\(\dfrac{32}{3.7}+\dfrac{6}{7.41}+\dfrac{9}{41.10}+\dfrac{1}{51.10}+\dfrac{19}{51.14}\)

\(=\dfrac{32}{21}+\dfrac{6.5}{7.5.41}+\dfrac{9.5}{41.10.5}+\dfrac{5}{51.10.5}+\dfrac{19.5}{51.14.5}\)

\(=\dfrac{32}{21}+5.\left(\dfrac{6}{35.41}+\dfrac{9}{41.50}+\dfrac{1}{51.50}+\dfrac{19}{51.70}\right)\)

\(=\dfrac{32}{21}+5.\left(\dfrac{41-35}{35.41}+\dfrac{50-41}{41.50}+\dfrac{51-50}{51.50}+\dfrac{70-51}{51.70}\right)\)

\(=\dfrac{32}{21}+5.\left(\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{51}+\dfrac{1}{51}-\dfrac{1}{70}\right)\)

\(=\dfrac{32}{21}+5.\left(\dfrac{1}{35}-\dfrac{1}{70}\right)\)

\(=\dfrac{32}{21}+5.\dfrac{1}{70}\)

\(=\dfrac{32}{21}+\dfrac{1}{14}\)

\(=\dfrac{32.2}{42}+\dfrac{1.3}{42}\)

\(=\dfrac{67}{42}\)

a) \(2^{3x-1}=16\)

\(\Leftrightarrow2^{3x-1}=2^4\)

\(\Leftrightarrow3x-1=4\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

b) \(\log_3\left(2x-5\right)=2\)

\(\Leftrightarrow\log_3\left(2x-5\right)=\log_33^2\)

\(\Leftrightarrow2x-5=9\)

\(\Leftrightarrow2x=14\)

\(\Leftrightarrow x=7\)

Bạn xem lại đề bài

Bài 12 :

a) \(16x^2-17x+1=0\left(1\right)\) 

Ta thấy : \(a+b+c=16-17+1=0\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{16}\end{matrix}\right.\)

b) \(2x^2-4x-6=0\left(1\right)\)

Ta thấy : \(a-b+c=2+4-6=0\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

c) \(2x^2-40x+38=0\left(1\right)\)

Ta thấy : \(a+b+c=2-40+38=0\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\x=19\end{matrix}\right.\)

d) \(1230x^2-5x-1235=0\left(1\right)\)

Ta thấy : \(a-b+c=1230+5-1235=0\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1235}{1230}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{247}{246}\end{matrix}\right.\)

Bài 17 :

a) \(x^2-\left(2m+1\right)x+m^2+m-6=0\left(1\right)\)

\(\Delta=\left(2m+1\right)^2-4\left(m^2+m-6\right)\)

\(\Delta=4m^2+4m+1-4m^2-4m+24\)

\(\Delta=25>0\)

\(Pt\left(1\right)\) luôn có 2 nghiệm phân biệt

b) Để \(\left(1\right)\) có 2 nghiệm âm phân biệt

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\S< 0\\P>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25>0,\forall m\\2m+1< 0\\m^2+m-6>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m< -\dfrac{1}{2}\\\left[{}\begin{matrix}m< -3\\m>2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow m< -3\)

Vậy \(m< -3\) thỏa mãn đề bài

c) \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-x_1x_2\)

\(\Rightarrow A=S^2-2P=\left(2m+1\right)^2-2\left(m^2+m-6\right)\)

\(\Rightarrow A=4m^2+4m+1-2m^2-2m+12\)

\(\Rightarrow A=2m^2+2m+13\)

\(\Rightarrow A=2\left(m^2+m+\dfrac{1}{4}\right)+13-\dfrac{1}{2}\)

\(\)\(\Rightarrow A=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{25}{2}\ge\dfrac{25}{2}\) \(\left(2\left(m+\dfrac{1}{2}\right)^2\ge0,\forall m\in R\right)\)

\(\Rightarrow GTNN\left(A\right)=\dfrac{25}{2}\left(tại.x=-\dfrac{1}{2}\right)\)

Ta có :

\(\left\{{}\begin{matrix}\dfrac{1}{5}=\dfrac{14}{70}\\\dfrac{3}{14}=\dfrac{15}{70}\end{matrix}\right.\)

\(\Rightarrow\dfrac{3}{14}>\dfrac{1}{5}\)

mà \(\dfrac{3}{14}< \dfrac{3}{13}< \dfrac{3}{12}< \dfrac{3}{11}< \dfrac{3}{10}\)

\(\Rightarrow S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}>\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}=\dfrac{5}{5}=1\left(1\right)\)

Ta lại có :

\(\dfrac{3}{10}< \dfrac{4}{10}\)

\(\Rightarrow S=\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}< \dfrac{4}{10}+\dfrac{4}{10}+\dfrac{4}{10}+\dfrac{4}{10}=\dfrac{20}{10}=2\left(2\right)\)

\(\left(1\right)\&\left(2\right)\Rightarrow1< S< 2\)

\(\Rightarrow dpcm\)

Áp dụng TCDTSBN, ta có :

\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+c+a+a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{c}{a+b}=\dfrac{1}{2}\\\dfrac{a}{b+c}=\dfrac{1}{2}\\\dfrac{b}{c+a}=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b}{c}=2\\\dfrac{b+c}{a}=2\\\dfrac{c+a}{b}=2\end{matrix}\right.\)

\(\Rightarrow S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=2+2+2=6\)

Vậy \(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=6\)

Em đánh nhầm \(x=0;y=0;y'=0\) sửa lại \(x=0;y=1;y'=0\)

Đáp án là C vì

- Tại \(x=0;y=0;y'=0\)

- \(y'\) đổi dấu từ âm sang dương ở \(x\in(-\infty;2)\)