\(a=3+3^2+3^3+...+3^{2008}\)

\(3a=3^2+3^3+3^4+...+3^{2009}\)

\(3a-a=\left(3^2+3^3+3^4+...+3^{2009}\right)-\left(3+3^2+3^3+...+3^{2008}\right)\)

\(2a=3^{2009}-3\)

\(2a+3=3^x=3^{2009}-3+3=3^{2009}\)

\(x=2009\)

\(\dfrac{3x}{5}=\dfrac{2y}{3}\Leftrightarrow\dfrac{3x}{5}.\dfrac{1}{6}=\dfrac{2y}{3}.\dfrac{1}{6}\Leftrightarrow\dfrac{3x}{30}=\dfrac{2y}{18}\Leftrightarrow\dfrac{x}{10}=\dfrac{y}{9}\Leftrightarrow\dfrac{x^2}{100}=\dfrac{y^2}{81}\)Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{100}=\dfrac{y^2}{81}=\dfrac{x^2-y^2}{100-81}=\dfrac{38}{19}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=2.100=200\\y^2=2.81=162\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{200}\\y=\pm\sqrt{162}\end{matrix}\right.\)

\(2x^2+3x-1=0\)

\(\Rightarrow2x^2+3x+1=2\)

\(\Rightarrow2x^2+2x+x+1=2\)

\(\Rightarrow2x\left(x+1\right)+1\left(x+1\right)=2\)

\(\Rightarrow\left(2x+1\right)\left(x+1\right)=2\)

Xét ước

Đặt: \(\sqrt{x}=t\left(t\ge0\right)\)

Nên \(\sqrt{8+t}+\sqrt{5-t}=5\)

\(\Rightarrow\sqrt{8+t}+\sqrt{5-t}-5=0\)

\(\Rightarrow\left(\sqrt{8+t}-3\right)+\left(\sqrt{5-t}-2\right)=0\)

\(\Rightarrow\dfrac{t-1}{\sqrt{8+t}+3}+\dfrac{t-1}{\sqrt{5-t}+2}=0\Leftrightarrow\left(t-1\right)\left(\dfrac{1}{\sqrt{8+t}+3}+\dfrac{1}{\sqrt{5-t}+2}\right)=0\)

\(t=1\Leftrightarrow x=1\left(tm\right)\)

\(\left|x-1,5\right|+\left|2,5-x\right|\ge\left|x-1,5+2,5-x\right|=1>0\)

Không có giá trị \(x\) thỏa mãn

\(\dfrac{x+1}{9}+\dfrac{x+1}{8}=\dfrac{x+1}{11}+\dfrac{x+1}{12}\)

\(\Rightarrow\dfrac{x+1}{9}+\dfrac{x+1}{8}-\dfrac{x+1}{11}-\dfrac{x+1}{12}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{11}-\dfrac{1}{12}\right)=0\)

Vì \(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{11}-\dfrac{1}{12}\ne0\Leftrightarrow x=-1\)

\(a=\sqrt{27}-3\sqrt{48}+2\sqrt{108}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(a=3\sqrt{3}-3\sqrt{48}+\sqrt{216}-2+\sqrt{3}\)

\(a=3\sqrt{3}-3\sqrt{48}+3\sqrt{24}-2+\sqrt{3}\)

\(a=3\left(\sqrt{3}-\sqrt{48}+\sqrt{24}+1\right)-2\)

Tính cái trong ngoặc là \(ok\).Em lười đi lấy máy tính lắm

\(b=3\sqrt{2}\left(\sqrt{50}-2\sqrt{18}+\sqrt{98}\right)\)

\(b=3\sqrt{100}-3\sqrt{72}+3\sqrt{196}\)

\(b=3\left(\sqrt{100}-\sqrt{72}+\sqrt{196}\right)\)(Tính trong ngoặc)

\(\sqrt{\left(3-\sqrt{10}\right)^2}=3-\sqrt{10}\)

\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=4-\sqrt{5}+1-\sqrt{5}=5-2\sqrt{5}\)
\(\sqrt{\left(2-\sqrt{7}\right)^2}=2-\sqrt{7}\)

\(\sqrt{\left(4-\sqrt{13}\right)^2}+\sqrt{\left(2-\sqrt{13}\right)^2}=4-\sqrt{13}+2-\sqrt{13}=6-2\sqrt{13}\)

Đặt:

\(\dfrac{a}{10}=\dfrac{b}{8}=\dfrac{c}{6}=t\Leftrightarrow\left\{{}\begin{matrix}a=10t\\b=8t\\c=6t\end{matrix}\right.\)

\(M=\dfrac{a+2b-3c}{a-2b+3c}=\dfrac{10t+16t-18t}{10t-16t+18t}=\dfrac{8t}{12t}=\dfrac{8}{12}=\dfrac{2}{3}\)

\(A=\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\)

\(A^2=\left(\sqrt{5+\sqrt{21}}^2+2\sqrt{\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}+\sqrt{5-\sqrt{21}}^2\right)\)

\(A^2=5+\sqrt{21}+\sqrt{4\left(5+\sqrt{21}\right)\left(5-\sqrt{21}\right)}+5-\sqrt{21}\)

\(A^2=10+\sqrt{4.\left(25-21\right)}\)

\(A^2=10+\sqrt{4.4}=10+4=14\)

\(A=\sqrt{14}\)

Tương tự