Tìm biểu thức ngắn gọn hơn cho tích sau đây:
Pn=\(\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2n-1\right)^2}\right)\)
Biết rằng nó đúng với mọi n>=1 và CM bằng phương pháp quy nạp toán học
Tìm biểu thức ngặn gọn hơn cho tích sau đây:
\(P_n=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)......\left(1-\frac{4}{\left(2n-1\right)^2}\right)\)
Biết rằng nó đúng với n>=1 và chứng minh bằng phương pháp quy nạp toán học
23 chuyên đề giải 1001 bài toán sơ cấp ,mk nhớ có trog quyển này
Tìm biểu thức ngăn hơn cho biểu thức sau:
\(P_{_{ }n}=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2n-1\right)^2}\right)\)
Biết rằng nó đúng với n>=1 và chúng chứng minh bằng phương pháp quy nạp toán học.
Ta sẽ chứng minh với \(n\ge1\)thì \(P_n=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2n-1\right)^2}\right)=\frac{-2n-1}{2n-1}\)
Với \(n=1\)mệnh đề đúng vì \(1-4=-3=\frac{-2.1-1}{2.1-1}\)
Giả sử mệnh đề đúng với \(n=k\)tức là \(\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2k-1\right)^2}\right)=\frac{-2k-1}{2k-1}\)
Ta sẽ chứng minh mệnh đề đúng với \(n=k+1\)tức là chứng minh \(\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2k+1\right)^2}\right)=\frac{-\left(2k+3\right)}{2k+1}\)
Thật vậy \(\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)...\left(1-\frac{4}{\left(2k-1\right)^2}\right)\left(1-\frac{4}{\left(2k+1\right)^2}\right)=\frac{-2k-1}{2k-1}.\frac{\left(2k-1\right)\left(2k+3\right)}{\left(2k+1\right)^2}\)
\(=\frac{-\left(2k+1\right)}{2k-1}.\frac{\left(2k-1\right)\left(2k+3\right)}{\left(2k+1\right)^2}=\frac{-\left(2k+3\right)}{2k+1}.\)
Theo nguyên lý quy nạp, mệnh đề đúng với mọi \(n\ge1\)
\(\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right)....\left(1-\frac{4}{\left(2n-1\right)^2}\right)\)Với n>=1 (Rút gọn)
\(A=\left(\frac{1^2-2^2}{1^2}\right)\left(\frac{3^2-2^2}{3^2}\right)\left(\frac{5^2-2^2}{5^2}\right)...\left(\frac{\left(2n-1\right)^2-2^2}{\left(2n-1\right)^2}\right)\)
\(=\frac{-1\cdot3}{1^2}\cdot\frac{1\cdot5}{3^2}\cdot\frac{3\cdot7}{5^2}...\cdot\frac{\left(2n-3\right)\left(2n+1\right)}{\left(2n-1\right)^2}=-\frac{1}{1}\cdot\frac{2n+1}{2n-1}=-\frac{2n+1}{2n-1}\)
tìm giá trị của các biểu thức sau
P=\(\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
Q=\(\left(\frac{2}{25}-1,0008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(P=\left(\dfrac{-1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{6}:2\)
\(=\left(\dfrac{1}{2}+\dfrac{3}{5}\right):3+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=\dfrac{11}{10}\cdot\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{22}{60}+\dfrac{15}{60}=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-\dfrac{126}{125}\right)\cdot\dfrac{7}{4}:\left[\dfrac{-119}{36}\cdot\dfrac{36}{17}\right]\)
\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left(-7\right)\)
\(=\dfrac{116}{125}\cdot\dfrac{7}{4}\cdot\dfrac{1}{7}=\dfrac{29}{125}\)
Tính giá trị biểu thức(giút gọn biểu thức)
A=\(\left(\left(\frac{2}{193}-\frac{3}{386}\right)\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{2001}+\frac{11}{4002}\right)\cdot\frac{2001}{25}+\frac{9}{2}\right)\)
\(B=\left(1+2+3+4+.....+100\right)\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)\cdot\left(\frac{6}{3}\cdot12-2,1\cdot3,6\right)\)
C=\(\frac{2\cdot8^4\cdot27^2+4\cdot69}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(F=1-\frac{1}{1+\frac{2}{1-\frac{3}{1-4}}}\)
ai làm đúng nhanh dễ hiểu thì mk tick cho
4) Tính giá trị biểu thức sau:
P = \(\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
Q = \(\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(P=\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(P=\left(-1,1\right):\left(-3\right)+\frac{1}{3}+\frac{1}{6}:\left(-2\right)\)
\(P=\frac{11}{30}+\frac{1}{3}+\left(-\frac{1}{12}\right)\)
\(P=\frac{37}{60}\)
\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right).2\frac{2}{17}\right]\)
\(Q=\left(-0,928\right):\frac{4}{7}:\left[\left(-\frac{119}{36}\right).2\frac{2}{17}\right]\)
\(Q=\left(-1,624\right):\left(-\frac{245}{36}\right)\)
\(Q=\frac{1044}{4375}\)
Thu gọn biểu thức sau :
a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
Rút gọn biểu thức :
a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{16}\right)...\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(10+1\right).\left(10^2+1\right)\left(10^3+1\right)...\left(10^{2n}+1\right)\)
Câu hỏi: Rút gọn biểu thức A = \(\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(\left(2k\right)^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(\left(2k-1\right)^4+\frac{1}{4}\right)}\) (k thuộc N*)