Giai pt $\frac{4x^2+16}{x^2+6}-\frac{3}{x^2+1}=\frac{5}{x^2+3}+\frac{7}{x^2+5}$ 07/07/2021 Bởi Mackenzie Giai pt $\frac{4x^2+16}{x^2+6}-\frac{3}{x^2+1}=\frac{5}{x^2+3}+\frac{7}{x^2+5}$
Đáp án: $S = \left\{ { \pm \sqrt 2 } \right\}$ Giải thích các bước giải: Ta có: $\begin{array}{l}\dfrac{{4{x^2} + 16}}{{{x^2} + 6}} – \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\ \Leftrightarrow \dfrac{{4\left( {{x^2} + 6} \right) – 8}}{{{x^2} + 6}} – \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\ \Leftrightarrow 4 – \dfrac{8}{{{x^2} + 6}} – \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\ \Leftrightarrow \dfrac{8}{{{x^2} + 6}} + \dfrac{3}{{{x^2} + 1}} + \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}} – 4 = 0\\ \Leftrightarrow \left( {\dfrac{8}{{{x^2} + 6}} – 1} \right) + \left( {\dfrac{3}{{{x^2} + 1}} – 1} \right) + \left( {\dfrac{5}{{{x^2} + 3}} – 1} \right) + \left( {\dfrac{7}{{{x^2} + 5}} – 1} \right) = 0\\ \Leftrightarrow \dfrac{{2 – {x^2}}}{{{x^2} + 6}} + \dfrac{{2 – {x^2}}}{{{x^2} + 1}} + \dfrac{{2 – {x^2}}}{{{x^2} + 3}} + \dfrac{{2 – {x^2}}}{{{x^2} + 5}} = 0\\ \Leftrightarrow \left( {2 – {x^2}} \right)\left( {\dfrac{1}{{{x^2} + 6}} + \dfrac{1}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 3}} + \dfrac{1}{{{x^2} + 5}}} \right) = 0\\ \Leftrightarrow 2 – {x^2} = 0\left( {do:\dfrac{1}{{{x^2} + 6}} + \dfrac{1}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 3}} + \dfrac{1}{{{x^2} + 5}} > 0,\forall x} \right)\\ \Leftrightarrow {x^2} = 2\\ \Leftrightarrow x = \pm \sqrt 2 \end{array}$ Vậy tập nghiệm của phương trình là: $S = \left\{ { \pm \sqrt 2 } \right\}$ Bình luận
`(4x^2+16)/(x^2+6) – 3/(x^2+1) = 5/(x^2+3) + 7/(x^2+5)` `⇔ ((4x^2+16)(x^2+1)(x^2+3)(x^2+5)-3(x^2+6)(x^2+3)(x^2+5))/((x^2+6)(x^2+1)(x^2+3)(x^2+5))=(5(x^2+6)(x^2+1)(x^2+5)+7(x^2+6)(x^2+1)(x^2+3))/((x^2+6)(x^2+1)(x^2+3)(x^2+5))` `⇒ (4x^2+16)(x^2+1)(x^2 + 5) – 3(x^2+6)(x^2+3)(x^2+5) = 5(x^2+6)(x^2+1)(x^2+5)+7(x^2+6)(x^2+1)(x^2+3)` `⇔ 4x^8 + 49x^6 + 194x^4 + 239x^2 – 30 = 12x^6 + 130x^4 + 394x^2 + 276` `⇔ 4x^8 + 37x^6 + 64x^4 – 155x^2 – 306 = 0` Đặt `x^2 = t(t \ge 0)` `⇔ 4t^4 + 37t^3 + 64t^2 – 155t – 306 = 0` `⇔ (t-2)(4t^3 + 45t^2 + 154t + 153) = 0` `⇔`\(\left[ \begin{array}{l}t-2=0\\4t^3+45t^2+154t+153=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}t=2(TM)\\t ∈ \mathbb{R}\end{array} \right.\) `⇔ x^2 = 2` `⇔ x = \pm\sqrt{2}` Vậy `S = {\pm\sqrt{2}}` Bình luận
