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

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

⇔ \(\frac{7}{6}-x=\frac{5}{6}+1\)

⇔ \(\frac{7}{6}-x=\frac{11}{6}\)

⇔ \(-x=\frac{11}{6}-\frac{7}{6}\)

⇔ \(-x=\frac{2}{3}\)

⇔ \(x=\frac{-2}{3}\)

Vậy tập nghiệm của pt là S = \(\left\{\frac{-2}{3}\right\}\)

A, \(\frac{x+3}{2}\)-\(\frac{x-1}{3}\)=\(\frac{x+5}{6}\)+1

⇔ \(\frac{3\left(x+3\right)}{6}\)-\(\frac{2\left(x-1\right)}{6}\)=\(\frac{x+5}{6}\)+\(\frac{6}{6}\)

⇔ 3x+9-2x+2=x+5+6

⇔ 3x-2x-x=5+6-9-2

⇔0x=0 (luôn đúng với mọi x)

Vậy phương trình có vô số nghiêm:S=R

cộng 1 vào từ số hạng ( hai vế cùng 3 số hạng=> không đổi)

Tử số còn lại x

\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}\)

\(\Leftrightarrow\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)x=0\)

cái (...) khác không=> x =0 là nghiệm duy nhất

Ta có

\(\frac{x-19}{19}+\frac{x-20}{20}+\frac{x-21}{21}=\frac{x-17}{17}+\frac{x-16}{16}+\frac{x-15}{15}\)

\(\Leftrightarrow\frac{x}{19}-1+\frac{x}{20}-1+\frac{x}{21}-1=\frac{x}{17}-1+\frac{x}{16}-1+\frac{x}{15}-1\)

\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}-3=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}-3\)

\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}\)

\(\Rightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}-\frac{x}{17}-\frac{x}{16}-\frac{x}{15}=0\)

\(\Leftrightarrow x\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)=0\)

Vì \(\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)\ne0\)

Nên phương trình chỉ co nghiệm duy nhất là x=0

Vậy x=0

d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7

\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>x^2+11x+28=54

=>x^2+11x-26=0

=>(x+13)(x-2)=0

=>x=2 hoặc x=-13

e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)

=>x-258=0

=>x=258

\(\left(\frac{x+1}{39}+1\right)+\left(\frac{x+2}{38}+1\right)=\left(\frac{x+3}{37}+1\right)+\left(\frac{x+4}{36}+1\right)\)

\(\Leftrightarrow\frac{x+40}{39}+\frac{x+40}{38}-\frac{x+40}{37}-\frac{x+40}{36}=0\)

\(\Leftrightarrow\left(x+40\right)\left(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\right)=0\)

<=> x+40=0 (vì \(\frac{1}{39}+\frac{1}{38}-\frac{1}{37}-\frac{1}{36}\ne\)0)

<=> x=-40

Vậy x=-40

Ta có:

\(x=35\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=36\\x+2=27\\2x-1=69\\x-1=34\end{matrix}\right.\) (1)

Thay (1) vào biểu thức ta được:

\(x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x-1\right)x^2-\left(x-1\right)x+15\)

\(=x^5-x^5-x^4+x^4+2x^3-2x^3+x^2-x^2+x+15\)

\(=x+15\)

\(=35+15\)

\(=50\)

50