cho 1/a+ 1/b+ 1/c = 2 va 2/ab – 1/c^2 = 4 . tinh P = (a + 2b +c )^2020 09/11/2021 Bởi Lydia cho 1/a+ 1/b+ 1/c = 2 va 2/ab – 1/c^2 = 4 . tinh P = (a + 2b +c )^2020
Đáp án: $P=1$ Giải thích các bước giải: Ta có: $\dfrac1a+\dfrac1b+\dfrac1c=2$ $\to \dfrac1a+\dfrac1b=2-\dfrac1c$ $\to (\dfrac1a+\dfrac1b)^2=(2-\dfrac1c)^2$ $\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac{2}{ab}=4-\dfrac4c+\dfrac{1}{c^2}$ $\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac{2}{ab}-\dfrac{1}{c^2}=4-\dfrac4c$ $\to \dfrac1{a^2}+\dfrac1{b^2}+4=4-\dfrac4c$ vì $\dfrac{2}{ab}-\dfrac{1}{c^2}=4$ $\to \dfrac1{a^2}+\dfrac1{b^2}=-\dfrac4c$ $\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac4c=0$ $\to \dfrac1{a^2}+\dfrac1{b^2}+4(2-\dfrac1a-\dfrac1b)=0$ vì $\dfrac1a+\dfrac1b+\dfrac1c=2$ $\to (\dfrac1{a^2}-\dfrac4{a}+4)+(\dfrac{1}{b^2}-\dfrac4b+4)=0$ $\to (\dfrac1a-2)^2+(\dfrac1b-2)^2=0$ $\to \dfrac1a-2=\dfrac1b-2=0$ $\to a=b=\dfrac12\to c=-\dfrac12$ $\to a+2b+c=1$ $\to P=(a+2b+c)^{2020}=1^{2020}=1$ Bình luận
Đáp án: $P=1$
Giải thích các bước giải:
Ta có:
$\dfrac1a+\dfrac1b+\dfrac1c=2$
$\to \dfrac1a+\dfrac1b=2-\dfrac1c$
$\to (\dfrac1a+\dfrac1b)^2=(2-\dfrac1c)^2$
$\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac{2}{ab}=4-\dfrac4c+\dfrac{1}{c^2}$
$\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac{2}{ab}-\dfrac{1}{c^2}=4-\dfrac4c$
$\to \dfrac1{a^2}+\dfrac1{b^2}+4=4-\dfrac4c$ vì $\dfrac{2}{ab}-\dfrac{1}{c^2}=4$
$\to \dfrac1{a^2}+\dfrac1{b^2}=-\dfrac4c$
$\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac4c=0$
$\to \dfrac1{a^2}+\dfrac1{b^2}+4(2-\dfrac1a-\dfrac1b)=0$ vì $\dfrac1a+\dfrac1b+\dfrac1c=2$
$\to (\dfrac1{a^2}-\dfrac4{a}+4)+(\dfrac{1}{b^2}-\dfrac4b+4)=0$
$\to (\dfrac1a-2)^2+(\dfrac1b-2)^2=0$
$\to \dfrac1a-2=\dfrac1b-2=0$
$\to a=b=\dfrac12\to c=-\dfrac12$
$\to a+2b+c=1$
$\to P=(a+2b+c)^{2020}=1^{2020}=1$