Tìm x, biết \(\eqalign{& a)\,\sqrt {1,69} .\left( {2\sqrt x + \sqrt {{{81} \over {121}}} } \right) = {{13} \over {10}} \cr & b)\,\left( {2{x^2} – 3} \right)\left( {3{x^2} – {1 \over {0,12}}} \right)\left( {{x^2} + 1} \right) = 0 \cr & c)\,x – 5\sqrt x = 0 \cr & d)\,2{x^7} = 3{x^9} \cr} \)

Tìm x, biết

\(\eqalign{& a)\,\sqrt {1,69} .\left( {2\sqrt x + \sqrt {{{81} \over {121}}} } \right) = {{13} \over {10}} \cr & b)\,\left( {2{x^2} – 3} \right)\left( {3{x^2} – {1 \over {0,12}}} \right)\left( {{x^2} + 1} \right) = 0 \cr & c)\,x – 5\sqrt x = 0 \cr & d)\,2{x^7} = 3{x^9} \cr} \)

A. \(a)\,{1 \over {121}}\)

\(b)\,x = \pm \sqrt {{3 \over 2}} \) hoặc \(x = \pm {5 \over 3}\)

c)x = 0 hoặc x = 25

x = 0 hoặc \(x = \pm \sqrt {{2 \over 3}} \)

B. \(a)\,{1 \over {122}}\)

\(b)\,x = \pm \sqrt {{3 \over 2}} \) hoặc \(x = \pm {4 \over 3}\)

c)x = 0 hoặc x = 25

x = 0 hoặc \(x = \pm \sqrt {{2 \over 3}} \)

C. \(a)\,{1 \over {121}}\)

\(b)\,x = \pm \sqrt {{3 \over 2}} \) hoặc \(x = \pm {5 \over 3}\)

c)x = 0 hoặc x = 20

x = 0 hoặc \(x = \pm \sqrt {{2 \over 5}} \)

D. \(a)\,{1 \over {121}}\)

\(b)\,x = \pm \sqrt {{5 \over 2}} \) hoặc \(x = \pm {5 \over 3}\)

c)x = 1 hoặc x = 25

x = 0 hoặc \(x = \pm \sqrt {{2 \over 3}} \)

Hướng dẫn

Chọn đáp án là: A

Phương pháp giải:

+ Tìm x theo thứ tự thực hiện phép tính.

\(\eqalign{ & {\rm{a)}}\sqrt {1,69} .\left( {2\sqrt x + \sqrt {{{81} \over {121}}} } \right) = {{13} \over {10}} \cr & 1,3.\left( {2\sqrt x + {9 \over {11}}} \right) = {{13} \over {10}} \cr & {{13} \over {10}}.\left( {2\sqrt x + {9 \over {11}}} \right) = {{13} \over {10}} \cr & \left( {2\sqrt x + {9 \over {11}}} \right) = 1 \cr & 2\sqrt x = 1 – {9 \over {11}} \cr & 2\sqrt x = {2 \over {11}} \cr & \sqrt x = {1 \over {11}} \cr & x = {1 \over {121}} \cr} \) \(\eqalign{& {\rm{b)}}\left( {2{x^2} – 3} \right)\left( {3{x^2} – {1 \over {0,12}}} \right) = 0 \cr & \matrix{\matrix{+ ){\rm{TH1:}}2{x^2} – 3 = 0 \hfill \cr 2{x^2} = 3 \hfill \cr {x^2} = {3 \over 2} \hfill \cr x = \pm \sqrt {{3 \over 2}} \hfill \cr \hfill \cr} & \matrix{+ ){\rm{TH2}}:3{x^2} – {1 \over {0,12}} = 0 \hfill \cr 3{x^2} = {1 \over {0,12}} \hfill \cr {x^2} = {1 \over {0,36}} \hfill \cr x = \pm \sqrt {{1 \over {0,36}}} = \pm {1 \over {0,6}} = \pm {5 \over 3} \hfill \cr} \cr } \cr} \)

\(\eqalign{& c)\,\,x – 5\sqrt x = 0 \cr & \sqrt x .\left( {\sqrt x – 5} \right) = 0 \cr & \Rightarrow \sqrt x = 0;\sqrt x – 5 = 0 \cr & + )\sqrt x = 0 \Rightarrow x = 0 \cr & + )\sqrt x – 5 = 0 \Rightarrow \sqrt x = 5 \Rightarrow x = 25 \cr} \) \(\eqalign{& d)\,2{x^7} = 3{x^9} \cr & 2{x^7} – 3{x^9} = 0 \cr & {x^7}\left( {2 – 3{x^2}} \right) = 0 \cr & \Rightarrow {x^7} = 0;2 – 3{x^2} = 0 \cr & + ){x^7} = 0 \Rightarrow x = 0 \cr & + )2 – 3{x^2} = 0 \Rightarrow 3{x^2} = 2 \Rightarrow {x^2} = {2 \over 3} \Rightarrow x = \pm \sqrt {{2 \over 3}} \cr} \)