Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath

Ta có:

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}\)  \(=\frac{1}{18}\)

\(\Leftrightarrow\)\(\frac{1}{\left(x+4\right)\left(x+5\right)}\) \(+\frac{1}{\left(x+5\right)\left(x+6\right)}\) \(+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

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

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x-26=0\Leftrightarrow\hept{\begin{cases}x_1=2\\x_2=-13\end{cases}}\)

Vậy nghiệm của phương trình là {2;-13}

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)(điều kiện: \(x\ne\left\{-4;-5;-6;-7\right\}\) )

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

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow54=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)(thỏa mãn ĐKXĐ)

Vậy tập nghiệm của pt là: \(S=\left\{-13;2\right\}\)

Lâu lắm không làm nhể

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)

Dùng công thứ \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

Khi đó \(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54\)

\(\Rightarrow\hept{\begin{cases}x+4=6\\x+7=9\end{cases}}\)hoặc \(\hept{\begin{cases}x+4=-6\\x+7=-9\end{cases}}\)

Suy ra \(x=3\)hoặc \(x=-3\)

Thêm đk: \(x\in Z\)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\frac{1}{\left(x^2+4x\right)+\left(4x+20\right)}+\frac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\frac{1}{\left(x^2+6x\right)+\left(7x+42\right)}=\frac{1}{18}\)

\(\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

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

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

\(\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)

đến đây b tự lập bảng nhé~

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\left(x\ne-4;-5;-6;-7;-8\right)\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

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

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)

vậy x=2; x=-13

Bài làm:

đkxđ: \(x\ne\left\{-4;-5;-6;-7\right\}\)

Ta có: \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

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

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)

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

x² + 9x + 20 = ( x + 4 )( x + 5 )

x² + 11x + 30 = ( x + 5 )( x + 6 )

x² + 13x + 42 = ( x + 6 )( x + 7 )

=> Phương trình đã cho đưa về dạng :

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\left(1\right)\)

ĐKXĐ : \(x\ne\left\{-4;-5;-6;-7\right\}\)

\(\left(1\right)\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+7}{\left(x+4\right)\left(x+7\right)}-\frac{x+4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28-54=0\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x^2+13x\right)-\left(2x+26\right)=0\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-13\\x=2\end{cases}\left(tmđk\right)}\)

Vậy S = { -13 ; 2 }

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

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

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+7}{\left(x+4\right)\left(x+7\right)}-\frac{x+4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=18\cdot3=54\)

\(\Rightarrow x^2+11x+28-54=0\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

Vậy x \(\in\left\{-13;2\right\}\)

\(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

ĐKXĐ: x khác -4;-5;-6;-7

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow3.18=x^2+11x+28\)

\(\Rightarrow x^2+11x-26=0\)

\(\Rightarrow\left(x-2\right).\left(x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}\left(tm\right)}\)

Vậy...

\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{3}{54}\)

\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

Ta có \(\Delta=11^2+4.26=225,\sqrt{\Delta}=15\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+15}{2}=2\\x=\frac{-11-15}{2}=-13\end{cases}}\)

Vậy tập nghiệm S =  {2;-13}

Mày nhìn cái chóa j

bn giải nhanh nhỉ quang duy

giải giùm ra kết quả cho tui mừng coi

<=>1/(x+4)(x+5)+1/(x+5)(x+6)+1/(x+6)(x+7)=1/18

<=>1/(x+4)-1/(x+5)+1/(x+5)-1/(x+6)+1/(x+6)-1/(x+7)=1/18

<=>1/(x+4)-1/(x+7)=1/18

<=>3/(x+4)(x+7)=1/18

<=>(x+4)(x+7)=54

<=>x²+11x+28=54

<=>x²+11x-16=0

<=>x²+11x+121/4-185/4=0

đến đây chắctự làm đc