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

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

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

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

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

Từ đó, bạn tính ra nhá! Hơi dài, ai có cách nào ngắn hơn thì nói với mình nha!

minh ko biet

mình ko bít

ĐK : \(\left(x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)

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

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

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

\(\Rightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\)

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

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

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

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

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

Đk:\(\left(x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)

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

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

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

\(\Rightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\)

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

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

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

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

Vậy pt có tập nghiệm là S={2,-13}

Đk:

Vậy pt có tập nghiệm là S={2,-13}

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

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

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

dit me may

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

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

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

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

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

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

\(\Rightarrow x^2+11x+\frac{121}{4}-\frac{9}{4}=54\)

\(\Rightarrow\left(x+\frac{11}{2}\right)^2=\frac{225}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{11}{2}=\sqrt{\frac{225}{4}}\\x+\frac{11}{2}=-\sqrt{\frac{225}{4}}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{11}{2}=\frac{25}{2}\\x+\frac{11}{2}=-\frac{25}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-18\end{cases}}\)

loại trừ bớt đi rùi tìm x 

\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{x.\left(x+2\right)}=\frac{24}{35}\)

\(\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{24}{35}\)

\(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)

\(\frac{3}{10}-\frac{3}{2x+4}=\frac{24}{35}\)

\(\frac{3}{2x+4}=\frac{-27}{70}\)

tự làm nốt

\(\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+....+\frac{3}{x\left(x+2\right)}=\frac{24}{35}\)

\(\Leftrightarrow3\left(\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+....+\frac{1}{x\left(x+2\right)}\right)=\frac{24}{25}\)

\(\Leftrightarrow3\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{24}{35}\)

\(\Leftrightarrow3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{24}{35}\)

\(\Leftrightarrow3\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)

\(\Leftrightarrow\frac{3}{5}-\frac{3}{x+2}=\frac{24}{35}\)

\(\Leftrightarrow\frac{3}{x+2}=\frac{3}{5}-\frac{24}{35}\)

\(\Leftrightarrow\frac{3}{x+2}=\frac{21-24}{35}\)

\(\Leftrightarrow\frac{3}{x+2}=\frac{3}{-35}\)

\(\Rightarrow x+2=-35\Leftrightarrow x=-35-2\Leftrightarrow x=-37\)

Vậy \(x=-37\)

2.[1/42+1/56+1/72+...+1/x.(x+1)]=2/9

1/6.7+1/7.8+1/8.9+....+1/x.(x+1)=1/9

1/6-1/7+1/7-1/8+1/8-1/9+.....+1/x-1/x+1=1/9

1/6-1/x+1=1/9

1:(x+1)=1/6-1/9

x+1=1:(1/18)

x+1=18

x=18-1

x=17

Vậy x=17

Chúc em học tốt

Ủng hộ anh nha^^

2/42 + 2/56 + 2/72 + ... + 2/x.(x+1) = 2/9

2.[1/42 + 1/56 + 1/72 + ... + 1/x.(x+1)] = 2/9

1/6.7 + 1/7.8 + 1/8.9 + ... + 1/x.(x+1) = 2/9 : 2

1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + ... + 1/x - 1/x+1 = 2/9 . 1/2

1/6 - 1/x+1 = 1/9

1/x+1 = 1/6 - 1/9

1/x+1 = 6/36 - 4/36

1/x+1 = 2/36 = 1/18

=> x+1=18

=> x=18-1

=> x=17

Vậy x=17

Đặt VT là A ta có:

\(A=2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x\left(x+1\right)}\right)\)

\(=2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

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

Thay A vào ta có:\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

\(\Rightarrow x+1=18\)

\(\Rightarrow x=17\)

ta xét VT=\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{x\left(x+1\right)}\right)\)

                =\(2\left(\frac{7-6}{6\cdot7}+\frac{8-7}{7\cdot8}+...+\frac{\left(x+1\right)-x}{x\left(x+1\right)}\right)=2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

                =\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)\)= 2*1/9

=> \(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

<=> \(\frac{1}{x+1}=\frac{1}{18}\)

<=> x+1=18

=> x=17

tớ làm khác nhưng kết quả thì giống