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Bài 2:

\(\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}\)

=>\(10\left(x^2-2y^2\right)=7\left(x^2+y^2\right)\)

=>\(10x^2-20y^2=7x^2+7y^2\)

=>\(3x^2=27y^2\)

=>\(x^2=9y^2\)

\(x^4y^4=81\)

=>\(x^2y^2=9\)

=>\(9y^2\cdot y^2=9\)

=>\(y^4=1\)

=>y=1 hoặc y=-1

TH1: y=1

\(x^2=9y^2=9\cdot1^2=9\)

=>x=3 hoặc x=-3

TH2: y=-1

\(x^2=9y^2=9\cdot\left(-1\right)^2=9\)

=>x=3 hoặc x=-3

D = x⁶ - x⁴yz + x³ yz² - x³y²z + x³y²z - z⁶ + 2018 

=> D = -z⁶ + x³ yz² + ( - x⁴) yz + x⁶ + 2018 

=> D = - ( z⁶ - x³ yz²  + x⁴yz - x⁶ - 2018 ) 

.... :) 

a ) 

\(3\left|2x-1\right|+1=\left(-2\right)^2-3\left(-2\right)^3\)

\(\Rightarrow3\left|2x-1\right|+1=4-3.-8\)

\(\Rightarrow3\left|2x-1\right|+1=4-\left(-24\right)\)

\(\Rightarrow3\left|2x-1\right|+1=28\)

\(\Rightarrow3\left|2x-1\right|=28-1\)

\(\Rightarrow3\left|2x-1\right|=27\)

\(\Rightarrow\left|2x-1\right|=27:3\)

\(\Rightarrow\left|2x-1\right|=9\)

\(\Rightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}\Rightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

b ) 

\(x^2\left(x+2\right)+4\left(x+2\right)=0\)

\(\Rightarrow\left(x^2+4\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+4=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-4\left(L\right)\\x=-2\end{cases}\Rightarrow}x=-2}\)

Vậy \(x=-2\)

~ Ủng hộ nhé 

c ) 

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