a, 223 - 6 . ( x + 5 ) =49
b, 4x3 + 12 =120
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a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
a) \(0,6+\dfrac{2}{3}=\dfrac{6}{10}+\dfrac{2}{3}=\dfrac{3}{5}+\dfrac{2}{3}=\dfrac{9}{15}+\dfrac{10}{15}=\dfrac{19}{15}\)
b) \(-\dfrac{5}{12}+0,75=-\dfrac{5}{12}+\dfrac{75}{100}=-\dfrac{5}{12}+\dfrac{3}{4}=-\dfrac{5}{12}+\dfrac{9}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
c) \(\dfrac{1}{3}-\left(-0,4\right)=\dfrac{1}{3}+\dfrac{4}{10}=\dfrac{1}{3}+\dfrac{2}{5}=\dfrac{5}{15}+\dfrac{6}{15}=\dfrac{11}{15}\)
d) \(1\dfrac{3}{5}+\dfrac{5}{6}=\dfrac{8}{5}+\dfrac{5}{6}=\dfrac{48}{40}+\dfrac{25}{30}=\dfrac{73}{30}\)
\(Y:\dfrac{7}{5}=\dfrac{5}{6}+\dfrac{5}{12}=\dfrac{15}{12}\\ Y=\dfrac{15}{12}\times\dfrac{7}{5}=\dfrac{7}{4}\)
\(y:\dfrac{7}{5}=\dfrac{5}{6}+\dfrac{5}{12}\\ \Leftrightarrow y:\dfrac{7}{5}=\dfrac{5\times2}{6\times2}+\dfrac{5}{12}\\ \Leftrightarrow y:\dfrac{7}{5}=\dfrac{10}{12}+\dfrac{5}{12}\\ \Leftrightarrow y:\dfrac{7}{5}=\dfrac{15}{12}\\ \Leftrightarrow y:\dfrac{7}{5}=\dfrac{5}{4}\\ \Leftrightarrow y=\dfrac{5}{4}\times\dfrac{7}{5}\\ \Leftrightarrow y=\dfrac{7}{4}\)
\(\Rightarrow C\)
a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
a: =>1/3*4+1/4*5+...+1/x(x+1)=10/39
=>1/3-1/4+...+1/x-1/x+1=10/39
=>1/3-1/(x+1)=10/39
=>1/(x+1)=13/39-10/39=3/39=1/13
=>x+1=13
=>x=12
b: =>15x=150
=>x=10
c: =>120-5x=45
=>5x=75
=>x=15
a) \(\dfrac{3}{8}+\dfrac{15}{-25}+\dfrac{3}{5}\)
\(=\dfrac{-9}{40}+\dfrac{3}{5}\)
\(=\dfrac{3}{8}\)
b) \(\dfrac{-5}{18}+\dfrac{23}{45}-\dfrac{9}{10}\)
\(=\dfrac{7}{30}-\dfrac{9}{10}\)
\(=\dfrac{-2}{3}\)
c) \(\dfrac{-5}{12}+\dfrac{15}{18}-2,25\)
\(=\dfrac{5}{12}-2,25\)
\(=\dfrac{-11}{6}\)
d) \(\dfrac{5}{6}+\dfrac{2}{3}-0,5\)
\(=\dfrac{3}{2}-0,5\)
\(=1\)
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
\(a,223-6\left(x+5\right)=49\\ 6\left(x+5\right)=223-49\\ 6\left(x+5\right)=174\\ x+5=174:6\\ x+5=29\\ x=29-5\\ x=24\\ ----\\ b,4x^3+12=120\\ 4x^3=120-12\\ 4x^3=108\\ x^3=108:4\\ x^3=27\\ Mà:3^3=27\\ Nên:x^3=3^3\\ Vậy:x=3\)
a. 223 - 6. (x + 5) = 49
6. (x + 5) = 223 - 49
6. (x + 5) = 174
x + 5 = 174 : 6
x + 5 = 29
x = 29 - 5
x = 24