\(ĐKXĐ:x\ne5,8\)

\(\frac{6}{x-5}+\frac{x+2}{x-8}=\frac{18}{\left(x-5\right)\left(8-x\right)}-1\)

\(\Rightarrow\frac{6}{x-5}+\frac{x+2}{x-8}=-\frac{18}{\left(x-5\right)\left(x-8\right)}-1\)

\(\Rightarrow6\left(x-8\right)+\left(x+2\right)\left(x-5\right)=-18-\left(x-5\right)\left(x-8\right)\)

\(\Rightarrow x^2+3x-58=-x^2+13x-58\)

\(\Rightarrow2x^2-10x=0\)

\(\Rightarrow2x\left(x-5\right)=0\)

\(\Rightarrow x\in\left\{0,5\right\}\)

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

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 = \(\left\{\frac{-2}{3}\right\}\)

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)

Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)

Lấy (4) trừ (3) ta có:

\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

\(\frac{7}{8}x-5\left(x-9\right)=\frac{20x-1,5}{6}\)

\(\Leftrightarrow\frac{7}{8}x-5x+45=\frac{10x}{3}-\frac{1}{4}\)

\(\Leftrightarrow\frac{-33}{8}x+45=\frac{10x}{3}-\frac{1}{4}\)

\(\Leftrightarrow\frac{-33}{8}x-\frac{10}{3}x=-\frac{1}{4}-45\)

\(\Leftrightarrow\frac{-179}{24}x=-\frac{181}{4}\)

\(\Leftrightarrow x=\frac{1086}{179}\)

\(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Rightarrow\frac{7}{8}x-5x+45=\frac{20x}{6}+\frac{1}{4}\)

\(\Rightarrow\frac{7}{8}x-\frac{40}{8}x+45=\frac{10x}{3}+\frac{1}{4}\)

\(\Rightarrow\frac{-33}{8}x+45=\frac{10x}{3}+\frac{1}{4}\)

\(\Rightarrow\frac{-33}{8}x-\frac{10x}{3}=\frac{1}{4}-45\)

\(\Rightarrow\frac{-179}{24}x=\frac{-179}{4}\)

\(\Rightarrow x=6\)

Vậy phương trình có 1 nghiệm là 6

\(\Leftrightarrow56\left(x+1\right)+63\left(x+2\right)=72\left(x+3\right)+84\left(x+4\right)\)

\(\Leftrightarrow56\left(x+1\right)+63\left(x+2\right)-72\left(x+3\right)-84\left(x+4\right)=0\)

\(\Leftrightarrow-37x-370=0\Leftrightarrow x=-10\)

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

Vậy $x = -10$

b.

\(\dfrac{x+5}{x-1}-\dfrac{x+1}{x-3}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Leftrightarrow\dfrac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Rightarrow x^2+2x-15-x^2+1=0\\ \Leftrightarrow2x-14=0\\ \Leftrightarrow x=7\)

Vậy x = 7

b.

\(\dfrac{x+5}{x-1}-\dfrac{x+1}{x-3}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Leftrightarrow\dfrac{x^2+2x-15}{\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-8}{\left(x-1\right)\left(x-3\right)}\\ \Rightarrow x^2+2x-15-x^2+1=-8\\ \Leftrightarrow2x-14=-8\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\)

\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

=> x =  - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)

\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)

\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)

mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)

nên x+100=0

hay x=-100

Vậy: S={-100}

          \(\left(x+2\right)\left(x+5\right)\left(x-6\right)\left(x-9\right)=280\)

 \(\Leftrightarrow\)\(\left(x^2-4x-12\right)\left(x^2-4x-45\right)-280=0\)

Đặt      \(x^2-4x-12=t\)    ta có:

                   \(t\left(t-33\right)-280=0\)

        \(\Leftrightarrow\)\(t^2-33t-280=0\)

        \(\Leftrightarrow\)\(t^2-40t+7t-280=0\)

        \(\Leftrightarrow\)\(\left(t-40\right) \left(t+7\right)=0\)

        \(\Leftrightarrow\)\(\orbr{\begin{cases}t-40=0\\t+7=0\end{cases}}\)

Đến đây bn thay trở lại và tìm   x   nhé! chúc bn hok tốt