Bài 2
\(a,\)\(\left(x^2+7\right)\left(x^2-49\right)< 0\)
Vì \(x^2+7>0\)\(\Rightarrow x^2-49< 0\)
\(\Rightarrow\left(x-7\right)\left(x+7\right)< 0\)
\(...\)
Bài 2:
a) \(\left(x^2+7\right).\left(x^2-49\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x^2+7< 0\\x^2-49>0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+7>0\\x^2-49< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2< -7\\x^2>49\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}x^2>-7\\x^2< 49\end{cases}}\)
\(\Leftrightarrow-7< x^2< 49\)
Mà \(x^2\ge0\)và \(x^2\)là 1 SCP
\(\Rightarrow x^2\in\left\{1;4;9;16;25;36\right\}\)
\(\Rightarrow x\in\left\{1;2;3;4;5;6\right\}\)
Vậy \(x\in\left\{1;2;3;4;5;6\right\}\)
Bài 1:
\(S=\frac{1+3+3^2+...+3^{2000}}{1-3^{2001}}\)
Đặt \(A=1+3+3^2+...+3^{2000}\)
\(\Rightarrow3A=3+3^2+...+3^{2001}\)
\(\Rightarrow3A-A=\left(3+3^2+...+3^{2001}\right)-\left(1+3+3^2+...+3^{2000}\right)\)
\(\Rightarrow2A=3^{2001}-1\)
\(\Rightarrow A=\frac{3^{2001}-1}{2}\)
\(\Rightarrow A=3^{2001}.2-2\)
\(\Rightarrow S=\frac{2.\left(3^{2001}-1\right)}{1-3^{2001}}\)
Bài 2;
c) \(xy-2x+3y=0\)
\(\Rightarrow x.\left(y-2\right)+3y=0\)
\(\Rightarrow x.\left(y-2\right)+3y-6=0-6\)
\(\Rightarrow x.\left(y-2\right)+3.\left(y-2\right)=-6\)
\(\Rightarrow\left(x+3\right).\left(y-2\right)=-6\)
Lập bảng tìm tiếp
Bài 2
d) \(\frac{1}{x}+\frac{y}{2}=\frac{5}{8}\)
\(\Leftrightarrow\frac{1}{x}=\frac{5}{8}-\frac{y}{2}\)
\(\Leftrightarrow\frac{1}{x}=\frac{5}{8}-\frac{4y}{8}\)
\(\Leftrightarrow\frac{1}{x}=\frac{5-4y}{8}\)
\(\Leftrightarrow x.\left(5-4y\right)=8\)
Lập bảng tìm tiếp
\(S=\frac{1+3+\cdot\cdot\cdot+3^{2000}}{1-3^{2001}}\)
\(A=1+3+3^2+\cdot\cdot\cdot+3^{2000}\)
\(3A=3+3^2+3^3+\cdot\cdot\cdot+3^{2001}\)
\(\Rightarrow3A-A=2A=\left(3+\cdot\cdot\cdot+3^{2001}\right)-(1+\cdot\cdot\cdot+3^{2000})\)
\(\Rightarrow2A=3^{2001}-1\)
\(2S=\frac{2A}{1-3^{2001}}=\frac{3^{2001}-1}{1-3^{2001}}\)
\(-2S=\frac{-\left(3^{2001}-1\right)}{1-3^{2001}}=1\)
\(S=1:\left(-2\right)=\frac{-1}{2}\)
Bài 1 làm theo bạn Nguyễn Tuấn Thảo nhé bạn NGuyễn anh đức
