1+1=??????? Em
K
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6 tháng 8 2021
3x4 + x2 - 4 = 0
Đặt t = x2 ( a ≥ 0 ) pt đã cho trở thành 3t2 + t - 4 = 0
Dễ thấy pt trên có a + b + c = 0 nên có hai nghiệm t1 = 1 (nhận) ; t2 = c/a = -4/3 (loại)
=> x2 = 1 <=> x = ±1
F
22 tháng 1 2020
Bài giải
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b, \(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
M
22 tháng 1 2020
Ta có :
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b,
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
16 tháng 10 2020
a) \(1+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(=\frac{16}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\)
\(=\frac{23}{16}\)
b) \(2-\frac{1}{8}-\frac{1}{12}-\frac{1}{16}\)
\(=\frac{96}{48}-\frac{6}{48}-\frac{4}{48}-\frac{3}{48}\)
\(=\frac{83}{48}\)
c) \(\frac{4}{99}\cdot\frac{18}{5}\div\frac{12}{11}+\frac{3}{5}\)
\(=\frac{4\cdot18\cdot11}{99\cdot5\cdot12}+\frac{3}{5}\)
\(=\frac{4\cdot9\cdot2\cdot11}{9\cdot11\cdot5\cdot4\cdot3}+\frac{3\cdot3}{3\cdot5}\)
\(=\frac{2}{15}+\frac{9}{15}=\frac{11}{15}\)
d) \(\left(1-\frac{3}{4}\right)\left(1+\frac{1}{3}\right)\div\left(1-\frac{1}{3}\right)\)
\(=\frac{1}{4}\cdot\frac{4}{3}\div\frac{2}{3}\)
\(=\frac{1\cdot4\cdot3}{4\cdot3\cdot2}=\frac{1}{2}\)
MK
0
14 tháng 7 2019
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Ta có: \(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right).x=\frac{3}{4}\)
\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right).x=2.\frac{3}{4}\)
\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right).x=\frac{3}{2}\)
\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right).x=\frac{3}{2}\)
\(\left(1-\frac{1}{101}\right).x=\frac{3}{2}\)
\(\frac{100}{101}.x=\frac{3}{2}\)
\(x=\frac{3}{2}:\frac{100}{101}\)
\(x=\frac{303}{200}\)
1 + 1 = 2 hok tốt
2 nhé
^_^