a: 3x+9=2x-11

=>3x-2x=-11-9

=>x=-20

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

=>4(2x-3)-20=5(2-x)

=>8x-12-20=10-5x

=>8x-32=10-5x

=>13x=42

hay x=42/13

b. `|x + 1| + |2x - 3| = |3x - 2|`

Ta có: \(\left|x+1\right|+\left|2x-3\right|\ge\left|x+1+2x-3\right|=\left|3x-2\right|\)

\(\Leftrightarrow\left|3x-2\right|=\left|3x-2\right|\) (luôn đúng với mọi x)

Vậy phương trình có vô số nghiệm.

(x-1)(2x^2-8)=0

\(\Leftrightarrow\left(x-1\right)\left(2x^2-8\right)=0\\ \left(2x^3-8x-2x^2+8\right)=0\)

\(\Leftrightarrow2x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow x=1;x=\dfrac{8}{2}\)

3x^2-8x+5=0

áp dụng công thức bậc 2 ta có:

\(x=\dfrac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4.3.5}}{2.3}\)

\(\Rightarrow x=\dfrac{5}{3};x=1\)

(7x-1).2x-7x+1=0

\(\Leftrightarrow\left(7x-1\right)\left(2x-1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{7};x=\dfrac{1}{2}\)

d: \(\Leftrightarrow\left(4x+2\right)\left(x-1\right)-\dfrac{1}{2}x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+2-\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-\dfrac{7}{2}x+2\right)=0\)

=>x=1 hoặc x=4/7

e: \(\Leftrightarrow2\left(5x-2\right)=3\left(5-3x\right)\)

=>10x-4=15-9x

=>19x=19

hay x=1

f: \(\Leftrightarrow\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}\)

=>2x-1+x-1=1

=>3x-2=1

hay x=1(loại)

g: =>1+3x-6=3-x

=>3x-5-3+x=0

=>4x-8=0

=>x=2(loại)

\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)

\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)

\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)

\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)

\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)

\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)

\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)

\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)

\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)

\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)

\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)

\(< =>5x-35=32x-8< =>32x-5x=-35+8\)

\(< =>27x=-27< =>x=-1\)

\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)

\(< =>12=6x-3< =>6x=12+3\)

\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)

\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)

\(< =>8x-4=9x+3< =>9x-8x=-4-3\)

\(< =>9x-8x=-7< =>x=-7\)

\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)

\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)

\(< =>12x-7x=14-4< =>5x=10\)

\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)

\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)

\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)

\(< =>6x-4x=4+6< =>2x=10\)

\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)

\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)

\(< =>\left(x+1\right)\left(x+1\right)=3.3\)

\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)

\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)

\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)

\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)

a, \(\frac{5}{6}=\frac{x-1}{x}\Leftrightarrow5x=6x-6\Leftrightarrow-x=-6\Leftrightarrow x=6\)

b, \(\frac{1}{2}=\frac{x+1}{3x}\Leftrightarrow3x=2x+2\Leftrightarrow x=2\)

c, \(\frac{3}{x+2}=\frac{5}{2x+1}\)ĐKXĐ : \(x\ne-2;-\frac{1}{2}\)

\(\Leftrightarrow6x+3=5x+10\Leftrightarrow x=7\)

a/ \(2x+\frac{1}{7}=\frac{1}{3}\)

=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)

=> \(2x=\frac{4}{21}\)

=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)

b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)

=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)

=> \(x-\frac{1}{2}=\frac{4}{27}\)

=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)

c/ \(\left(x-5\right)^2+4=68\)

=> \(\left(x-5\right)^2=68-4=64\)

=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)

d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)

=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)

e) \(5x+2=3x+8\)

=> \(5x-3x=8-2=6\)

=> \(2x=6\)

=> \(x=6:2=3\)

f/ \(26-\left(5-2x\right)=27\)

=> \(5-2x=26-27=-1\)

=> \(2x=5-\left(-1\right)=5+1=6\)

=> \(x=6:2=3\)

g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)

=> \(4x-8-2x+6=4\)

=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)

=> \(2x+-2=4\)

=> \(2x=4+2=6\)

=> \(x=6:2=3\)

h/ \(\left(x+3\right)^3:3-1=-10\)

=> \(\left(x+3\right)^3:3=-10+1=-9\)

=> \(\left(x+3\right)^3=-9.3=-27\)

=> \(x+3=-3\)

=> \(x=-3-3=-6\)

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