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2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt
a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)
\(x=\dfrac{8}{13}-\dfrac{3}{4}\)
\(x=-\dfrac{7}{52}\)
b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
c, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(2x-\dfrac{1}{7}=0\)
\(x-\dfrac{1}{7}=0:2\)
\(x-\dfrac{1}{7}=0\)
\(x=0-\dfrac{1}{7}\)
\(x=\dfrac{1}{7}\)
d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)
\(1:x=\dfrac{2}{5}\)
\(x=1:\dfrac{2}{5}\)
\(x=\dfrac{5}{2}\)
a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)
c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)
\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)
\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)
vậy \(x=0;x=\dfrac{1}{7}\)
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
a/ \(\dfrac{\dfrac{-5}{12}}{\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|}=\dfrac{\dfrac{-4}{9}}{\dfrac{8}{15}}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\left(-\dfrac{5}{12}\right).\left(\dfrac{8}{15}\right)\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|.\left(-\dfrac{4}{9}\right)=\dfrac{-2}{9}\)
\(\Leftrightarrow\left|\dfrac{2}{3}x+\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\\dfrac{2}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=0\\\dfrac{2}{3}x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy ....
\(1,\dfrac{2x+4}{7}=\dfrac{4x-2}{15}=\dfrac{2.\left(2x+4\right)}{2.7}=\dfrac{4x+8}{14}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{2x+4}{7}=\dfrac{4x-2}{15}==\dfrac{4x+8}{14}=\dfrac{\left(4x+8\right)-\left(4x-2\right)}{14-15}=\dfrac{10}{-1}=-10\)
\(\Rightarrow\dfrac{2x+4}{7}=-10\)
\(\Rightarrow2x+4=-10.7=-70\)
\(\Rightarrow2x=-70+4=-66\)
\(\Rightarrow x=-66:2=-33\)
Vậy \(x=-33\)
\(2,\dfrac{2x+3}{5}=\dfrac{7x-3}{15}=\dfrac{7.\left(2x+3\right)}{7.5}=\dfrac{2.\left(7x-3\right)}{2.15}=\dfrac{14x+21}{35}=\dfrac{14x-6}{30}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{2x+3}{5}=\dfrac{14x+21}{35}=\dfrac{14x-6}{30}=\dfrac{\left(14x+21\right)-\left(14x-6\right)}{35-30}=\dfrac{29}{5}\)
\(\Rightarrow\dfrac{2x+3}{5}=\dfrac{29}{5}\)
\(\Rightarrow2x+3=29\)
\(\Rightarrow2x=29-3=26\)
\(\Rightarrow x=26:2=13\)
\(3,\dfrac{11x-2}{7x+5}=\dfrac{11}{8}\)
\(\Rightarrow\dfrac{11x-2}{11}=\dfrac{7x+5}{8}=\dfrac{7.\left(11x-2\right)}{7.11}=\dfrac{11.\left(7x+5\right)}{8.11}=\dfrac{77x-14}{77}=\dfrac{77x+55}{88}=\dfrac{\left(77x+55\right)-\left(77x-14\right)}{88-77}=\dfrac{69}{11}\)
\(\Rightarrow\dfrac{11x-2}{11}=\dfrac{69}{11}\)
\(\Rightarrow11x-2=69\)
\(\Rightarrow11x=69+2=71\)
\(\Rightarrow x=\dfrac{71}{11}\)
`4/5 - [(-5/2) - ( -3/4 + 2/8)]`
`= 4/5 -[(-5/2 ) + 3/4 + 2/8)]`
`= 4/5 + 5/2 + 3/4 + 2/8`
`= 4/5 + 5/2 + 3/4 + 1/4`
`= (4/5 + 5/2) + (3/4 + 1/4)`
`= 33/10 + 1`
`= 43/10 = 4,3`