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e) \(5^{x+2}=625\\ \Leftrightarrow5^{x+2}=5^4\\ \Leftrightarrow x+2=4\\ \Leftrightarrow x=2\)
Vậy...
g) \(\left(2x-1\right)^3=-8\\ \Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\\ \Leftrightarrow2x-1=-2\\ \Leftrightarrow x=-\frac{1}{2}\)
VẬy...
Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
x+(-31/12)^2=(49/12)^2-x
x+x=(49/12)^2-(-31/12)^2
tính x
từ x tìm ra y
b)x(x-y):[y(x-y)]=3/10:(-3/50)=...
=>x/y=... =>x=...;y=...
\(\frac{7}{8}.(\frac{2}{12}+\frac{4}{10})\)
\(\Rightarrow\frac{7}{8}.(\frac{10+24}{60})\)
\(\Rightarrow\frac{7}{8}.\frac{34}{60}=\frac{238}{480}\)
bt2
\(2.x-\frac{5}{4}=\frac{20}{15}\)
\(\Leftrightarrow2x=\frac{20}{15}+\frac{5}{4}\)
\(\Leftrightarrow2x=\frac{80+75}{60}\)
\(\Leftrightarrow2x=2,5\)
\(\Leftrightarrow x=1,25\)
.7/8.(1/6+2/5)=7/8.17/30=119/240
3/2-5/6:1/4+\(\sqrt{4}\)=3/2-10/3+2=1/6
2x=20/15+5/4
2x=31/12
x=31/12:2
x=31/24
ko bt nha thông cảm
a) \(\frac{5}{6}\)x - \(\frac{3}{8}\)x - 10 = 12
=> \(\left(\frac{5}{6}-\frac{3}{8}\right)\)x = 12 + 10
=> \(\frac{11}{24}\)x = 22
=> x = 22 : \(\frac{11}{24}\)
=> x = 48
Vậy x = 48.
b) (\(\left|x\right|\) - \(\frac{1}{8}\)) . \(\left(-\frac{1}{8}\right)^5\) = \(\left(-\frac{1}{8}\right)^7\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^7\) : \(\left(-\frac{1}{8}\right)^5\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^{7-5}\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\frac{1}{64}\)
=> \(\left|x\right|\) = \(\frac{1}{64}\) + \(\frac{1}{8}\)
=> \(\left|x\right|\) = \(\frac{9}{64}\)
=> x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)
Vậy x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)
hên quá làm đúng hì hì, cảm ơn nhen, hết sợ bị sai ồi
\(x\cdot3\frac{1}{10}=\left(x+8\right)\cdot\frac{31}{12}\)
\(x\cdot\frac{31}{10}=x\cdot\frac{31}{12}+8\cdot\frac{31}{12}\)
\(x\cdot\frac{31}{10}-8\cdot\frac{31}{12}=x\cdot\frac{31}{12}\)
\(\frac{62}{3}=x\cdot\frac{31}{10}-x\cdot\frac{31}{12}=x\left(\frac{31}{10}-\frac{31}{12}\right)\)
\(\frac{62}{3}=x\cdot\frac{161}{60}\)
\(x=\frac{62}{3}:\frac{161}{60}=\frac{62\cdot60}{3\cdot161}\)
Phần còn lại bạn tính giúp mình nha!