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25 tháng 10 2016

x = -100

8 tháng 6 2019

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∈{\(-\frac{12}{7}\)}

8 tháng 6 2019

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

19 tháng 3 2020

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) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)

\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{4}{15}\)

21 tháng 6 2017

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

11 tháng 8 2019

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)

\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1}{2009}\)

11 tháng 8 2019

1,

\(| x - \frac{2}{7} | = \frac{-1}{5}.\frac{-5}{7}\)

\(|x- \frac{2}{7}|=\frac{1}{7}\)

<=> \(x- \frac{2}{7} = \frac{1}{7} => x= \frac{3}{7} \)

Và \(x - \frac{2}{7} =\frac{-1}{7} => x= \frac{1}{7}\)

Học tốt

9 tháng 8 2017

Ta có : 3x + 3x + 2 = 810

=> 3x(1 + 32) = 810

=> 3x.10 = 810

=> 3x = 81

=> 3x = 34

=> x = 4

12 tháng 8 2017

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