Bài 1

a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)

                y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)

                          y= \(\frac{-3}{14}\)

b.5^x + 5^x+2=650

5^x . 1 + 5^x + 5²=650

5^x(1+25)=650

5^x.26=650

5^x=25

x=2

1:2:3:4:5:7:9:10

\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

mk hạng 8 huhu

mik hạng 9: Nhân Mã

mất thời gian vậy

Dài zậy trời

1) 40 + 15 + (-10) + (-15)                     2) -13 + (-750) + (-17) + 750                    3) (35 - 17) + (17 + 120 - 35)

= 40 + 15 - 10 - 15                               = -13 - 750 - 17 + 750                               = 35 - 17 + 17 + 120 - 35

= (40 - 10) + (15 - 15)                          = (-13 - 17) + (-750 + 750)                        = (35 - 35) + (-17 + 17) + 120

= 30                                                     = -30                                                          = 120

4) (55 + 45 + 15) - (15 - 55 + 45)               5) -(12 + 21 - 23) - (23 - 21 + 10)             6) (2020 - 79 + 15) - (-79 + 15)

= 55 + 45 + 15 - 15 + 55 - 45                     = -12 -21 + 23 - 23 + 21 - 10                    = 2020 - 79 + 15 + 79 - 15

= (45 - 45) + (15 - 15) + (55 + 55)              = (-12 - 10) + (-21 + 21) + (23 - 23)          = 2020 + (-79 + 79) + (15 - 15)

= 110                                                          = -22                                                         = 2020

7) -(515 - 80 + 91) - (2010 + 80 - 91)                     8) 25 - (-17) + 24 - 12               9) 235 - (34 + 135) - 100        

= -515 + 80 - 91 - 2010 - 80 + 91                           = 25 + 17 + 24 - 12                   = 235 - 34 - 135 - 100

= (-515 -2010) + (80 - 80) + (-91 + 91)                   = 54                                           = -34

= -2525       

10) (13 + 49) - (13 - 135 + 49)

= 13 + 49 - 13 + 135 - 49

= (13 - 13) + (49 - 49) +135

= 135                          

tự tính 

~HT~

Đề mk ko hiểu gì cả 

sao ko hiểu

1 ) 10 \(⋮\) n

=> n \(\in\) Ư ( 10 )

Ư ( 10 ) = { 1 , 2 , 5 , 10 }

Vậy n \(\in\) { 1 ; 2 ; 5 ; 10 }

2 ) 12 : \(⋮\) ( n - 1 )

=> n - 1 \(\in\) Ư ( 12 )

=> Ư ( 12 ) = { 1 ; 12 ; 2 ; 6 ; 3 ; 4 }

Vậy n \(\in\) { 2 , 13 , 3 , 7 , 4 , 5 }

3 ) 20 \(⋮\) ( 2n + 1 )

=> 2n + 1 \(\in\) Ư ( 20 )

=> Ư ( 20 ) = { 1 ; 20 ; 2 ; 10 ; 4 ; 5 }

Các trường hợp loại , vì n \(\in\) N

Vậy n thuộc { 0 , 2 }

a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)

b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)

\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)

c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)

\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)

e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)

f) \(\left|12-x\right|-7=5\)

th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)

th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)

i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)

k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)

a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)

b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)

Làm nốt nha

1+2+3+...+x = (1 + x) * x / 2 = 666

(1 + x) * x = 666 * 2 = 1332

1332 / x - 1 = x <=> x * (x + 1) = 1332

=> x = 36