a)\(3x-2=2x+5\\ \Leftrightarrow x=7\\ \Rightarrow S=\left\{7\right\}\)

b)\(2\left(x+3\right)< 4x-\left(2+x\right)\)

\(\Leftrightarrow2x+6< 4x-2-x\\ \Leftrightarrow-x< -8\\ \Leftrightarrow x>8\)

\(S=\left\{x\in R|x>8\right\}\)

c)\(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)

\(\Leftrightarrow6x^2-x-2=10x^2-11x-8\\ \Leftrightarrow4x^2-10x-6=0\\ \Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow S=\left\{-\dfrac{1}{2};3\right\}\)

d)\(\dfrac{x+6}{5}-\dfrac{x-2}{3}< 2\)

\(\Leftrightarrow\dfrac{3x+18-5x+10-30}{15}< 0\\ \Leftrightarrow\dfrac{-2x-2}{15}< 0\\ \Leftrightarrow-2x-2< 0\left(15>0\right)\\ \Leftrightarrow x>-1\)

Vậy \(S=\left\{x\in R|x>-1\right\}\)

e)\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\left(x\ne2;x\ne-2\right)\)

\(\Leftrightarrow\dfrac{\left(x-2\right)^2-3\left(x+2\right)-2\left(x-11\right)}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x-5\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\\ \Leftrightarrow S=\left\{4;5\right\}\)

Tick hộ nha 😘

\(-9x^4+3x^2+2\\ =-9x^4+6x^2-3x^2+2\\ =-3x^2\left(3x^2-2\right)-\left(3x^2-2\right)\\ =-\left(3x^2-2\right)\left(3x^2+1\right)\)

\(\left(xy+1\right)^2-\left(x+y\right)^2\\ =\left(xy+1-x-y\right)\left(xy+1+x+y\right)\\ \)

\(5a^2-14ab-3b^2\\ =5a^2-15ab-ab-3b^2\\ =5a\left(a-3b\right)-b\left(a-3b\right)\\ =\left(5a-b\right)\left(a-3b\right)\)

a) Khi \(x=-1\)

\(A=\dfrac{2\left(-1\right)-1}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}\)

b)

\(B=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\\ B=\dfrac{3x+15+x-3-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{2x+6}{\left(x+3\right)\left(x-3\right)}\\ B=\dfrac{2}{x-3}\)

c)\(P=\dfrac{B}{A}=\dfrac{2}{x-3}:\dfrac{2x-1}{x-3}=\dfrac{2}{2x-1}\)

Để \(P< 0\)

\(\Leftrightarrow\dfrac{2}{2x-1}< 0\\ \Leftrightarrow2x-1< 0\left(2>0\right)\\ \Leftrightarrow x< \dfrac{1}{2}\)

Ta có BM//CN nên BMNC là hình thang

Xét tam giác ABC cân tại A có AH là đường cao nên cũng là trung tuyến

Xét hình thang BMNC có BH=HC và AH//BM//CN nên AM=AN (đường trung bình của hình thang BMNC)

\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ \Rightarrow A=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\\ \Rightarrow A=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{16}{2}+\dfrac{17}{2}\\ \Rightarrow A=\dfrac{1}{2}\left(2+3+4+...+17\right)=76\)

Na đâu ra vậy?

\(KHCO_3\) thể hiện tính bazơ: \(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)

\(KHCO_3\) thể hiện tính axit: \(KHCO_3+KOH\rightarrow K_2CO_3+H_2O\)