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19 tháng 11 2020

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

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

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

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

\(\Leftrightarrow\frac{\left(x+6\right)+\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)\(\Leftrightarrow\frac{2x+10}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{2\left(x+7\right)+\left(x+4\right)}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3x+18}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3\left(x+6\right)}{\left(x+4\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

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

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

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

\(\Leftrightarrow x^2+11x+30.25=56.25\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+5.5\right)^2=\left(7.5\right)^2\\\left(x+5.5\right)^2=\left(-7.5\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+5.5=7.5\\x+5.5=-7.5\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=7.5-5.5=2\\x=-7.5-5.5=-13\end{cases}}\)

\(\text{Vậy x }\in\left\{2;-13\right\}\)

19 tháng 11 2020

\(\text{nhớ tích cho mk nha Thanh Do}\)

24 tháng 11 2016

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

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

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

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

19 tháng 9 2016

minh ko biet

19 tháng 9 2016

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}

 

14 tháng 6 2020

Đk:(x≠−4;x≠−5;x≠−6;x≠−7)(x≠−4;x≠−5;x≠−6;x≠−7)

⇒1(x+4)(x+5)+1(x+5)(x+6)+1(x+6)(x+7)=118⇒1(x+4)(x+5)+1(x+5)(x+6)+1(x+6)(x+7)=118

⇒1x+4−1x+5+1x+5−1x+6+1x+6−1x+7=118⇒1x+4−1x+5+1x+5−1x+6+1x+6−1x+7=118

⇒1x+4−1x+7=118⇒1x+4−1x+7=118

⇒3x2+11x+28=118⇒3x2+11x+28=118

⇔x2+11x+28=54⇔x2+11x+28=54

⇒x2+11x−26=0⇒x2+11x−26=0

⇒(x−2)(x+13)=0⇒(x−2)(x+13)=0

⇒[x=2x=−13⇒[x=2x=−13

Vậy pt có tập nghiệm là S={2,-13}

25 tháng 12 2019

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

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

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

25 tháng 12 2019

dit me may

5 tháng 6 2017

phân tích mẫu thành nhân tử r áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) sau đó rút gọn quy đồng

5 tháng 6 2017

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\) \(\left(ĐKXĐ:x\ne0;x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)

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

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

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

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

\(\Leftrightarrow\frac{\left(x^2+13x+42\right)+\left(x^2+11x+28\right)+\left(x^2+9x+20\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\frac{3\left(x^2+11x+30\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=18.3\left(x^2+11x+30\right)\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=54\left(x+5\right)\left(x+6\right)\)

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

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

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

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

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

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

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

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

3 tháng 3 2017

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

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

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

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

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

\(\Rightarrow x=2\)

3 tháng 3 2017

nếu x<0 thì có lẽ là x=-2

23 tháng 4 2018

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

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

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

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

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

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

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

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

23 tháng 4 2018

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

\(\Rightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{x+5}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Rightarrow\dfrac{18\left(x+7\right)}{18\left(x+5\right)\left(x+7\right)}-\dfrac{18\left(x+5\right)}{18\left(x+5\right)\left(x+7\right)}=\dfrac{\left(x+5\right)\left(x+7\right)}{18\left(x+5\right)\left(x+7\right)}\\ \Rightarrow18x+126-18x-90=x^2+5x+7x+35\\ \Leftrightarrow x^2+12x+35=36\\ \Leftrightarrow x^2+12x-1=0\\ \Leftrightarrow x^2+12x+36-37=0\\ \Leftrightarrow\left(x^2+12x+36\right)-37=0\\ \Leftrightarrow\left(x+6\right)^2-37=0\\ \Leftrightarrow\left(x+6+\sqrt{37}\right)\left(x+6-\sqrt{37}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+6+\sqrt{37}=0\\x+6-\sqrt{37}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6-\sqrt{37}\\x=\sqrt{37}-6\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\sqrt{37}-6;-\sqrt{37}-6\right\}\)

16 tháng 9 2015

bn giải nhanh nhỉ quang duy

16 tháng 9 2015

giải giùm ra kết quả cho tui mừng coi

18 tháng 3 2016

3,867400402

27 tháng 12 2019

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

10 tháng 3 2019

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

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

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

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

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

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

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

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

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

Vậy x \(\in\left\{-13;2\right\}\)