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Trừ cả hai vế đi 7,8x
25 = 0,2x
x = 25 : 0,2
x = 125
8x = 7,8x + 25
=> 8x - 7,8x = 25
=> 0,2x = 25
=> x = 25 : 0,2
=> x = 125
\(\frac{25^x}{5^{x+y}}=125\)
\(\frac{5^2^x}{5^{x+y}}=125\)
\(5^{2x-x-y}=125\)
\(5^{x-y}=125\)
\(5^{x-y}=5^3\)
\(x-y=3\)
\(x=3+y\)
a) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\Rightarrow\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{3}\)
\(\Rightarrow x=1\)
\(x-25\%=\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
c) \(-\frac{5}{6}+\frac{8}{3}+-\frac{29}{6}\le x\le-\frac{1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
Ta có: \(\left(x-3\right)^2=25\)
\(\Leftrightarrow\left(x-3\right)^2=5^2\left(x\inℕ^∗\right)\)
\(\Leftrightarrow x-3=5\)
\(\Leftrightarrow x=5+3\)
\(\Leftrightarrow x=8\)
hok tốt!!
mik làm nhắng ko bt đúng ko
\(2^{x+1}-2^x=32\)
\(2^x.2^1-2^x.1=32\)
\(2^x\left(2^1-1\right)=32\)
\(2^x.1=32\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
\(\frac{-15}{25}\)= \(\frac{-3}{5}\)
\(\frac{a}{b}\)= \(\frac{-3}{5}\)Suy ra 5a= -3b (*)
\(b-a=32\)
\(a=b-32\)(**)
\(5.\left(b-32\right)\)=\(-3b\)
\(5b-160=-3b\)
\(5b+3b=160\)
\(b=160:8=20\)
\(\frac{2}{2}.\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{2}{x+\left(x+1\right)}\right)=\frac{12}{25}\)
\(2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{12}{25}\)
\(2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{12}{25}\)
\(2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{12}{25}\)
\(2.\left(\frac{1}{4}-\frac{1}{x+1}\right)=\frac{12}{25}\)
\(\frac{1}{2}-\frac{2}{x+1}=\frac{12}{25}\)
\(\frac{2}{x+1}=\frac{1}{2}-\frac{12}{25}\)
\(\frac{2}{x+1}=\frac{1}{50}\)
(x+1).1=50.2
x+1=100
x=100-1
x=99
Lời giải:
$8x=7,8x+25$
$8x-7,8x=25$
$0,2x=25$
$x=25:0,2=125$