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=\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\\ =>\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \left(=\right)\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\\ gpt=>x=2\)
Câu hỏi của Duong Thi Nhuong TH Hoa Trach - Phong GD va DT Bo Trach - Toán lớp 8 | Học trực tuyến
a)
ĐKXĐ: x khác -4;-5;-6;-7
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+20}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{8}\\ \Rightarrow x^2+11x+28=24\\ \Leftrightarrow x^2+11x+4=0\)
ta có: \(\Delta=11^2-4.1.4=105>0\) nên phương trình có 2 nghiệm phân biệt.
\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-11-\sqrt{105}}{2}\\x_2=\dfrac{-11+\sqrt{105}}{2}\end{matrix}\right.\)
Đk:\(x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\)
\(\Rightarrow x^2+11x+28=54\)
\(\Rightarrow x^2+11x-26=0\)
\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)
Vậy....
giải hộ với mấy chế ơi, tôi rút gọn mak nó ra cái jj ý _0_
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 \(S=\left\{2;-13\right\}\)
# Bài 1
* Ta cm BĐT sau \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\) (1) bằng cách biến đổi tương đương
* Với \(x,y>0\) áp dụng (1) ta có
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{\left(\sqrt{x}\right)^2}+\dfrac{1}{\left(\sqrt{y}\right)^2}\ge\dfrac{1}{2}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\)
Mà \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\)
\(\Rightarrow\) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)^2\le1\) \(\Leftrightarrow\) \(0< \dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\le1\) (I)
* Ta cm BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) với \(a,b>0\) (2)
Áp dụng (2) với x , y > 0 ta có
\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\ge\dfrac{4}{\sqrt{x}+\sqrt{y}}\) (II)
* Từ (I) và (II) \(\Rightarrow\) \(\dfrac{4}{\sqrt{x}+\sqrt{y}}\le1\)
\(\Leftrightarrow\) \(\sqrt{x}+\sqrt{y}\ge4\)
Dấu "=" xra khi \(x=y=4\)
Vậy min \(\sqrt{x}+\sqrt{y}=4\) khi \(x=y=4\)
a) \(\sqrt{\dfrac{9x^2}{25}}+\dfrac{1}{5}x\) (x<0)
=\(\dfrac{-3x}{5}+\dfrac{x}{5}\) (vì x<0)
=\(\dfrac{-2x}{5}\)
b)2xy\(\sqrt{\dfrac{9x^2}{y^6}}-\sqrt{\dfrac{49x^2}{y^2}}\) (x<0 , y>0)
=2xy\(\dfrac{-3x}{y^3}+\dfrac{7x}{y}\)(vì x<y<0)
=\(\dfrac{-6x}{y^2}+\dfrac{7xy}{y^2}\)
=\(\dfrac{7xy-6x}{y^2}\)
c) \(\dfrac{1}{ab}\sqrt{a^6\left(a-b\right)^2}\) (a<b<0)
=\(\dfrac{1}{ab}\sqrt{a^6}\sqrt{\left(a-b\right)^2}\)
=\(\dfrac{1}{ab}\left(-a^3\right)\left(b-a\right)\) (vì a<b<0)
=\(\dfrac{\left(a-b\right)a^3}{a-b}\)
=a3
Cảm ơn bạn Thu Trang nhiều nhé, sau này có gì giúp đỡ nhau nha. 
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{18\left(x+7-x-4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(18.3=\left(x+4\right)\left(x+7\right)\)
\(x^2+11x+28-54=0\)
\(x^2+11x-26=0\)
\(\left(x-2\right)\left(x+13\right)=0\)
\(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Theo đề x < 0 nên x = -13