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Ta có \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}\)
\(=\frac{5}{14}\)
Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{1}{6}-\frac{1}{22}\)
\(=\frac{4}{33}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(B=\frac{1}{2}-\frac{1}{7}\)
\(B=\frac{5}{14}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)
Lời giải:
a)
$\frac{2}{12}+\frac{7}{-42}=\frac{1}{6}+\frac{-1}{6}=0$
b)
$\frac{24}{40}+\frac{-14}{35}=\frac{3}{5}+\frac{-2}{5}=\frac{1}{5}$
c)
$\frac{-14}{49}+\frac{12}{30}=\frac{-2}{7}+\frac{2}{5}=\frac{4}{35}$
d)
$\frac{6}{-21}+\frac{-9}{36}=\frac{-2}{7}+\frac{-1}{4}=\frac{-15}{28}$
a) \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow x^2\left(-1\right)=-36\)
\(\Rightarrow x^2=6^2\)
\(\Rightarrow x=\pm6\)
Vậy \(x=\left[\begin{matrix}6\\-6\end{matrix}\right.\)
b) \(\frac{12}{x+1}=\frac{5}{30}\)
\(\Rightarrow\left(x+1\right)5=12.30\)
\(\Rightarrow5x+5=360\)
\(\Rightarrow5x=355\)
\(\Rightarrow x=355:5\)
\(\Rightarrow x=71\)
Vậy \(x=71.\)
c) \(\frac{x+42}{5}=\frac{5x+2}{20}\)
\(\Rightarrow\left(x+42\right)20=5\left(5x+2\right)\)
\(\Rightarrow20x+840=25x+10\)
\(\Rightarrow20x-25x=-840+10\)
\(\Rightarrow-5x=-830\)
\(\Rightarrow x=166\)
Vậy \(x=166.\)
a, \(\frac{-x}{4}=\frac{-9}{x}\Rightarrow-x.x=-9.4\)
\(\Rightarrow-x.x=-36=-6.6\Rightarrow x=6\)
Vậy x = 6
b, \(\frac{12}{x+1}=\frac{5}{30}=\frac{1}{6}\Rightarrow x+1=12.6\)
\(\Rightarrow x+1=72\Rightarrow x=72-1=71\)
Vậy x = 71
c, \(\frac{x+42}{5}=\frac{5x+2}{20}\Rightarrow20\left(x+42\right)=5\left(5x+2\right)\)
\(\Rightarrow20x+840=25x+10\Rightarrow20x-25x=-840+10\)
\(\Rightarrow\left(20-25\right)x=-830\Rightarrow-5x=-830\)
\(\Rightarrow x=-830\div\left(-5\right)\Rightarrow x=166\)
Vậy x = 166
Bài1
a) 25/42 - 20/63 =5/18
b) 9/50 - 13/75 - 1/6 = -4/25
c) 2/15 - 2/65 - 4/39 = 0
Bài2
a) x + 7/12 =17/18-1/9 b) 29/30 - (18/23 + x)=7/69
x + 7/12 = 5/6 18/23 + x =29/30 - 7/69
x =5/6 - 7/12 18/23 +x = 199/230
x = 1/4 x = 199/230 - 18/23
x= 19/230
b) Đặt B = A : C ta có:
\(A=\frac{5^3}{6}+\frac{5^3}{12}+\frac{5^3}{20}+\frac{5^3}{42}+\frac{5^3}{56}+\frac{5^3}{72}+\frac{5^3}{90}\)
\(A=5^3.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=5^3.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=5^3.\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{5^3.2}{5}\)
\(A=5^2.2\)
\(\Rightarrow A=50\)
\(C=\frac{1124.2247-1123}{1124+1123.2247}\)
\(C=\frac{\left(1123+1\right).2274-1123}{1123.2247+1124}\)
\(C=\frac{1123.2247-2247-1123}{1123.2247+1124}\)
\(C=\frac{1123.2247+1124}{1123.2247+1124}=1\)
\(\Rightarrow B=50:1=50\)
Vậy B = 50
Bài 1 mik học xong quên hết òi (mấy bài kia là hok biết luôn :V)