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Bài 1:
a) f(x)-g(x)=2x3-x2+5+2x3-x2-1-2x
=4x3-2x2-2x-1
Bậc: 3
b) f(x)+g(x)=2x3-x2+5-2x3+x2+1+2x
=2x+6
Để f(x)+g(x)=0 thì 2x+6=0
=>2x=-6
=>x=-3
Bài 2:
a) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{7}=0\)
\(\Rightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{1}{7}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{7}\Rightarrow x=\dfrac{-2}{35}\\x+\dfrac{1}{5}=\dfrac{-1}{7}\Rightarrow x=\dfrac{-12}{35}\end{matrix}\right.\)
b) \(2\left(x+1\right)+4^2=2^4\)
\(\Rightarrow2x+2+16=16\)
\(\Rightarrow2x+2=0\)
\(\Rightarrow2x=-2\)
\(\Rightarrow x=-1\)
a)\(-\frac{3}{2}-2x+\frac{3}{4}=2\)
\(\Leftrightarrow-2x=2+\frac{3}{2}-\frac{3}{4}=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{4}:\left(-2\right)=-\frac{11}{8}\)
b , pt <=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\) <=> \(\frac{3}{5}x\) = -4 <=> x = \(\frac{-20}{3}\)
c , \(\frac{2}{3x}-\frac{3}{12}\) = \(\frac{4}{5}-\left(\frac{7}{x}-x\right)\) (ĐK : x khác 0 )
<=> \(\frac{23}{3x}\) - x =\(\frac{21}{20}\) <=>\(-3x^2-\frac{21}{20}.3x+23=0\Leftrightarrow\left[{}\begin{matrix}x=\\x=\end{matrix}\right.\)...
a: =>|5/4x-7/2|=|5/8x+3/5|
=>5/4x-7/2=5/8x+3/5 hoặc 5/4x-7/2=-5/8x-3/5
=>5/8x=41/10 hoặc 15/8x=29/10
=>x=164/25 hoặc x=116/75
b: =>3:|x/4-2/3|=6-21/5=9/5
=>|1/4x-2/3|=5/3
=>1/4x-2/3=5/3 hoặc 1/4x-2/3=-5/3
=>1/4x=7/3 hoặc 1/4x=-1
=>x=28/3 hoặc x=-4
c: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(2x-x-9\right)\left(2x+x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)
e: =>|2x-7|=2x-7
=>2x-7>=0
=>x>=7/2
