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\(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}\\ =\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{7}{10}\\ =\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{49}{70}\\ =\dfrac{27}{70}\\ 3,5-\left(-\dfrac{2}{7}\right)\\ =\dfrac{7}{2}+\dfrac{2}{7}\\ =\dfrac{49}{14}+\dfrac{4}{14}\\ =\dfrac{53}{14}\\ \left(-3\right).\left(-\dfrac{7}{12}\right)\\ =\dfrac{21}{12}\\ =\dfrac{7}{4}\)
a) Ta có: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
\(\Leftrightarrow9\cdot5^x=9\cdot5^6\)
\(\Leftrightarrow5^x=5^6\)
hay x=6
b) Ta có: \(2^{2x+1}+4^{x+3}=264\)
\(\Leftrightarrow4^x\cdot2+4^x\cdot64=264\)
\(\Leftrightarrow4^x=4\)
hay x=1
a) \(\left|x\right|=3,5\)
\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
b) \(\left|1-x\right|+0,73=3\)
\(\Rightarrow\left|1-x\right|=3-0,73\)
\(\Rightarrow\left|1-x\right|=2,27\)
\(\Rightarrow\left[{}\begin{matrix}1-x=2,27\\1-x=-2,27\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1,27\\x=3,27\end{matrix}\right.\)
c) \(\dfrac{-5}{8}+x=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{4}{9}-\left(\dfrac{-5}{8}\right)\)
\(\Rightarrow x=\dfrac{77}{72}\)
Bài 2:
a:
1: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
hay \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(a,\frac{x+15}{x}=\frac{4}{3}\Rightarrow4x=3x+45\Leftrightarrow x=45\)
\(b,\frac{7,5-x}{3,5+x}=\frac{5}{6}\Rightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Rightarrow x=2,5\)
\(c,\frac{x-20}{x-10}=\frac{x+40}{x+70}\Rightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)
\(\Leftrightarrow x^2+50x-1400=x^2+30x-400\)
\(\Leftrightarrow20x=1000\)
\(\Rightarrow x=50\)
a. \(\frac{\left(x+15\right)}{x}=\frac{4}{3}\Leftrightarrow4x=3\left(x+15\right)\Leftrightarrow4x=3x+45\Leftrightarrow x=45\)
Vậy x=45
b. \(\frac{7,5-x}{3,5+x}=\frac{5}{6}\Leftrightarrow5\left(3,5+x\right)=6\left(7,5-x\right)\Leftrightarrow17,5+5x=45-6x\Leftrightarrow11x=27,5\Leftrightarrow x=2,5\)
Vậy x=2,5
c. \(\frac{x+20}{x-10}=\frac{x+40}{x+70}\Leftrightarrow\left(x+40\right)\left(x-10\right)=\left(x+20\right)\left(x+70\right)\)
\(\Leftrightarrow x^2+30x-400=x^2+90x+1400\Leftrightarrow-60x=-30\Leftrightarrow x=-30\)
Vậy x=-30
\(\left(\frac{2}{5}\right)^6:\left(\frac{2}{5}\right)^4=\left(\frac{2}{5}\right)^2=\frac{4}{25}\)
\(\left(\frac{3}{16}\right)^2:\left(\frac{9}{8}\right)^2=\frac{1}{12}\)
\(\left(\frac{2}{7}-\frac{1}{2}\right)^2=\frac{9}{196}\)
A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10
A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10
A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10
A = 0 + (3\(x-3x\)) - 10
A = 0 - 10
A = - 10