(X+y)^2-25= 3x^2-6xy+3y^2= (X^4-2x^3+4x^2-8x):(x^2+4)= (X^2+2x^3+10x-25):(x^2+5)= 07/08/2021 Bởi Alexandra (X+y)^2-25= 3x^2-6xy+3y^2= (X^4-2x^3+4x^2-8x):(x^2+4)= (X^2+2x^3+10x-25):(x^2+5)=
Đáp án: $\begin{array}{l} + ){\left( {x + y} \right)^2} – 25 = {\left( {x + y} \right)^2} – {5^2} = \left( {x + y – 5} \right)\left( {x + y + 5} \right)\\ + )3{x^2} – 6xy + 3{y^2} = 3\left( {{x^2} – 2xy + {y^2}} \right) = 3{\left( {x – y} \right)^2}\\ + )\left( {{x^4} – 2{x^3} + 4{x^2} – 8x} \right):\left( {{x^2} + 4} \right)\\ = \left[ {{x^3}\left( {x – 2} \right) + 4x\left( {x – 2} \right)} \right]:\left( {{x^2} + 4} \right)\\ = \left( {x – 2} \right)\left( {{x^3} + 4x} \right):\left( {{x^2} + 4} \right)\\ = \left( {x – 2} \right).x.\left( {{x^2} + 4} \right):\left( {{x^2} + 4} \right)\\ = \left( {x – 2} \right).x\\ + )\frac{{{x^4} + 2{x^3} + 10x – 25}}{{{x^2} + 5}}\\ = \frac{{\left( {{x^4} – 25} \right) + \left( {2{x^3} + 10x} \right)}}{{{x^2} + 5}}\\ = \frac{{\left( {{x^2} + 5} \right)\left( {{x^2} – 5} \right) + 2x\left( {{x^2} + 5} \right)}}{{{x^2} + 5}}\\ = \frac{{\left( {{x^2} + 5} \right)\left( {{x^2} – 5 + 2x} \right)}}{{{x^2} + 5}}\\ = {x^2} + 2x – 5\end{array}$ Bình luận
Đáp án: $(x+y)^{2}$ -25 =$(x+y)^{2}$ – $5^{2}$ =$(x+y-5)^{2}$ * $(x+y+5)^{2}$ Giải thích các bước giải: Bình luận
Đáp án:
$\begin{array}{l}
+ ){\left( {x + y} \right)^2} – 25 = {\left( {x + y} \right)^2} – {5^2} = \left( {x + y – 5} \right)\left( {x + y + 5} \right)\\
+ )3{x^2} – 6xy + 3{y^2} = 3\left( {{x^2} – 2xy + {y^2}} \right) = 3{\left( {x – y} \right)^2}\\
+ )\left( {{x^4} – 2{x^3} + 4{x^2} – 8x} \right):\left( {{x^2} + 4} \right)\\
= \left[ {{x^3}\left( {x – 2} \right) + 4x\left( {x – 2} \right)} \right]:\left( {{x^2} + 4} \right)\\
= \left( {x – 2} \right)\left( {{x^3} + 4x} \right):\left( {{x^2} + 4} \right)\\
= \left( {x – 2} \right).x.\left( {{x^2} + 4} \right):\left( {{x^2} + 4} \right)\\
= \left( {x – 2} \right).x\\
+ )\frac{{{x^4} + 2{x^3} + 10x – 25}}{{{x^2} + 5}}\\
= \frac{{\left( {{x^4} – 25} \right) + \left( {2{x^3} + 10x} \right)}}{{{x^2} + 5}}\\
= \frac{{\left( {{x^2} + 5} \right)\left( {{x^2} – 5} \right) + 2x\left( {{x^2} + 5} \right)}}{{{x^2} + 5}}\\
= \frac{{\left( {{x^2} + 5} \right)\left( {{x^2} – 5 + 2x} \right)}}{{{x^2} + 5}}\\
= {x^2} + 2x – 5
\end{array}$
Đáp án:
$(x+y)^{2}$ -25 =$(x+y)^{2}$ – $5^{2}$ =$(x+y-5)^{2}$ * $(x+y+5)^{2}$
Giải thích các bước giải: