CỘNG 2 VÀO MỖI VẾ

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

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

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

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

\(\text{ma}:\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

=> x + 0 = 10

=> x       = 0 -10

=> x       = -10

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

Cộng cả 2 vế với 2 ta được :

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

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

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

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

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

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(x\ne1\right)\)

\(\Leftrightarrow\frac{1}{x-1}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{x^2+x+1}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4x-4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{3x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Rightarrow3x=0\)

=> x=0 (tmđk)
Vậy x=0

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

\(=\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)

\(\Rightarrow5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)\left(x-2\right)=16\left(x-2\right)\left(x-4\right)\)

\(\Leftrightarrow5x^2-35x+60+5x^2-20x+20=16x^2-96x+128\)

\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)

\(\Leftrightarrow-6x^2+41x-48=0\)

......

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

\(\Leftrightarrow\frac{x-3}{x-2}+\frac{x-2}{x-4}=\frac{16}{5}\)

\(\Leftrightarrow\frac{5\left(x-3\right)\left(x-4\right)+5\left(x-2\right)^2}{5\left(x-2\right)\left(x-4\right)}=\frac{16.\left(x-2\right)\left(x-4\right)}{5\left(x-2\right)\left(x-4\right)}\)

\(\Rightarrow5x^2-20x-15x+60+5x^2-20x+20=16x^2-64x-32x+128\)

\(\Leftrightarrow10x^2-55x+80=16x^2-96x+128\)

\(\Leftrightarrow6x^2-41x+48=0\)

\(\Leftrightarrow x=\frac{16}{3};x=\frac{3}{2}\)

1) Hình như đề bị sai rồi bạn.

Thông thường pt đã cho sẽ là \(\frac{2x}{x-2}-\frac{5}{x-3}=\frac{5}{x^2-5x+6}\)

Ta thấy \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

Nên ĐKXĐ là \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

pt đã cho \(\Leftrightarrow\frac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{5\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{2x^2-6x-5x+10}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)\(\Rightarrow2x^2-11x+5=0\)(*)

Ta có \(\Delta=\left(-11\right)^2-4.2.5=81>0\)nên pt (*) có 2 nghiệm phân biệt:

\(\orbr{\begin{cases}x_1=\frac{-\left(-11\right)+\sqrt{81}}{2.2}=5\left(nhận\right)\\x_2=\frac{-\left(-11\right)-\sqrt{81}}{2.2}=\frac{1}{2}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{1}{2};5\right\}\)

2) Nhận thấy \(3x^2-27=3\left(x^2-9\right)=3\left(x-3\right)\left(x+3\right)\)nên ĐKXĐ ở đây là \(x\ne\pm3\)

pt đã cho \(\Leftrightarrow\frac{1}{3\left(x-3\right)\left(x+3\right)}+\frac{3}{4}=1+\frac{1}{x-3}\)

\(\Leftrightarrow\frac{1}{3\left(x-3\right)\left(x+3\right)}-\frac{3\left(x+3\right)}{3\left(x-3\right)\left(x+3\right)}=\frac{1}{4}\)

\(\Leftrightarrow\frac{1-3x-9}{3x^2-27}=\frac{1}{4}\)\(\Rightarrow-12x-32=3x^2-27\)\(\Leftrightarrow3x^2+12x+5=0\)(#)

Nhận thấy \(\Delta'=6^2-3.5=21>0\)

Vậy pt (#) có 2 nghiệm phân biệt \(\orbr{\begin{cases}x_1=\frac{-12+\sqrt{21}}{3}\left(nhận\right)\\x_2=\frac{-12-\sqrt{21}}{3}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{-12\pm\sqrt{21}}{3}\right\}\)

Theo bài ra ,ta có : 

\(\frac{x+1}{x-2}-\frac{1}{x}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow\frac{x+1}{x-2}-\frac{1}{x}=\frac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\left(ĐKXĐ:x\ne0;x\ne2;x\ne-2\right)\)

Quy đồng và khử mẫu ta được 

\(x\left(x+1\right)\left(x+2\right)-\left(x-2\right)\left(x+2\right)=2x\left(x^2+2\right)\)

\(\Leftrightarrow\left(x^2+x\right)\left(x+2\right)-\left(x-2\right)\left(x+2\right)=2x^3+4x\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+x-x+2\right)=2x^3+4x\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+2\right)=2x^3+4x\)

\(\Leftrightarrow x^3+2x+2x^2+4=2x^3+4x\)

\(\Leftrightarrow x^3-2x^3+2x^2+2x-4x+4=0\)

\(\Leftrightarrow-x^3+2x^2-2x+4=0\)

\(\Leftrightarrow-\left(x^3-2x^2+2x-4\right)=0\)

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

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

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

\(\Leftrightarrow2-x=0\)( Vì x² + 2 luôn luôn > 2 với mọi x ) 

\(\Leftrightarrow x=2\)(Không TMĐKXĐ) ( Loại )

Vậy S={rỗng}

Chúc bạn học tốt =))

Mình sẽ k cho bạn nào nhanh nhất nhé <3

\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)

\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)