1,x/5=5/6+-19/30

x/5=25/30+-19/30

x/5=6/30

x/5=1/5 =>x=1

2, x=1/2+-2/3

x=3/6+-4/6

x=-1/6

3, x/30=3/5+7/30

x/30=18/30+7/30

x/30=25/30 => x=25

May Mắn

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

a, X=\(\frac{1}{4}\)

a, X=\(\frac{1}{4}\)                                                                                                                                                                 

b, X=1

 Mình không biết.

a) 

\(A=\left(\frac{19}{24}-\frac{7}{24}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)\)

\(A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}\)

\(A=\frac{1}{3}\)

\(B=\left(\frac{7}{12}-\frac{5}{12}\right)+\left(\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\right)\)

\(B=\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{1}{4}-\frac{3}{7}\)

\(B=\frac{5}{4}-\frac{3}{7}\)

\(B=\frac{23}{28}\)

b)

\(x=A-B\)

\(x=\frac{1}{3}-\frac{23}{28}\)

\(x=\frac{-41}{84}\)

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)

a: 0-|x+1|=5

=>|x+1|=0-5=-5<0(vô lý)

=>x∈∅

b: \(2-\left|\frac34-x\right|=\frac{7}{12}\)

=>\(\left|x-\frac34\right|=2-\frac{7}{12}=\frac{17}{12}\)

=>\(\left[\begin{array}{l}x-\frac34=\frac{17}{12}\\ x-\frac34=-\frac{17}{12}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{17}{12}+\frac{9}{12}=\frac{26}{12}=\frac{13}{6}\\ x=-\frac{17}{12}+\frac34=-\frac{17}{12}+\frac{9}{12}=-\frac{8}{12}=-\frac23\end{array}\right.\)

c: \(2\left|\frac12x-\frac13\right|-\frac32=\frac14\)

=>\(\left|2\left(\frac12x-\frac13\right)\right|=\frac32+\frac14=\frac74\)

=>\(\left|x-\frac23\right|=\frac74\)

=>\(\left[\begin{array}{l}x-\frac23=\frac74\\ x-\frac23=-\frac74\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac74+\frac23=\frac{21+8}{12}=\frac{29}{12}\\ x=-\frac74+\frac23=\frac{-21+8}{12}=-\frac{13}{12}\end{array}\right.\)

d: \(\left|x-\frac13\right|=\frac56\)

=>\(\left[\begin{array}{l}x-\frac13=\frac56\\ x-\frac13=-\frac56\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac56+\frac13=\frac76\\ x=-\frac56+\frac13=-\frac36=-\frac12\end{array}\right.\)

e: \(\frac34-2\left|2x-\frac23\right|=2\)

=>\(2\left|2x-\frac23\right|=\frac34-2=-\frac54<0\) (vô lý)

=>x∈∅

f: \(\frac{2x-1}{2}=\frac{5+3x}{3}\)

=>3(2x-1)=2(3x+5)

=>6x-3=6x+10

=>-3=10(vô lý)

=>x∈∅

a)5^x+2=625

<=>5^x+2=5⁴

<=>x+2=4

<=>x=2

b)

 \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}....\frac{30}{62}.\frac{31}{64}=2^x\)

  \(\frac{1.2.3.4.....30.31}{4.6.8.10....62.64}=2^x\)

 \(\frac{1.2.3.4.5....30.31}{2.2.2.3.2.4.2.5.....2.31.64}=2^x\)

 \(\frac{1.2.3.4.5.....30.31}{\left(2.2.2....2.2\right).\left(2.3.4.5....30.31\right).64}=2^x\)

 \(2.2.2.2.2.....2.64=2^x\)

 \(2^{31}.2^6=2^x\)

\(2^{37}=2^x\)

=> \(x=37\)