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Bài làm
= \(\frac{18.25+9.45.2+3.27.6}{100-99+98-97+96-95+.....+2-1}\)
= \(\frac{18.25+9.2.45+3.6.27}{50}\)
= \(\frac{18.25+18.45+18.27}{50}\)
= \(\frac{18.\left(25+45+27\right)}{50}\)
= \(\frac{18.97}{50}\)
= \(\frac{1746}{50}\)
\(\frac{1}{5.8}\)\(+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=3.\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)
\(\Leftrightarrow x+3=103\)
\(\Leftrightarrow x\)\(=103-3\)
\(\Leftrightarrow x\)\(=100\)
Vậy x = 100
~~~~~~~Hok tốt~~~~~~~~
ta có \(\frac{1}{5.8}+\frac{1}{8.11}+...\frac{1}{x.\left(x+3\right)}\)\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)\)\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}=\frac{1}{103}\)
\(\Rightarrow x+3=103\)
\(\Rightarrow x=100\)
nhớ k nha
Mong mọi người giải được. Chúc các bạn may mắn!
Câu 1: Viết vào chỗ chấm:
a) 98- 42 = 140 - 84
b) 1 x 2 x 3=1+2.+.3
c) 0x0x0+0+0
d) 100-50=50+50-100+50
Câu 2: Tìm x:
Đề bài: 5x + 4x x 2 = 50+50+20+20 + 2 + 2
5x+4x x 2=144
5x+4x=144/2
5x+4x=72
x(5+4)=72
9x=72
=>x=72/9=8
Câu 3: Viết biểu thức sau thành tích của 2 thừa số:
Đề bài: 12+21+13+13+14+41+15+51+16+61=257 x 1
Câu 4: Tính giá trị của a:
Đề bài: a x 2 -20 + a x 6 = a x 3 + a
a x (-18)+a x 6 =a x (3+1)
a x (-18+6)=a x 4
a x (-12)=a x 4
=>a=0
Lời giải :
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
ko chép lại đề :
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ......... + \(\frac{1}{98}\)- \(\frac{1}{99}\)+ \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
\(\frac{2}{5}×\frac{3}{4}+\frac{6}{15}:\frac{4}{9}×5\)
\(=\frac{3}{10}+\frac{6}{15}×\frac{9}{4}×5\)
\(=\frac{3}{10}+\frac{9}{10}×5\)
\(=\frac{3}{10}+\frac{9}{2}\)
\(=\frac{3}{10}+\frac{45}{10}\)
\(=\frac{3+45}{10}\)
\(=\frac{48}{10}=\frac{24}{5}=4,8\)
\(\frac{3}{10}+\frac{9}{10}\cdot5=\frac{3}{10}+\frac{45}{10}=\frac{48}{10}=\frac{24}{5}\)
khong can ghi de bai nha
bài 1
a,
32 + 68 :17 x 5 - 29
= 32 + 20 -29
= 52 - 29
= 23
b,
15 x 48 - 30 x 24 - 125
= 720 - 720 -125
= 0-125
a,
32 + 68 :17 x 5 - 29
= 32 + 20 -29
= 52 - 29
= 23
b,
15 x 48 - 30 x 24 - 125
= 720 - 720 -125
= 0-125
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{101}-\frac{1}{103}\)
\(A=\frac{1}{3}-\frac{1}{103}\)
\(A=\frac{100}{309}\)
\(A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}+\frac{2}{101\times103}\)
\(A=1\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}+\frac{1}{101}-\frac{1}{103}\right)\)
\(A=1\times\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(A=1\times\frac{100}{309}\)
\(A=\frac{100}{309}\)
Kết quả: 55%