Giải Pt a,(x-3) ³-2(x-1)=x(x-2) ²-5x ² b,x(x+3) ²-3x=(x+2) ³+1 07/12/2021 Bởi Eva Giải Pt a,(x-3) ³-2(x-1)=x(x-2) ²-5x ² b,x(x+3) ²-3x=(x+2) ³+1
Đáp án: \(\begin{array}{l}a.x = \frac{{25}}{{21}}\\b.x = – \frac{3}{2}\end{array}\) Giải thích các bước giải: \(\begin{array}{l}a.{\left( {x – 3} \right)^3} – 2\left( {x – 1} \right) = x{\left( {x – 2} \right)^2} – 5{x^2}\\ \to {x^3} – 9{x^2} + 27x – 27 – 2x + 2 = x\left( {{x^2} – 4x + 4} \right) – 5{x^2}\\ \to {x^3} – 9{x^2} + 25x – 25 = {x^3} – 4{x^2} + 4x – 5{x^2}\\ \to 21x = 25\\ \to x = \frac{{25}}{{21}}\\b.x{\left( {x + 3} \right)^2} – 3x = {\left( {x + 2} \right)^3} + 1\\ \to x\left( {{x^2} + 6x + 9} \right) – 3x = {x^3} + 6{x^2} + 12x + 8 + 1\\ \to {x^3} + 6{x^2} + 9x – 3x = {x^3} + 6{x^2} + 12x + 9\\ \to 6x = – 9\\ \to x = – \frac{3}{2}\end{array}\) Bình luận
Đáp án:
\(\begin{array}{l}
a.x = \frac{{25}}{{21}}\\
b.x = – \frac{3}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.{\left( {x – 3} \right)^3} – 2\left( {x – 1} \right) = x{\left( {x – 2} \right)^2} – 5{x^2}\\
\to {x^3} – 9{x^2} + 27x – 27 – 2x + 2 = x\left( {{x^2} – 4x + 4} \right) – 5{x^2}\\
\to {x^3} – 9{x^2} + 25x – 25 = {x^3} – 4{x^2} + 4x – 5{x^2}\\
\to 21x = 25\\
\to x = \frac{{25}}{{21}}\\
b.x{\left( {x + 3} \right)^2} – 3x = {\left( {x + 2} \right)^3} + 1\\
\to x\left( {{x^2} + 6x + 9} \right) – 3x = {x^3} + 6{x^2} + 12x + 8 + 1\\
\to {x^3} + 6{x^2} + 9x – 3x = {x^3} + 6{x^2} + 12x + 9\\
\to 6x = – 9\\
\to x = – \frac{3}{2}
\end{array}\)