Toán Giải phương trình: $\frac{x^{2}+x }{\sqrt{x^{2}+x+10 } }$+2=$\sqrt{ x^{2}+x+4}$ 24/07/2021 By Eliza Giải phương trình: $\frac{x^{2}+x }{\sqrt{x^{2}+x+10 } }$+2=$\sqrt{ x^{2}+x+4}$
Đáp án: Giải thích các bước giải: Làm vầy luôn cho gọn , đặt $; t = x² + x$ $ PT ⇔ t + 2\sqrt{t + 10} = \sqrt{t + 4}.\sqrt{t + 10}$ $ ⇒ t² + 4t\sqrt{t + 10} + 4(t + 10) = t² + 14t + 40$ $ ⇔ 4t\sqrt{t + 10} – 10t = 0$ $ ⇔ 2t(2\sqrt{t + 10} – 5) = 0$ TH1 $: t = 0 ⇔ x(x + 1) = 0 ⇔ x = 0; x = -1 (TM)$ TH2 $: 2\sqrt{t + 10} – 5 = 0 ⇔ 2\sqrt{t + 10} = 5$ $ ⇔ 4t + 40 = 25 ⇔ 4t + 15 = 0$ $ ⇔ 4x² + 4x + 15 = 0 ⇔ (2x + 1)² + 14 = 0$(VN) Trả lời
$\begin{array}{*{20}{l}} {\dfrac{{{x^2} + x}}{{\sqrt {{x^2} + x + 10} }} + 2 = \sqrt {{x^2} + x + 4} }\\ {{\rm{ \;}} \Leftrightarrow \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 10} }} = \sqrt {{x^2} + x + 4} {\rm{ \;}} – 2}\\ {{\rm{ \;}} \Leftrightarrow \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 10} }} = \dfrac{{{x^2} + x + 4 – 4}}{{\sqrt {{x^2} + x + 4} {\rm{ \;}} + 2}}}\\ {{\rm{ \;}} \Leftrightarrow \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 10} }} – \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 4} {\rm{ \;}} + 2}} = 0}\\ {{\rm{ \;}} \Leftrightarrow x\left( {x + 1} \right)\left( {\underbrace {\dfrac{1}{{\sqrt {{x^2} + x + 10} }} – \dfrac{1}{{\sqrt {{x^2} + x + 4} {\rm{ \;}} + 2}}}_{ > 0}} \right) = 0}\\ {{\rm{ \;}}}\\ {} \end{array}$ $\begin{array}{l} \dfrac{1}{{\sqrt {{x^2} + x + 10} }} = \dfrac{1}{{\sqrt {{x^2} + x + 4} + 2}}\\ \Leftrightarrow \sqrt {{x^2} + x + 10} = \sqrt {{x^2} + x + 4} + 2\\ \Leftrightarrow {x^2} + x + 10 = {x^2} + x + 4 + 4 + 4\sqrt {\left( {{x^2} + x + 10} \right)\left( {{x^2} + x + 4} \right)} \\ \Leftrightarrow \sqrt {\left( {{x^2} + x + 10} \right)\left( {{x^2} + x + 4} \right)} = \dfrac{1}{2}\\ + \min {x^2} + x + 10 = \dfrac{{39}}{4}\\ + \min {x^2} + x + 4 = \dfrac{{15}}{4}\\ \sqrt {\left( {{x^2} + x + 10} \right)\left( {{x^2} + x + 4} \right)} > \sqrt {\dfrac{{39}}{4}.\dfrac{{15}}{4}} = \dfrac{{3\sqrt {65} }}{4} > \dfrac{1}{2}\end{array}$ Vậy $S = \left\{ {0; – 1} \right\}$ Trả lời
Đáp án:
Giải thích các bước giải:
Làm vầy luôn cho gọn , đặt $; t = x² + x$
$ PT ⇔ t + 2\sqrt{t + 10} = \sqrt{t + 4}.\sqrt{t + 10}$
$ ⇒ t² + 4t\sqrt{t + 10} + 4(t + 10) = t² + 14t + 40$
$ ⇔ 4t\sqrt{t + 10} – 10t = 0$
$ ⇔ 2t(2\sqrt{t + 10} – 5) = 0$
TH1 $: t = 0 ⇔ x(x + 1) = 0 ⇔ x = 0; x = -1 (TM)$
TH2 $: 2\sqrt{t + 10} – 5 = 0 ⇔ 2\sqrt{t + 10} = 5$
$ ⇔ 4t + 40 = 25 ⇔ 4t + 15 = 0$
$ ⇔ 4x² + 4x + 15 = 0 ⇔ (2x + 1)² + 14 = 0$(VN)
$\begin{array}{*{20}{l}} {\dfrac{{{x^2} + x}}{{\sqrt {{x^2} + x + 10} }} + 2 = \sqrt {{x^2} + x + 4} }\\ {{\rm{ \;}} \Leftrightarrow \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 10} }} = \sqrt {{x^2} + x + 4} {\rm{ \;}} – 2}\\ {{\rm{ \;}} \Leftrightarrow \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 10} }} = \dfrac{{{x^2} + x + 4 – 4}}{{\sqrt {{x^2} + x + 4} {\rm{ \;}} + 2}}}\\ {{\rm{ \;}} \Leftrightarrow \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 10} }} – \dfrac{{x\left( {x + 1} \right)}}{{\sqrt {{x^2} + x + 4} {\rm{ \;}} + 2}} = 0}\\ {{\rm{ \;}} \Leftrightarrow x\left( {x + 1} \right)\left( {\underbrace {\dfrac{1}{{\sqrt {{x^2} + x + 10} }} – \dfrac{1}{{\sqrt {{x^2} + x + 4} {\rm{ \;}} + 2}}}_{ > 0}} \right) = 0}\\ {{\rm{ \;}}}\\ {} \end{array}$
$\begin{array}{l} \dfrac{1}{{\sqrt {{x^2} + x + 10} }} = \dfrac{1}{{\sqrt {{x^2} + x + 4} + 2}}\\ \Leftrightarrow \sqrt {{x^2} + x + 10} = \sqrt {{x^2} + x + 4} + 2\\ \Leftrightarrow {x^2} + x + 10 = {x^2} + x + 4 + 4 + 4\sqrt {\left( {{x^2} + x + 10} \right)\left( {{x^2} + x + 4} \right)} \\ \Leftrightarrow \sqrt {\left( {{x^2} + x + 10} \right)\left( {{x^2} + x + 4} \right)} = \dfrac{1}{2}\\ + \min {x^2} + x + 10 = \dfrac{{39}}{4}\\ + \min {x^2} + x + 4 = \dfrac{{15}}{4}\\ \sqrt {\left( {{x^2} + x + 10} \right)\left( {{x^2} + x + 4} \right)} > \sqrt {\dfrac{{39}}{4}.\dfrac{{15}}{4}} = \dfrac{{3\sqrt {65} }}{4} > \dfrac{1}{2}\end{array}$
Vậy $S = \left\{ {0; – 1} \right\}$