x=3;-0,5;-2

a) \(\Delta=169-56=113>0\)

\(\hept{\begin{cases}x_1=\frac{13+\sqrt{113}}{14}\\x_2=\frac{13-\sqrt{113}}{14}\end{cases}}\)

b)  \(\Delta=25-4.3.60< 0\)

vô nghiệm 

Cac ban giup mk di mk dang can ngay bay gio .

\(2ab=0\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

Đã bảo bạn là đăng vào phần lớp 8 ấy. Sẽ có những người học cùng cấp giải cho bạn.

2ab =0 <=> a= 0 hoặc b=0 

Chúc bạn học tốt!!

Ta có: 2ab=0

nên \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

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

Kết quả:

1. \(-\frac{2}{3}\)

2. \(3\)

\(\Leftrightarrow\left(b-2\sqrt{2}\right)\left(b+2\sqrt{2}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}b\ge2\sqrt{2}\\b\le-2\sqrt{2}\end{matrix}\right.\)