\(\left(\frac{4}{5}:x\right)=\frac{-1}{125}\)
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Bài 1:
\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)
=\(\frac{67}{4}\)
\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)
=\(\frac{3}{7}-\frac{2}{3}\)
=\(-\frac{5}{21}\)
\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)
=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)
=\(\frac{5}{16}:\frac{7}{30}+1\)
=\(\frac{131}{56}\)
\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)
=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)
=\(\frac{3}{7}-\frac{13}{33}\)
=\(\frac{8}{231}\)
Bài đ làm giống hệt như bài c
Bài 2 :
\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)
Vậy x ∈{1;\(\frac{1}{3}\)}
\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)
=>\(\frac{19}{15}.x=\frac{19}{10}\)
=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)
Vậy x ∈ {\(\frac{3}{2}\)}
c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)
=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)
Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}
\(d,x-30\%.x=-1\frac{1}{5}\)
=\(70\%x=-\frac{6}{5}\)
=\(\frac{7}{10}.x=-\frac{6}{5}\)
=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)
Vậy x∈{\(-\frac{12}{7}\)}
Bài 2
a/
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)
b/ Đặt x làm thừa số chung rồi tính như bình thường
c/ Tương tự câu a
d/ Tương tự câu b
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
b) ( 2/5 )^ x-1 : ( 2/5 )^2 = ( 2/5 )^3
(2/5)^ x-1 = ( 2/5) ^3 . (2/5)^2
(2/5)^ x-1 = ( 2/5 ) ^ 6
=> x-1 = 6
=> x= 7
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)
\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)
\(=\frac{1}{2009}\)
Ta có : 3x + 3x + 2 = 810
=> 3x(1 + 32) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
ta có \(3^3+3^x+2=810\)
=>\(3^x\left(1+3^2\right)=810\)
=>\(3^x.10=810\)
=>\(3^x=81\)
=>\(3^x=3^4\)
=>x=4
Vậy x=4
x = -100