d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(\Leftrightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\Leftrightarrow\frac{x+2}{20}=\frac{4x-3}{36}\Leftrightarrow36x+72=80x-60\Leftrightarrow44x=132\Rightarrow x=2\)

\(\Leftrightarrow\frac{\frac{10x+x+2}{2}}{9}-\frac{\frac{x+3+75}{5}}{12}=x-2\)\(\Leftrightarrow\frac{11x+2}{18}-\frac{x+78}{60}=x-2\)\(\Leftrightarrow\left(\frac{11}{18}-\frac{1}{60}-1\right)x+\left(\frac{2}{18}-\frac{78}{60}+2\right)=0\).Giải típ nha, ko có Casio nên mk ko bấm

\(\Leftrightarrow5x+\frac{x+2}{18}-x-\frac{x+3+75}{60}+2=0\Leftrightarrow\left(5+\frac{1}{18}-1-\frac{1}{60}\right)x+\left(\frac{2}{18}-\frac{78}{60}+2\right)=0\).Giải típ nha

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)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

a)\(\frac{\frac{x-3}{5}+1}{4}=\frac{\frac{2x}{3}-\frac{1}{2}}{6}\)

\(\Rightarrow\frac{\frac{x-3+5}{5}}{4}=\frac{\frac{4x}{6}-\frac{3}{6}}{6}\)

\(\Rightarrow\frac{\frac{x+2}{5}}{4}=\frac{\frac{4x-3}{6}}{6}\)

\(\Rightarrow\frac{x+2}{5}:4=\frac{4x-3}{6}:6\)

\(\Rightarrow\frac{x+2}{5}.\frac{1}{4}=\frac{4x-3}{6}.\frac{1}{6}\)

\(\Rightarrow\frac{x+2}{20}=\frac{4x-3}{36}\)

\(\Rightarrow36.\left(x+2\right)=20.\left(4x-3\right)\)

\(\Rightarrow36x+72=80x-60\)

\(\Rightarrow72+60=80x-36x\)

\(\Rightarrow132=44x\)

\(\Rightarrow x=\frac{132}{44}=3\)

Vậy x=3

b)Bn làm tương tự phần a nha

Chúc bn học tốt

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Chúc bn học tốt

\(\Leftrightarrow\frac{3x-3+5}{20}=|\frac{4x-3}{36}|\)

\(\Leftrightarrow\frac{3x+2}{5}=|\frac{4x-3}{9}|\)

\(\Leftrightarrow\left(3x+2\right)9=5|4x-3|\)

ĐK: \(3x+2\ge0\Rightarrow x\ge\frac{-2}{3}\)

\(\Leftrightarrow\orbr{\begin{cases}5.\left(4x-3\right)=9.\left(3x+2\right)\\-5.\left(4x-3\right)=9.\left(3x+2\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}20x-15=27x+18\\-20x+15=27x+18\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}7x=-33\\-47x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-33}{7}\left(L\right)\\x=\frac{-3}{47}\end{cases}}}\)

Vậy x= - 3/47

b) \(\Leftrightarrow\frac{10x+x+2}{18}=\frac{x+3+75}{60}-2\)

\(\Leftrightarrow\frac{11x+2}{18}=\frac{x-42}{60}\)

\(\Leftrightarrow10.\left(11x+2\right)=3.\left(x-42\right)\)

\(\Leftrightarrow110x+20=3x-126\)

\(\Leftrightarrow107x=-146\Leftrightarrow x=\frac{-146}{107}\)