giải phương trình: $\frac{1}{(x+29)^2}$ +$\frac{1}{(x+30)^2}$ =$\frac{13}{36}$ 23/07/2021 Bởi Jade giải phương trình: $\frac{1}{(x+29)^2}$ +$\frac{1}{(x+30)^2}$ =$\frac{13}{36}$
$\begin{array}{l} Đặt x + 29 = a\\ \Rightarrow \dfrac{1}{{{{\left( {x + 29} \right)}^2}}} + \dfrac{1}{{{{\left( {x + 30} \right)}^2}}} = \dfrac{{13}}{{36}}\\ \Leftrightarrow \dfrac{1}{{{a^2}}} + \dfrac{1}{{{{\left( {a + 1} \right)}^2}}} = \dfrac{{13}}{{36}}\\ \Leftrightarrow 36\left( {{a^2} + {a^2} + 2a + 1} \right) = 13{a^2}\left( {{a^2} + 2a + 1} \right)\\ \Leftrightarrow 13{a^4} + 26{a^3} – 59{a^2} – 72a – 36 = 0\\ \Leftrightarrow 13{a^4} – 26{a^3} + 52{a^3} – 104{a^2} + 45{a^2} – 90a + 18a – 36 = 0\\ \Leftrightarrow \left( {a – 2} \right)\left( {13{a^3} + 52{a^2} + 45a + 18} \right) = 0\\ \Leftrightarrow \left( {a – 2} \right)\left( {a + 3} \right)\left( {13{a^2} + 13a + 6} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} a = 2\\ a = – 3 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x + 29 = 2\\ x + 29 = – 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = – 27\\ x = – 32 \end{array} \right.\\ \Rightarrow S = \left\{ { – 27 – 32} \right\} \end{array}$ Bình luận
Đáp án: Giải thích các bước giải: `1/(x+29)^2 +1/(x+30)^2 =13/36 (x \ne -29;-30)` Đặt `x+29=a(a \ne 0;-1)` Ta có `1/a^2 +1/(a+1)^2=13/36` `<=>[(a+1)^2 +a^2]/[a(a+1)]^2=13/36` `=>36(2a^2+2a+1)=13[a(a+1)]^2` `<=>13a^4+26a^3-59a^2-72a-36=0` `<=>(a-2)(a+3)(13a^2+13a+6)=0` `+)a-2=0` `=>a=2(t“/m)` `=>x=-27(t“m)` `+)a+3=0` `=>a=-3` `=>x=-32(t“/m)` `+)13a^2+13a+6=0` `Δ=b^2-4ac=13^2-4.13.6=-143<0` `=>` Phương trình vô nghiệm Vậy `x in{-27,-32}` Bình luận
$\begin{array}{l} Đặt x + 29 = a\\ \Rightarrow \dfrac{1}{{{{\left( {x + 29} \right)}^2}}} + \dfrac{1}{{{{\left( {x + 30} \right)}^2}}} = \dfrac{{13}}{{36}}\\ \Leftrightarrow \dfrac{1}{{{a^2}}} + \dfrac{1}{{{{\left( {a + 1} \right)}^2}}} = \dfrac{{13}}{{36}}\\ \Leftrightarrow 36\left( {{a^2} + {a^2} + 2a + 1} \right) = 13{a^2}\left( {{a^2} + 2a + 1} \right)\\ \Leftrightarrow 13{a^4} + 26{a^3} – 59{a^2} – 72a – 36 = 0\\ \Leftrightarrow 13{a^4} – 26{a^3} + 52{a^3} – 104{a^2} + 45{a^2} – 90a + 18a – 36 = 0\\ \Leftrightarrow \left( {a – 2} \right)\left( {13{a^3} + 52{a^2} + 45a + 18} \right) = 0\\ \Leftrightarrow \left( {a – 2} \right)\left( {a + 3} \right)\left( {13{a^2} + 13a + 6} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} a = 2\\ a = – 3 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x + 29 = 2\\ x + 29 = – 3 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = – 27\\ x = – 32 \end{array} \right.\\ \Rightarrow S = \left\{ { – 27 – 32} \right\} \end{array}$
Đáp án:
Giải thích các bước giải:
`1/(x+29)^2 +1/(x+30)^2 =13/36 (x \ne -29;-30)`
Đặt `x+29=a(a \ne 0;-1)`
Ta có
`1/a^2 +1/(a+1)^2=13/36`
`<=>[(a+1)^2 +a^2]/[a(a+1)]^2=13/36`
`=>36(2a^2+2a+1)=13[a(a+1)]^2`
`<=>13a^4+26a^3-59a^2-72a-36=0`
`<=>(a-2)(a+3)(13a^2+13a+6)=0`
`+)a-2=0`
`=>a=2(t“/m)`
`=>x=-27(t“m)`
`+)a+3=0`
`=>a=-3`
`=>x=-32(t“/m)`
`+)13a^2+13a+6=0`
`Δ=b^2-4ac=13^2-4.13.6=-143<0`
`=>` Phương trình vô nghiệm
Vậy `x in{-27,-32}`