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\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(x+\frac{103}{50}\right)\right]\cdot2=89\)
\(\left(1-\frac{1}{10}\right)\cdot100-\frac{5}{2}:\left(x+\frac{103}{50}\right)\cdot2=89\)
\(\frac{9}{10}\cdot100-\frac{5}{2}\cdot2:\left(x+\frac{103}{50}\right)=89\)
\(90-5\cdot\left(x+\frac{103}{50}\right)=89\)
\(5\cdot\left(x+\frac{103}{50}\right)=1\)
\(x+\frac{103}{50}=\frac{1}{5}\)
\(x=-\frac{93}{50}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=\dfrac{3}{4}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)=\dfrac{3}{4}\Leftrightarrow\dfrac{9}{10}\left(x-\dfrac{1}{3}\right)=\dfrac{3}{8}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{5}{12}\Leftrightarrow x=\dfrac{5}{12}+\dfrac{1}{3}=\dfrac{9}{12}=\dfrac{3}{4}\)
a)
Số số hạng của dãy trên là;
(n - 1) : 1 + 1 = n(số hạng)
Tổng dãy trên là:
(n + 1) x n : 2 = ? (tùy giá trị n)
b) Đặt A = 1x2 + 2x3 + 3x4 + ... + 99 x 100
3A= 3 x ( 1x2 + 2x3 + 3x4 + ... + 99 x 100)
3A = 1 x 2 x (3 - 0) + 2 x 3 x(4-1) + .....+99.100.(101 - 98)
3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + .......+ 99.100.101
3A = 99.100.101
A = \(\frac{\text{99.100.101}}{3}=333300\)
a, 1 + 2 + 3 + ... + n
= ( 1 + n) × n : 2
b, 1×2 + 2×3 + 3×4 + ... + 99×100
= 1/3 × ( 1×2×3 + 2×3×3 + 3×4×3 + ... + 99×100×3)
= 1/3 × [ 1×2×(3-0) + 2×3×(4-1) + 3×4×(5-2) + ... + 99×100×(101-98) ]
= 1/3 × ( 1×2×3 - 0×1×2 + 2×3×4 - 1×2×3 + 3×4×5 - 2×3×4 + ... + 99×100×101 - 98×99×100 )
= 1/3 × [ ( 1×2×3 + 2×3×4 + 3×4×5 + ... + 99×100×101) - ( 0×1×2 + 1×2×3 + 2×3×4 + ... + 98×99×100) ]
= 1/3 × ( 99×100×101 - 0×1×2)
= 1/3 × ( 99×100×101 - 0)
= 1/3 × 99×100×101
= 333 300
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+....+1/19-1/20
A=1-1/20
A=20/20-1/20
A=19/20
\(a)\) \(2x-5=21\)
\(\Leftrightarrow\) \(2x=21+5\)
\(\Leftrightarrow\) \(2x=26\)
\(\Leftrightarrow\) \(x=26:2\)
\(\Leftrightarrow\) \(=13\)
\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)
\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)
\(\Leftrightarrow\) \(x=\frac{1}{3}\)
Bài 1:
Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)
\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)
\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)
\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(A=\frac{2x19}{20}=\frac{19}{10}\)
Bài 2:
Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)
Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)
\(Bx100=\frac{9}{10}x100=90\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)
=> \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)
bài 1 đáp án là:19/10
2:147/50