\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)

\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)

\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)

\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)

\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)

\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)

\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)

\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)

Vì \(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\ne0\)

Nên x + 116 = 0

<=> x = -116

Bài 1:

Thay \(x=\frac{4}{3};y=-1\)vào biểu thức A, ta được:

\(A=\frac{4}{3}\cdot\left[3\cdot\frac{4}{3}-\left(-1\right)\right]-\left(3\cdot\frac{4}{3}+1\right)\left(-1\right)\)

\(A=\frac{20}{3}+5=\frac{35}{3}\)

Vậy khi \(x=\frac{4}{3};y=-1\)thì A=\(\frac{35}{3}\)

\(B=3\frac{1}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot5\frac{118}{119}-\frac{8}{39}\)

\(B=\frac{352}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot\frac{713}{119}-\frac{8}{39}=-\frac{412}{1071}\)

Điều kiện xác định: \(x\ne1;3\)

Với điều kiện xác định như trên:

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

\(\Leftrightarrow\frac{3\left(x-1\right)-2\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{3\left(x-1\right)-2\left(x-3\right)}{6}\)

\(\Leftrightarrow\frac{x+3}{\left(x-1\right)\left(x-3\right)}=\frac{x+3}{6}\)

\(\Leftrightarrow\left(x+3\right)\left(\frac{1}{\left(x-1\right)\left(x-3\right)}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x-1\right)\left(x-3\right)=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(tm\right)\\\left(x-4x+3-6=0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\pm\sqrt{7}\left(tm\right)\end{matrix}\right.\)

Vậy phương trình có 3 nghiệm \(x=-3\) hoặc \(x=2\pm\sqrt{7}\)

a, Đặt \(x=\frac{1}{117}\), \(y=\frac{1}{119}\) ta có:

\(A=\left(3+x\right)y-4x\left(5+1-y\right)-5xy+24x\)

\(=3y+xy-24x+4xy-5xy+24x\)

\(=3y\)

\(=\frac{3}{119}\)

b, Thay 8 bằng x + 1 ta có:\(B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...-x^3-x^2+x^2+x-5\)

\(=7-5\)

= 2

a) Đặt a = \(\frac{1}{117}\)và b = \(\frac{1}{119}\)

Theo đề ta có:

A = (3 + a) b - 4a ( 5+1-b)-5ab+24a

= 3b + ab - 20a -4a + 4ab - 5ab + 24a

= 3b

= 1.\(\frac{1}{119}\) = \(\frac{3}{119}\)

Vậy A = \(\frac{3}{119}\)

\(ĐKXĐ:x\ne\frac{5-\sqrt{13}}{2};x\ne\frac{5+\sqrt{13}}{2}\)

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

*) Xét x = 0 thì \(\frac{4x}{x^2+x+3}+\frac{5x}{x^2-5x+3}=0\)(Loại)

*) Xét \(x\ne0\)thì phương trình tương đương \(\frac{4}{x+\frac{3}{x}+1}+\frac{5}{x+\frac{3}{x}-5}=-\frac{3}{2}\)

Đặt \(x+\frac{3}{x}=t\)thì phương trình trở thành \(\frac{4}{t+1}+\frac{5}{t-5}=-\frac{3}{2}\)

\(\Leftrightarrow\frac{4t-20+5t+5}{\left(t+1\right)\left(t-5\right)}=-\frac{3}{2}\Leftrightarrow\frac{9t-15}{t^2-4t-5}=-\frac{3}{2}\)

\(\Leftrightarrow18t-30=-3t^2+12t+15\Leftrightarrow3t^2+6t-45=0\)

\(\Leftrightarrow3\left(t-3\right)\left(t+5\right)=0\Leftrightarrow\orbr{\begin{cases}t=3\\t=-5\end{cases}}\)

+) t = 3 thì \(x+\frac{3}{x}=3\Leftrightarrow\frac{x^2+3}{x}=3\Leftrightarrow x^2-3x+3=0\)

Mà \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}>0\forall x\)nên loại trường hợp t = 3

+) t = -5 thì \(x+\frac{3}{x}=-5\Leftrightarrow\frac{x^2+3}{x}=-5\Leftrightarrow x^2+5x+3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5+\sqrt{13}}{2}\\x=\frac{-5-\sqrt{13}}{2}\end{cases}}\)

Vậy phương trình có 2 nghiệm \(\left\{\frac{-5+\sqrt{13}}{2};\frac{-5-\sqrt{13}}{2}\right\}\)

Bài làm:

đkxđ: \(x\ne\left\{\frac{5+\sqrt{13}}{2};\frac{5-\sqrt{13}}{2}\right\}\)

+ Nếu x = 0:

\(Pt\Leftrightarrow0=-\frac{3}{2}\)(vô nghiệm)

+ Nếu x khác 0:

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

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

Đặt \(x+\frac{3}{x}=y\)

\(Pt\Leftrightarrow\frac{4}{y+1}+\frac{5}{y-5}=-\frac{3}{2}\)

\(\Leftrightarrow\frac{8\left(y-5\right)+10\left(y+1\right)}{2\left(y+1\right)\left(y-5\right)}=-\frac{3\left(y-5\right)\left(y+1\right)}{2\left(y+1\right)\left(y-5\right)}\)

\(\Rightarrow8y-40+10y+10=-3\left(y^2-4y-5\right)\)

\(\Leftrightarrow18y-30=-3y^2+12y+15\)

\(\Leftrightarrow3y^2+6y-45=0\)

\(\Leftrightarrow y^2+2y-15=0\)

\(\Leftrightarrow\left(y-3\right)\left(y+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y-3=0\\y+5=0\end{cases}}\Leftrightarrow\Leftrightarrow\orbr{\begin{cases}y=3\\y=-5\end{cases}}\)

Nếu: \(y=3\Leftrightarrow x+\frac{3}{x}=3\Leftrightarrow\frac{x^2+3}{x}=3\Leftrightarrow x^2+3=3x\)

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

\(\Leftrightarrow\left(x^2-3x+\frac{9}{4}\right)+\frac{3}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-\frac{3}{4}\)(vô lý)

=> không tồn tại x thỏa mãn

Nếu: \(y=-5\Leftrightarrow x+\frac{3}{x}=-5\Leftrightarrow\frac{x^2+3}{x}=-5\Leftrightarrow x^2+3=-5x\)

\(\Leftrightarrow x^2+5x+3=0\)

\(\Leftrightarrow\left(x^2+5x+\frac{25}{4}\right)-\frac{13}{4}=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}-\frac{\sqrt{13}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{13}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5-\sqrt{13}}{2}=0\\x+\frac{5+\sqrt{13}}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-5}{2}\\x=\frac{-5-\sqrt{13}}{2}\end{cases}}\)(thỏa mãn)

Vậy tập nghiệm của PT \(S=\left\{\frac{-5-\sqrt{13}}{2};\frac{\sqrt{13}-5}{2}\right\}\)

\(2\frac{1}{117}.3\frac{1}{119}-\frac{116}{117}.5\frac{118}{119}-\frac{3}{119}=\left(3-\frac{116}{117}\right)\cdot\left(4-\frac{118}{119}\right)-5\cdot\frac{116}{117}\cdot\frac{118}{119}-\frac{3}{119}\)

mình đang ngại mình làm đến đó bạn tự phá ngoại rồi đặt nhân tử chung nha