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

\(ĐK:x\le\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)

it's urgent, but i don't know how to save :)))

Thứ 7 tuần sau bạn ạ

Bài 1:

a: \(\left(5-\frac23+\frac37\right):\left(24\frac{4}{21}-25\frac{8}{21}\right)\)

\(=\left(\frac{105}{21}-\frac{14}{21}+\frac{9}{21}\right):\left(24+\frac{4}{21}-25-\frac{8}{21}\right)\)

\(=\frac{100}{21}:\left(-1-\frac{4}{21}\right)=\frac{100}{21}:\frac{-25}{21}=\frac{100}{-25}=-4\)

b: \(\left(2\frac46-\frac{7}{15}\right):\left(\frac23-\frac25\right)^2\)

\(=\left(2+\frac23-\frac{7}{15}\right):\left(\frac{10}{15}-\frac{6}{15}\right)^2\)

\(=\left(2+\frac{10}{15}-\frac{7}{15}\right):\left(\frac{4}{15}\right)^2=\left(2+\frac{3}{15}\right):\frac{16}{225}=\frac{11}{5}\cdot\frac{225}{16}\)

\(=\frac{11}{16}\cdot45=\frac{495}{16}\)

c: \(\left(\frac{13}{18}-\frac{1}{72}\right):\frac{1}{18}-\left(\frac49-\frac{25}{16}\right)\cdot\frac92\)

\(=\left(\frac{52}{72}-\frac{1}{72}\right)\cdot18-\left(\frac{64}{144}-\frac{100}{144}\right)\cdot\frac92\)

\(=\frac{51}{72}\cdot18+\frac{36}{144}\cdot\frac92=\frac{51}{4}+\frac14\cdot\frac92=\frac{111}{8}\)

d: \(\frac35:\left(-\frac{1}{15}-\frac16\right)+\frac35:\left(-\frac13-1\frac{1}{15}\right)\)

\(=\frac35:\left(-\frac{2}{30}-\frac{5}{30}\right)+\frac35:\left(-\frac{5}{15}-\frac{16}{15}\right)\)

\(=\frac35:\frac{-7}{30}+\frac35:\frac{-21}{15}=\frac35\cdot\frac{-30}{7}+\frac35\cdot\frac{-5}{7}=\frac35\left(-\frac{30}{7}-\frac57\right)\)

\(=\frac35\cdot\left(-5\right)=-3\)

e: \(4\left(-\frac12\right)^3-2\left(-\frac12\right)^2+3\cdot\left(-\frac12\right)+\left(-1\right)^{2002}\)

\(=4\cdot\frac{-1}{8}-2\cdot\frac14-\frac32+1\)

\(=-\frac12-\frac12-\frac32+1=-\frac32\)

f: \(\frac{2^4\cdot2^6}{\left(2^5\right)^2}-\frac{2^5\cdot15^3}{6^3\cdot10^2}\)

\(=\frac{2^{10}}{2^{10}}-\frac{2^5\cdot5^3\cdot3^3}{2^3\cdot5^3\cdot2^2\cdot5^2}=1-\frac{3^3}{5^2}=1-\frac{27}{25}=-\frac{2}{25}\)

Bài 3:

a: |x-5|=8

=>\(\left[\begin{array}{l}x-5=8\\ x-5=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=13\\ x=-5\end{array}\right.\)

b: \(\left|1-\frac23x\right|+\frac34-5\frac12=0\)

=>\(\left|\frac23x-1\right|+\frac34-\frac{11}{2}=0\)

=>\(\left|\frac23x-1\right|+\frac34-\frac{22}{4}=0\)

=>\(\left|\frac23x-1\right|=\frac{19}{4}\)

=>\(\left[\begin{array}{l}\frac23x-1=\frac{19}{4}\\ \frac23x-1=-\frac{19}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac23x=\frac{19}{4}+1=\frac{23}{4}\\ \frac23x=-\frac{19}{4}+1=-\frac{15}{4}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{23}{4}:\frac23=\frac{23}{4}\cdot\frac32=\frac{69}{8}\\ x=-\frac{15}{4}:\frac23=-\frac{15}{4}\cdot\frac32=\frac{-45}{8}\end{array}\right.\)

c: |9-7x|+7=26

=>|7x-9|=26-7=19

=>\(\left[\begin{array}{l}7x-9=19\\ 7x-9=-19\end{array}\right.\Rightarrow\left[\begin{array}{l}7x=19+9=28\\ 7x=-19+9=-10\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac{10}{7}\end{array}\right.\)

Ok bn

ok  

20 strong enough to carry that heavy box.

21 isn't hot enough to boil the kettle.

22 wasn't warm enough for us to go swimming.

23 swims skillfully.

24 a very good English learner.

25 swims well.

26 a very fast runner.

27 more interesting than plays.

28 more difficult than English.

29 as intelligent as Jill is

30 to have a friend like you

31 to smoke 20 cigarettes a day, but now he doesn't smoke any more.

32 enough to lift that heavy table.

hai tai dựng đứng

hàm răng trắng muốt

mấy câu kia anh chịu =))

cái đu...i dài ve vẩy