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1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
BÀI 1: Tính
1) -(-10)-(+14)=10-14=-4
2) -(+15)-12=-15-12=-27
3) (-11)-(-13)=-11+13=2
4) -(-13)-(-10)=13+10=23
5) -4-(+7)=-4-7=-11
Bài 2 : tìm x:
a)x+(-5)=-(-7)
x=7+(-5)
x=2
vậy x=2
b)x-8=-(+10)
x-8=10
x=10+8
x=18
vậy x=18
c)x-(-12)=14
x+12=14
x=14-12
x=2
vậy x=2
d)x-(+3)=-17
x-3=17
x=17+3
x=20
vậy x=20
e)x+20=-(-23)
x+20=23
x=23-20
x=3
vậy x=3
BÀI 3:tính
a)17-(-9)-(+25)
=17+9-25
=26-25
=1
b)-[-13]-15+(-20)
=13-15-20
=-2-20
=-22
c)-17-(-16)-23
=-17+16-23
=-1-23
=-24
d)-(-25)-(-14)+(-17)
=25+14-17
=39-17
=22
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1. -x+20 = -(-15)-8+13
=> -x=15-8+13-20
=> -x=0
=> x=0
2. -(-10)+x=-13+(-9)+(-6)
=> 10+x=-13-9-6
=> x = -13-9-6-10
=> x = -38
3. 8-(-12)+10=-(-14)-x
=> 8+12+10=14-x
=> x = 14-8-12-10
=> x = -16
4. -(+12)+(-x)-(-3)=5-(-7)
=> -12-x+3=5+7
=> -x=5+7+12-3
=> -x=21
=> x=-21
5. 14-x+(-10)=-(-9)+(+15)
=> 14-x-10=9+15
=> -x=9+15-14+10
=> -x=20
=> x=-20
6. 12-(-17)+(-3)=-5+x
=> 12+17-3+5=x
=> x=31
7. x-(-19)-(+32)=14-(+16)
=> x+19-32=14-16
=> x=14-16+32-19
=> x=11
8. x-|-15|-|7|=-(-9)+|-5|
=> x-15-7=9+5
=> x=9+5+7+15
=> x=36
9. 15-x+17=13-(-21)
=> 15-x+17=13+21
=> -x=13+21-15-17
=> -x=2
=> x=-2
10. -|-5|-(-x)+4=3-(-25)
=> -5+x+4=3+25
=> x=3+25-4+5
=> x=29
1) -x - 3 = 7
-x = 7 + 3
-x = 10
x = -10
2) x + 5 = -10
x = -10 - 5
x = -15
3) 2x - 7 = 713
2x = 713 + 7
2x = 720
x = 720 : 2
x = 360
4) -129 - (35 - x) = 55
35 - x = -129 - 55
35 - x = -184
x = 35 - (-184)
x = 219
5) 103 - x = 16 - (13 - 8)
103 - x = 16 - 5
103 - x = 11
x = 103 - 11
x = 192
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
\(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}+\frac{x}{13\cdot16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\left[1-\frac{1}{16}\right]=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}\cdot\frac{15}{16}=\frac{5}{2}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}:\frac{15}{16}\)
\(\Rightarrow\frac{x}{3}=\frac{5}{2}\cdot\frac{16}{15}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{1}\cdot\frac{8}{3}\)
\(\Rightarrow\frac{x}{3}=\frac{8}{3}\Leftrightarrow x=8\)