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1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
a,\(\left(x-\frac{7}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)
\(x-\frac{7}{9}=\frac{4}{9}\)
\(x=\frac{4}{9}+\frac{7}{9}\)
\(x=\frac{11}{9}\)
Vậy x=\(\frac{11}{9}\)
cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11)
TH1: x-7=0 => x=7 => 0^8=0^18 (TM)
TH2: x-7=1 => x=8 (TM)
TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại
KL: x = 7 hoặc x=8
( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
tới đây dễ òi
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
a,(x-7)x+1-(x-7)x+11=0
=>(x-7)x+1.[1-(x-7)10]=0
=>(x-7)x+1=0
=>x-7=0
=>x=7
hoặc 1-(x-7)10=1
=>(x-7)10=1
=>x-7=-1;1
=>x=8;6
vậy x=6;7;8
b,(x-1)2=36/49
=>x-1=6/7;-6/7
=>x=13/7;1/7
vậy x=1/7;13/7
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
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) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)