Cho`f(x) = 2x^2-3x^3+x+1; g(x) = 3x-2(x^3-x^2)-x^3-2; h(x) = 2x^2 + 1`. Tính `2f(-1)-1/4h(-1/2)`. 11/08/2021 Bởi Claire Cho`f(x) = 2x^2-3x^3+x+1; g(x) = 3x-2(x^3-x^2)-x^3-2; h(x) = 2x^2 + 1`. Tính `2f(-1)-1/4h(-1/2)`.
`f(-1)=2.(-1)^2-3.(-1)^3 +(-1)+1` `⇔f(-1)=2+3=5 ` `h(-1/2)=2.(-1/2)^2+1=1/2+1=3/2` `⇒2f(-1)-1/4h(-1/2)=2×5-3/2×1/4=(77)/8` Bình luận
Đáp án: \(\dfrac{{77}}{8}\) Giải thích các bước giải: \(\begin{array}{l}f\left( { – 1} \right) = 2.{\left( { – 1} \right)^2} – 3.{\left( { – 1} \right)^3} – 1 + 1\\ = 2 + 3 = 5\\ \to 2f\left( { – 1} \right) = 10\\h\left( { – \dfrac{1}{2}} \right) = 2.{\left( { – \dfrac{1}{2}} \right)^2} + 1 = \dfrac{3}{2}\\ \to – \dfrac{1}{4}h\left( { – \dfrac{1}{2}} \right) = – \dfrac{1}{4}.\dfrac{3}{2} = – \dfrac{3}{8}\\2f\left( { – 1} \right) – \dfrac{1}{4}h\left( { – \dfrac{1}{2}} \right) = 10 – \dfrac{3}{8} = \dfrac{{77}}{8}\end{array}\) Bình luận
`f(-1)=2.(-1)^2-3.(-1)^3 +(-1)+1`
`⇔f(-1)=2+3=5 `
`h(-1/2)=2.(-1/2)^2+1=1/2+1=3/2`
`⇒2f(-1)-1/4h(-1/2)=2×5-3/2×1/4=(77)/8`
Đáp án:
\(\dfrac{{77}}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
f\left( { – 1} \right) = 2.{\left( { – 1} \right)^2} – 3.{\left( { – 1} \right)^3} – 1 + 1\\
= 2 + 3 = 5\\
\to 2f\left( { – 1} \right) = 10\\
h\left( { – \dfrac{1}{2}} \right) = 2.{\left( { – \dfrac{1}{2}} \right)^2} + 1 = \dfrac{3}{2}\\
\to – \dfrac{1}{4}h\left( { – \dfrac{1}{2}} \right) = – \dfrac{1}{4}.\dfrac{3}{2} = – \dfrac{3}{8}\\
2f\left( { – 1} \right) – \dfrac{1}{4}h\left( { – \dfrac{1}{2}} \right) = 10 – \dfrac{3}{8} = \dfrac{{77}}{8}
\end{array}\)