Toán Lớp 8: 3) (x-y)(x+2y)^2+3y^2 , Bài 2 : 1)x(x-2)-x^2=4 , 2) (x-2)-x(x+2)= 16 , 3) (3x-5)x^2+(5x-2)(-3+x)+17x = 9
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Toán Lớp 8: 3) (x-y)(x+2y)^2+3y^2 , Bài 2 : 1)x(x-2)-x^2=4 , 2) (x-2)-x(x+2)= 16 , 3) (3x-5)x^2+(5x-2)(-3+x)+17x = 9
Comments ( 2 )
3)\left( {x – y} \right){\left( {x + 2y} \right)^2} + 3{y^2}\\
= \left( {x – y} \right)\left( {{x^2} + 4xy + {y^2}} \right) + 3{y^2}\\
= {x^3} + 4{x^2}y + x{y^2} – {x^2}y – 4x{y^2} – {y^3} + 3{y^2}\\
= {x^3} – {y^3} + 3{y^2} + 3{x^2}y – 3x{y^2}\\
B2)\\
1)x\left( {x – 2} \right) – {x^2} = 4\\
\Leftrightarrow {x^2} – 2x – {x^2} = 4\\
\Leftrightarrow – 2x = 4\\
\Leftrightarrow x = – 2\\
Vậy\,x = – 2\\
2)\left( {x – 2} \right) – x\left( {x + 2} \right) = 16\\
\Leftrightarrow x – 2 – {x^2} – 2x = 16\\
\Leftrightarrow {x^2} + x + 18 = 0\left( {vn} \right)\\
Vậy\,x \in \emptyset \\
3)\left( {3x – 5} \right){x^2} + \left( {5x – 2} \right)\left( { – 3 + x} \right) + 17x = 9\\
\Leftrightarrow 3{x^3} – 5{x^2} – 15x + 5{x^2} + 6 – 2x + 17x = 9\\
\Leftrightarrow 3{x^3} = 3\\
\Leftrightarrow {x^3} = 1\\
\Leftrightarrow x = 1\\
Vậy\,x = 1
\end{array}$