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6 tháng 3 2020

a) |2x + 1| - 19 = -7

=> \(\left|2x+1\right|=-7+19=12\)

=> \(\left[{}\begin{matrix}2x+1=12\\2x+1=-12\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=12-1=11\\2x=-12-1=-13\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{11}{2}\\x=-\frac{13}{2}\end{matrix}\right.\)

Vậy:............

b) -28 – 7. |- 3x + 15| = -70

=> \(\text{7. |- 3x + 15| = -28 - (-70) = -28 + 70 = 42}\)

=> \(\left|-3x+15\right|=42:7=6\)

=> \(\left[{}\begin{matrix}-3x+15=6\\-3x+15=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}-3x=6-15=-9\\-3x=-6-15=-6+\left(-15\right)=-21\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-9:\left(-3\right)=3\\x=x=-21:\left(-3\right)=7\end{matrix}\right.\)

Vậy:.....................

c) |18 – 2. |-x + 5|| = 12

=> \(\left[{}\begin{matrix}18-2.\left|-x+5\right|=12\\18-2.\left|-x+5\right|=-12\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2.\left|-x+5\right|=18-12=6\\2.\left|-x+5\right|=18-\left(-12\right)=18+12=30\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left|-x+5\right|=6:2=3\\\left|-x+5\right|=30:2=15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x+5=3\\-x+5=-3\end{matrix}\right.\\\left[{}\begin{matrix}-x+5=15\\-x+5=-15\end{matrix}\right.\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}-x=3-5=-2\\-x=-3-5=-8\end{matrix}\right.\\\left[{}\begin{matrix}-x=15-5=10\\-x=-15-5=-20\end{matrix}\right.\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\\\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\end{matrix}\right.\)

7 tháng 3 2020

thank bn ✰❤❤✔

bạn ghi lại đề đi bạn

25 tháng 1 2017

a)\(n+7⋮n+2\)

\(\Rightarrow\left(n+2\right)+5⋮n+2\)

\(\Rightarrow5⋮n+2\)

\(\Rightarrow n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta có bảng sau:

n+2 1 -1 5 -5
n -1 -3 3 -7

Vậy \(n\in\left\{-1;-3;3;-7\right\}\)

b)\(9-n⋮n-3\)

\(\Rightarrow6-\left(n-3\right)\)

\(\Rightarrow6⋮n-3\)

\(\Rightarrow n-3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

nếu n-3=1 thì n=4

nếu n-3=-1 thì n=2

nếu n-3=2 thì n=5

nếu n-3=-2 thì n=1

nếu n-3=3 thì n=6

nếu n-3=-3 thì n=0

nếu n-3=6 thì n=9

nếu n-3=-6 thì n=-3

Vậy \(n\in\left\{4;2;5;1;6;0;9;-3\right\}\)

25 tháng 1 2017

c)\(n^2+n+17⋮n+1\)

\(\Rightarrow n\left(n+1\right)+17⋮n+1\)

\(\Rightarrow17⋮n+1\)

\(\Rightarrow n+1\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)

nếu n+1=1 thì n=0

nếu n+1=-1 thì n=-2

nếu n+1=17 thì n=16

nếu n+1=-17 thì n=-18

Vậy \(n\in\left\{0;-2;16;-18\right\}\)

b: \(\Leftrightarrow n-2+5⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{3;1;7;-3\right\}\)

c: \(\Leftrightarrow n-3+4⋮n-3\)

\(\Leftrightarrow n-3\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{4;2;5;1;7;-1\right\}\)

d: \(\Leftrightarrow n-5+4⋮n-5\)

\(\Leftrightarrow n-5\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{6;4;7;3;9;1\right\}\)

e: \(\Leftrightarrow3n-3+4⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)

30 tháng 3 2018

các bạn ơi giúp mìh với mìh đag cần gấp ai nhanh và đúng thì mih tick cho

20 tháng 3 2017

a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)

=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)

=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)

=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)

b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)

=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)

=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)

c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)

=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)

= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)

d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)

=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)

=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)

e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)

=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)

21 tháng 7 2018

a, \(\dfrac{-7}{9}.2\dfrac{3}{4}\)

= \(\dfrac{-7}{9}.\dfrac{11}{4}\)

= \(\dfrac{-77}{36}\)

b, \(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-2}{5}\)

= \(\dfrac{2}{3}+\dfrac{-2}{15}\)

= \(\dfrac{10}{15}+\dfrac{-2}{15}\)

= \(\dfrac{-8}{15}\)

c , \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\)

= \(\dfrac{2}{3}-4.\dfrac{5}{4}\)

= \(\dfrac{2}{3}-5\)

= \(\dfrac{-13}{3}\)

d, \(\left(\dfrac{1}{-3}+\dfrac{5}{6}\right).11-7\)

= \(\dfrac{1}{2}\) . 11 - 7

= \(\dfrac{11}{2}-\dfrac{14}{2}\)

= \(\dfrac{-3}{2}\)

e, \(\dfrac{3}{4}.15\dfrac{1}{3}-\dfrac{3}{4}.43\dfrac{1}{3}\)

= \(\dfrac{3}{4}.\left(15\dfrac{1}{3}-43\dfrac{1}{3}\right)\)

= \(\dfrac{3}{4}.-28\)

= \(-21\)

28 tháng 11 2016

\(8-12x+6x^2-x^3\)

\(=\left(2-x\right)^3\)

\(125x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

\(x^2-xz-9y^2+3yz\)

\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y-z\right)\)

\(x^3-x^2-5x+125\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

\(12x^3+4x^2-27x-9\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x^2-9\right)\)

\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)

\(4x^4+4x^3-x^2-x\)

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

\(=x\left(x+1\right)\left(4x^2-1\right)\)

\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)

26 tháng 3

Là 00000000000000

2 tháng 5 2017

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)