5 bài tập Rút gọn biểu thức và tính giá trị biểu thức (có lời giải)

a) Ta có: $\sqrt {0{,}49{a^2}} = 0{,}7\left| a \right| = - 0{,}7a$ (do $a \le 0$).

b) Ta có: $\sqrt {{{\left( {\frac{a}{2}} \right)}^4}{{\left( {6 - 2a} \right)}^2}} = {\left( {\frac{a}{2}} \right)^2}\left| {6 - 2a} \right| = \frac{{{a^2}.2\left| {3 - a} \right|}}{4} = \frac{{{a^2}\left( {a - 3} \right)}}{2}$ (do $a > 3$).

c) Ta có: $\sqrt {19.76{{\left( {2 - a} \right)}^2}} = \sqrt {1444{{\left( {2 - a} \right)}^2}} = 38\left| {2 - a} \right| = 38\left( {a - 2} \right)$ (do $a > 2$).

d) Ta có: $\frac{1}{{a - b}}.\sqrt {{a^2}{{\left( {{a^2} - {b^2}} \right)}^2}} = \frac{1}{{a - b}}.\left| a \right|.\left| {{a^2} - {b^2}} \right| = \frac{1}{{a - b}}.a.\left( {a - b} \right)\left( {a + b} \right) = a\left( {a + b} \right)$

(do $a > b \ge 0$).

a) Ta có: $\sqrt {\frac{{2a}}{5}} .\sqrt {\frac{{5a}}{{18}}} = \sqrt {\frac{{10{a^2}}}{{90}}} = \sqrt {\frac{{{a^2}}}{9}} = \frac{a}{3}$ (do $a \ge 0$).

b) Ta có: $\sqrt {11a} .\sqrt {\frac{{99}}{a}} = \sqrt {11a.\frac{{99}}{a}} = \sqrt {1089} = 33$.

c) Ta có: $21a - \sqrt {11a} .\sqrt {44a} = 21a - \sqrt {484{a^2}} = 21a - 22a = - a$ (do $a \ge 0$)

d) Ta có: ${\left( {4 + a} \right)^2} - \sqrt {0{,}4} .\sqrt {160{a^2}} = {\left( {4 + a} \right)^2} - \sqrt {64{a^2}} = {\left( {4 + a} \right)^2} - 8\left| a \right| = 16 + {a^2} + 8a - 8\left| a \right|$

Nếu $a \ge 0$ thì ${\left( {4 + a} \right)^2} - \sqrt {0{,}4} .\sqrt {160{a^2}} = 16 + {a^2}$.

Nếu $a < 0$ thì ${\left( {4 + a} \right)^2} - \sqrt {0{,}4} .\sqrt {160{a^2}} = 16 + {a^2} + 16a$.

a) Ta có:$\sqrt {9{{\left( {4 + 20x + 25{x^2}} \right)}^2}} = \sqrt 9 .\sqrt {{{\left( {2 + 5x} \right)}^4}} = 3{\left( {2 + 5x} \right)^2}$.

Thay $x = - \sqrt 5$ vào biểu thức đã rút gọn, ta được:

$3{\left( {2 + 5x} \right)^2} = 3{\left( {2 - 5\sqrt 5 } \right)^2} = 3\left( {129 - 20\sqrt 5 } \right) \approx 252{,}836$.

b) Ta có:$\sqrt {2{a^2}\left( {2{b^2} - 12b + 18} \right)} = \sqrt {4{a^2}\left( {{b^2} - 6b + 9} \right)} = \sqrt {4{a^2}} .\sqrt {{{\left( {b - 3} \right)}^2}} = 2\left| a \right|.\left| {b - 3} \right|$

Thay $a = - 3,b = \sqrt 3$ vào biểu thức đã rút gọn, ta được:

$2\left| a \right|.\left| {b - 3} \right| = 2\left| { - 3} \right|.\left| {\sqrt 3 - 3} \right| = 6\left( {3 - \sqrt 3 } \right) \approx 7{,}608$.

a) Ta có: $\frac{y}{x}\sqrt {\frac{{{x^2}}}{{{y^4}}}} = \frac{y}{x}.\frac{{\left| x \right|}}{{{y^2}}} = \frac{y}{x}.\frac{x}{{{y^2}}} = \frac{1}{y}$( do $x > 0,y \ne 0$).

b) Ta có $2{y^2}\sqrt {\frac{{{x^4}}}{{4{y^2}}}} = 2{y^2}\sqrt {\frac{{{x^4}}}{{4{y^2}}}} = 2{y^2}\frac{{{x^2}}}{{2\left| y \right|}} = 2{y^2}.\frac{{{x^2}}}{{ - 2y}} = - {x^2}y$.

c) Ta có:. $5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}} = 5xy.\frac{{\sqrt {25{x^2}} }}{{\sqrt {{y^6}} }} = 5xy.\frac{{5\left| x \right|}}{{\left| {{y^3}} \right|}} = 5xy.\frac{{ - 5x}}{{{y^3}}} = - \frac{{25{x^2}}}{{{y^2}}}$ (do $x < 0,y > 0$).

d) $0{,}2{x^3}{y^3}\sqrt {\frac{{16}}{{{x^4}{y^8}}}} = 0{,}2{x^3}{y^3}.\frac{4}{{{x^2}{y^4}}} = \frac{{0{,}8x}}{y}.$

a) Ta có: $a{b^2}\sqrt {\frac{3}{{{a^2}{b^4}}}} = a{b^2}.\frac{{\sqrt 3 }}{{\left| {a{b^2}} \right|}} = a{b^2}.\frac{{\sqrt 3 }}{{ - a{b^2}}} = - \sqrt 3$ ( do $a < 0,b \ne 0$).

b) Ta có $\sqrt {\frac{{27{{\left( {a - 3} \right)}^2}}}{{48}}} = \sqrt {\frac{{9{{\left( {a - 3} \right)}^2}}}{{16}}} = \frac{{3\left| {a - 3} \right|}}{4} = \frac{{3\left( {a - 3} \right)}}{4}$ (do $a > 3$).

c) Ta có: $\sqrt {\frac{{9 + 12a + 4{a^2}}}{{{b^2}}}} = \sqrt {\frac{{{{\left( {3 + 2a} \right)}^2}}}{{{b^2}}}} = \frac{{\left| {3 + 2a} \right|}}{{\left| b \right|}} = \frac{{3 + 2a}}{{ - b}}$. (do$b < 0,a > - 1{,}5$).

d) $\left( {a - b} \right)\sqrt {\frac{{ab}}{{{{\left( {a - b} \right)}^2}}}} = \left( {a - b} \right)\frac{{\sqrt {ab} }}{{\sqrt {{{\left( {a - b} \right)}^2}} }} = \left( {a - b} \right)\frac{{\sqrt {ab} }}{{\left| {a - b} \right|}} = \left( {a - b} \right)\frac{{\sqrt {ab} }}{{b - a}} = - \sqrt {ab}$ (do $a < b < 0)$