-6(X-2)=|-8|.6-50
8(x - 7) - 6(x-2) = 8.6 - 50
8x - 56 - 6x +12 =48 -50
2x - 44 = -2
2x = 42
x=21
3/ 2(4X-8)-7(3+X)=|-4|(3-2)
8x-16 - 21 - 7x = 4.1
x-37=4
x=41
4/ 12(X-4)=6(x-2)-16(X+3)=7|-4|
12x - 48 = 6x - 12 - 16x -48 =7.4
12x - 48 = 28
12x=76
x=19/3
5/ 4(X-5)-7(5-X)+10(5-X)=-3
4x - 20 -35 +7x + 50 -10x = -3
x - 5 = -3
x = -2
Chúc bạn học tốt!
LG
19 tháng 5 2016
X = ( 1+10 ) * 10 / 2 = 11 * 10 / 2 = 110 / 2 = 55
X= (2+20) * 10 /2 = 110
X= (1+19)*10/2= 100
19 tháng 5 2016
1+2+3+4+5+6+7+8+9+10= (10+1)X10 : 2 = 55
2+4+6+8+10+12+14+16+18+20 = (20+2)x10 : 2 = 110
1+3+5+7+9+11+13+15+17+19 = (19+1) x 10 : 2 = 100
1.
\(\left(x+\dfrac{1}{3}\right)^2=16\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=4\\x+\dfrac{1}{3}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=\dfrac{-13}{3}\end{matrix}\right.\)
2.
\(x+\dfrac{3}{10}=\dfrac{7}{5}\\ x=\dfrac{7}{5}-\dfrac{3}{10}\\ x=\dfrac{11}{10}\)
\(\left(x+\dfrac{1}{3}\right)^2=16\)
\(\left[{}\begin{matrix}\left(x+\dfrac{1}{3}\right)^2=4^2\\\left(x+\dfrac{1}{3}\right)^2=\left(-4\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+\dfrac{1}{3}=4\\x+\dfrac{1}{3}=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4-\dfrac{1}{3}\\x=-4-\dfrac{1}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{11}{3}\\x=\dfrac{-13}{3}\end{matrix}\right.\)