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14 tháng 4 2019

\(\frac{3}{4}-\frac{1}{4}x-1=40\%\times\frac{-5}{3}\)

\(\Leftrightarrow\frac{1}{4}\left(3-x\right)-1=\frac{40}{100}\times\frac{-5}{3}\)

\(\Leftrightarrow\frac{1}{4}\left(3-x\right)-1=\frac{2}{5}\times\frac{-5}{3}\)

\(\Leftrightarrow\frac{1}{4}\left(3-x\right)-\frac{15}{15}=\frac{-10}{15}\)

\(\Leftrightarrow\frac{1}{4}\left(3-x\right)=\frac{-10}{15}+\frac{15}{15}\)

\(\Leftrightarrow\frac{1}{4}\left(3-x\right)=\frac{5}{15}\)

\(\Leftrightarrow3-x=\frac{5}{15}\div\frac{1}{4}\)

\(\Leftrightarrow3-x=\frac{5}{15}\times4\)

\(\Leftrightarrow\frac{45}{15}-x=\frac{20}{15}\)

\(\Leftrightarrow x=\frac{45}{15}-\frac{20}{15}\)

\(\Leftrightarrow x=\frac{25}{15}=\frac{5}{3}\)

3 tháng 7 2019

a)

\(5x\left(4-8x\right)+40\left(x^2-1\right)=3\\ \Leftrightarrow20x-40x^2+40x^2-40=3\\ \Leftrightarrow20x-40=3\\ \Leftrightarrow20\left(x-2\right)=3\\ \Leftrightarrow x-2=\frac{3}{20}\\ \Leftrightarrow x=\frac{43}{20}\)

b)

\(\left(4x-5\right)\left(7-8x\right)+4x\left(3+8x\right)=4\\ \Leftrightarrow28x-32x^2-35+40x+12x+32x^2=4\\ \Leftrightarrow80x-35=4\\ \Leftrightarrow80x=39\\ \Leftrightarrow x=\frac{39}{80}\)

3 tháng 7 2019

a) \(5x\left(4-8x\right)+40\left(x^2-1\right)=3\)

\(\Leftrightarrow\)\(20x-40x^2+40x^2-40=3\)

\(\Leftrightarrow20x-40=3\)

\(\Leftrightarrow20x=43\)

\(\Leftrightarrow x=\frac{43}{20}\)

b, \(\left(4x-5\right)\left(7-8x\right)+4x\left(3+8x\right)=4\)

\(\Leftrightarrow28x-32x^2-35+40x+12x+32x^2=4\)

\(\Leftrightarrow80x-35=4\)

\(\Leftrightarrow80x=39\)

\(\Leftrightarrow x=\frac{39}{80}\)

a: \(\Leftrightarrow2x^2+4-x^2+\dfrac{3}{2}=-3+4x^2-\dfrac{4}{3}x^2+1\)

\(\Leftrightarrow x^2+\dfrac{11}{2}=\dfrac{8}{3}x^2-2\)

\(\Leftrightarrow x^2\cdot\dfrac{-5}{3}=-\dfrac{15}{2}\)

\(\Leftrightarrow x^2=\dfrac{9}{2}\)

hay \(x\in\left\{\dfrac{3\sqrt{2}}{2};-\dfrac{3\sqrt{2}}{2}\right\}\)

b: \(\Leftrightarrow\left|x\right|-4-2+\left|x\right|-\dfrac{1}{3}\left|x\right|+5=0\)

\(\Leftrightarrow\left|x\right|\cdot\dfrac{5}{3}=1\)

hay \(x\in\left\{\dfrac{3}{5};-\dfrac{3}{5}\right\}\)

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

Bài 1: 

b: \(=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)

\(=8x^2+40x+50+48x^2-3\)

\(=56x^2+40x+47\)

Bài 2: 

b: \(\Leftrightarrow3x-3+9x-18=2x-6+4x-4\)

=>12x-21=6x-10

=>6x=11

hay x=11/6

17 tháng 6 2017

\(a,A=-1+3-5+7-9+...-2013+2015-2017=\left(-1+3\right)+\left(-5+7\right)+...+\left(-2013+2015\right)-2017\)\(=2+2+..+2-2017\)

\(=2.504-2017=-1009\)

\(b,B=2-4+6-8+...+2014-2016+2018\)\(=2+\left(-4+6\right)+\left(-8+10\right)+...+\left(-2016+2018\right)==2+2+...+2\)\(=2+503.2=1008\)