\(\dfrac{x+2}{-4}=\dfrac{-9}{x+2}\)(ĐKXĐ: x<>-2)

=>\(\left(x+2\right)^2=\left(-9\right)\cdot\left(-4\right)=36\)

=>\(\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

sos

1. Recycling is a good activity to keep our planet clean and safe to live.

2. First, we can recycple plastic bags or bottles because they are polluted for the environment.

3. If we do that, we can protect the environment.

4. Second, we can also recycle waste paper, magazines to produce new books or (and) notebooks for students.

5. Third, in the future, we can (also) recycle glass and glass bottle to make new products such as flower vases.

6. If we recycle these things, we can make the world a better place to live.

#hoctot!

tick for me pls ^^

1. Recycling/ good activitity/ keep our planet/ clean/ safe/ to live

Recycling is a good activity that keeps our planet clean and safe to live

2. First,/we can/ recycle/ plastic bags/ bottles/ because/ they/ pollute/ the environment.

First, we can recycle plastic bags and bottles because they will pollute the environment

3. If/ we/ do that/, we / protect/ environment.

If we do that, we will protect the environment

4. Second/ we/ also/ recycle/ waste paper, magazines/ to produce/ new book/ notebooks/ for students.

Second, we also recycle waste paper, magazines to produce new books and notebooks for students

5. Third, in the future/ we/ recycle glass/ glass bottle/ to make/ new products/ such/ flower vases.

Third, in the future we will recycle glass and glass bottle to make new products such as flower vases

6. If/ we/ recycle/ these things/ we/ make /the world/ better place/ live.

If we recycle these things, we will make the world a better place to live

\(\dfrac{1}{4}-\left(\dfrac{3}{4}+x\right)=2\)

=>\(x+\dfrac{3}{4}=\dfrac{1}{4}-2=-\dfrac{7}{4}\)

=>\(x=-\dfrac{7}{4}-\dfrac{3}{4}=-\dfrac{10}{4}=-\dfrac{5}{2}\)

sos giúp mình với

a: B nằm giữa A và C

=>BA+BC=AC

=>BC+4=9

=>BC=5(cm)

b: I là trung điểm của AB

=>\(AI=BI=\dfrac{AB}{2}=\dfrac{4}{2}=2\left(cm\right)\)

Vì I nằm giữa A và B

mà B nằm giữa A và C

nên I nằm giữa A và C

=>AI+IC=AC

=>IC+2=9

=>IC=7(cm)

a: \(\dfrac{7}{12}x+\dfrac{3}{8}=\dfrac{1}{6}\)

=>\(\dfrac{7}{12}x=\dfrac{1}{6}-\dfrac{3}{8}=\dfrac{4}{24}-\dfrac{9}{24}=-\dfrac{5}{24}\)

=>\(x=-\dfrac{5}{24}:\dfrac{7}{12}=-\dfrac{5}{24}\cdot\dfrac{12}{7}=\dfrac{-5}{14}\)

b: \(\dfrac{x}{15}=\dfrac{3}{5}+\dfrac{-2}{3}\)

=>\(\dfrac{x}{15}=\dfrac{9}{15}-\dfrac{10}{15}=-\dfrac{1}{15}\)

=>x=-1

c: \(\dfrac{x^2-4}{15}=\dfrac{1}{3}\)

=>\(x^2-4=\dfrac{15}{3}=5\)

=>\(x^2=9\)

=>\(x\in\left\{3;-3\right\}\)

d: \(\dfrac{x}{3}-\dfrac{1}{4}=-\dfrac{5}{6}\)

=>\(\dfrac{x}{3}=\dfrac{1}{4}+\dfrac{-5}{6}=\dfrac{3}{12}-\dfrac{10}{12}=-\dfrac{7}{12}\)

=>\(x=-\dfrac{7}{12}\cdot3=-\dfrac{7}{4}\)

e: \(3\dfrac{3}{4}-2\dfrac{5}{6}=\dfrac{4}{x}\)

=>\(\dfrac{4}{x}=\dfrac{15}{4}-\dfrac{17}{6}=\dfrac{45}{12}-\dfrac{34}{12}=\dfrac{11}{12}\)

=>\(x=4\cdot\dfrac{12}{11}=\dfrac{48}{11}\)

f: ĐKXĐ: x<>0

\(\dfrac{x}{4}=\dfrac{25}{x}\)

=>\(x^2=25\cdot4=100\)

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

g: ĐKXĐ: x<>0

\(\dfrac{x-1}{14}=\dfrac{4}{x}\)

=>\(x\left(x-1\right)=14\cdot4=56\)

=>\(x^2-x-56=0\)

=>(x-8)(x+7)=0

=>\(\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

h: ĐKXĐ: x<>0

\(\dfrac{-x}{9}=\dfrac{-4}{x}\)

=>\(x^2=4\cdot9=36\)

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

a

3,2 . x - (4/5 + 2/3) : 11/3 = 7/20

3,2 . x -22/15 : 11/3         = 7/20

3,2 . x -2/5                      = 7/20

3,2 . x                             = 7/20 + 2/5

3,2 . x                             = 3/4

        x                             = 3/4 : 3,2

        x                             = 15/64

Vậy x = 15/64

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\(\left(2x+\dfrac{2}{3}\right)^2=\dfrac{16}{25}\)

\(\Rightarrow\left(2x+\dfrac{2}{3}\right)^2=\left(\pm\dfrac{4}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{2}{3}=\dfrac{4}{5}\\2x+\dfrac{2}{3}=-\dfrac{4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{2}{15}\\2x=-\dfrac{22}{15}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{15}\\x=-\dfrac{11}{15}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{15};-\dfrac{11}{15}\right\}\).

a: \(A=\dfrac{7}{1\cdot3}+\dfrac{7}{3\cdot5}+...+\dfrac{7}{49\cdot51}\)

\(=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{49\cdot51}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{51}\right)=\dfrac{7}{2}\cdot\dfrac{50}{51}=\dfrac{175}{51}\)

b: \(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)

\(=\dfrac{5}{28}+\dfrac{5}{70}+...+\dfrac{5}{700}\)

\(=\dfrac{5}{4\cdot7}+\dfrac{5}{7\cdot10}+...+\dfrac{5}{25\cdot28}\)

\(=\dfrac{5}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}\cdot\dfrac{6}{28}=\dfrac{5}{3}\cdot\dfrac{3}{14}=\dfrac{5}{14}\)