Đáp án:
$S = \left\{ { \pm \sqrt 2 } \right\}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\dfrac{{4{x^2} + 16}}{{{x^2} + 6}} – \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\
\Leftrightarrow \dfrac{{4\left( {{x^2} + 6} \right) – 8}}{{{x^2} + 6}} – \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\
\Leftrightarrow 4 – \dfrac{8}{{{x^2} + 6}} – \dfrac{3}{{{x^2} + 1}} = \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}}\\
\Leftrightarrow \dfrac{8}{{{x^2} + 6}} + \dfrac{3}{{{x^2} + 1}} + \dfrac{5}{{{x^2} + 3}} + \dfrac{7}{{{x^2} + 5}} – 4 = 0\\
\Leftrightarrow \left( {\dfrac{8}{{{x^2} + 6}} – 1} \right) + \left( {\dfrac{3}{{{x^2} + 1}} – 1} \right) + \left( {\dfrac{5}{{{x^2} + 3}} – 1} \right) + \left( {\dfrac{7}{{{x^2} + 5}} – 1} \right) = 0\\
\Leftrightarrow \dfrac{{2 – {x^2}}}{{{x^2} + 6}} + \dfrac{{2 – {x^2}}}{{{x^2} + 1}} + \dfrac{{2 – {x^2}}}{{{x^2} + 3}} + \dfrac{{2 – {x^2}}}{{{x^2} + 5}} = 0\\
\Leftrightarrow \left( {2 – {x^2}} \right)\left( {\dfrac{1}{{{x^2} + 6}} + \dfrac{1}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 3}} + \dfrac{1}{{{x^2} + 5}}} \right) = 0\\
\Leftrightarrow 2 – {x^2} = 0\left( {do:\dfrac{1}{{{x^2} + 6}} + \dfrac{1}{{{x^2} + 1}} + \dfrac{1}{{{x^2} + 3}} + \dfrac{1}{{{x^2} + 5}} > 0,\forall x} \right)\\
\Leftrightarrow {x^2} = 2\\
\Leftrightarrow x = \pm \sqrt 2
\end{array}$
Vậy tập nghiệm của phương trình là: $S = \left\{ { \pm \sqrt 2 } \right\}$
`(4x^2+16)/(x^2+6) – 3/(x^2+1) = 5/(x^2+3) + 7/(x^2+5)`
`⇔ ((4x^2+16)(x^2+1)(x^2+3)(x^2+5)-3(x^2+6)(x^2+3)(x^2+5))/((x^2+6)(x^2+1)(x^2+3)(x^2+5))=(5(x^2+6)(x^2+1)(x^2+5)+7(x^2+6)(x^2+1)(x^2+3))/((x^2+6)(x^2+1)(x^2+3)(x^2+5))`
`⇒ (4x^2+16)(x^2+1)(x^2 + 5) – 3(x^2+6)(x^2+3)(x^2+5) = 5(x^2+6)(x^2+1)(x^2+5)+7(x^2+6)(x^2+1)(x^2+3)`
`⇔ 4x^8 + 49x^6 + 194x^4 + 239x^2 – 30 = 12x^6 + 130x^4 + 394x^2 + 276`
`⇔ 4x^8 + 37x^6 + 64x^4 – 155x^2 – 306 = 0`
Đặt `x^2 = t(t \ge 0)`
`⇔ 4t^4 + 37t^3 + 64t^2 – 155t – 306 = 0`
`⇔ (t-2)(4t^3 + 45t^2 + 154t + 153) = 0`
`⇔`\(\left[ \begin{array}{l}t-2=0\\4t^3+45t^2+154t+153=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}t=2(TM)\\t ∈ \mathbb{R}\end{array} \right.\)
`⇔ x^2 = 2`
`⇔ x = \pm\sqrt{2}`
Vậy `S = {\pm\sqrt{2}}`