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1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
a/ 2x - 10 - [3x - 14 - (4 - 5x) - 2x] = 2
=> 2x - 10 - (3x - 14 - 4 + 5x - 2x) = 2
=> 2x - 10 - 3x + 14 + 4 - 5x + 2x = 2
=> -4x + 6 = 0
=> -4x = -6
=> x = 3/2
b/ \(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+1\right)=4,5\)
\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-1-\frac{9}{2}=0\)
\(\Rightarrow\frac{17}{24}x-\frac{17}{2}=0\)
\(\Rightarrow\frac{17}{24}x=\frac{17}{2}\)
\(\Rightarrow x=12\)
a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)
=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)
=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)
b) \(2x-\left|x+1\right|=\frac{1}{2}\)
=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))
=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)
=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
a)\(2\left|2x-3\right|=\frac{1}{2}\)
\(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{8}\\x=\frac{11}{8}\end{matrix}\right.\)
Vậy....
b)\(7,5-3\left|5-2x\right|=-4,5\)
\(\Leftrightarrow\left|5-2x\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)
VẬy...
c)\(\left|3x-4\right|+\left|5-2x\right|=0\)
Có: \(\left|3x-4\right|\ge0với\forall x\\ \left|5-2x\right|\ge0với\forall x\)
\(\Rightarrow\left[{}\begin{matrix}3x-4=0\\5-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\\x=\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow x\in\varnothing\)