Question

Bài 13 (trang 11 SGK Toán 9 Tập 1)

Rút gọn các biểu thức sau:

a) $2\sqrt{a^2}-5a$ với $a<0$  ;             b) $\sqrt{25a^2}+3a$ với $a \le 0$;

c) $\sqrt{9a^4}+3a^2$  ;                             d) $5\sqrt{4a^6}-3a^3$ với $a<0$.

Answer 1

a, \(2\sqrt{a^2}-5a=2\left|a\right|-5a\)do a < 0 

\(=-2a-5a=-7a\)

b, \(\sqrt{25a^2}+3a=\sqrt{\left(5a\right)^2}+3a=\left|5a\right|+3a\)do \(a\le0\)

TH1 : \(-5a+3a=-2a\)với \(a< 0\)

hoặc TH2 : \(5+3=8\)

c, \(\sqrt{9a^4}+3a^2=\sqrt{\left(3a^2\right)^2}+3a^2=\left|3a^2\right|+3a^2\)

\(=3a^2+3a^2=6a^2\)do \(3>0;a^2\ge0\forall a\Rightarrow3a^2\ge0\forall a\)

d, \(5\sqrt{4a^6}-3a^3=5\sqrt{\left(2a^3\right)^2}-3a^3\)

\(=5\left|2a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)do \(a< 0\Rightarrow a^3< 0\)

Answer 2

a) \(2\sqrt{a^2}-5a\)=2\(|a|\)-5a = -2a-5a=-7a

b) \(\sqrt{25a^2}\) +3a = 5\(|a|\) + 3a=5a+3a=8a.

c) \(\sqrt{9a^4}\) + 3\(a^2\)=6\(a^2\)

d) \(5\sqrt{4a^6}\) - 3\(a^3\)=-13\(a^3\)

Answer 3

a, 2|a| - 5a = -2a - 5a = -7a

b, 5|a|+3a = 5a + 3a = 8a

c, |3a^2| + 3a^2 = 3a^2 + 3a^2 = 6a^2

d, 5|2a^3| - 3a^3 = 5. (-2a^3) - 3a^3 = -10a^3- 3a^3=-13a^3

Answer 4

a) \(2\sqrt{a^2}-5a=2\left(-a\right)-5a(do.a< 0nên\sqrt{a^2}=-a)=-7a\)

b) \(\sqrt{25a^2}+3a=5a+3a\left(do.a\le0nên\sqrt{25a^2}=5a\right)=8a\)

c) \(\sqrt{9a^4}+3a^2=3a^2+3a^2=6a^2\)

d) \(5\sqrt{4a^6}-3a^3=5.\left(-2a^3\right)-3a^3=-13a^3\)

Answer 5

Answer 6

Answer 7

a) $2\sqrt{a^2}-5a$

$=2|a|-5a$

$=-2a-5a=-7a$ (do $a<0$ nên $|a|=-a$).

b) $\sqrt{25a^2}+3a$

$=5|a|+3a$

$=5a+3a=8a$ (do $a\ge 0$ nên $|a|=a$).

c) $\sqrt{9a^4}+3a^2$

$=\sqrt{(3a^2{)}^2}+3a^2$

$=|3a^2|+3a^2$ 

$=3a^2+3a^2=6a^2$ (do $a^2\ge 0$ với mọi a nên $|3a^2|=3a^2$).

d) $5\sqrt{4a^6}-3a^3$

$=5\sqrt{(2a^3{)}^2}-3a^3$

$=5|2a^3|-3a^3$

$=5.(-2a^3)-3a^3=-10a^3-3a^3=-13a^3$ (do $a<0$ $\Rightarrow$ $2a^3<0$ nên $|2a^3|=–2a^3$).

Answer 8

Answer 9

Answer 10

a) $2\sqrt{a^2}-5a$

$=2|a|-5a$

$=-2a-5a=-7a$ (do $a<0$ nên $|a|=-a$)

b) $\sqrt{25a^2}+3a$

$=5|a|+3a$

$=5a+3a=8a$ (do $a\ge 0$ nên $|a|=a$)

c) $\sqrt{9a^4}+3a^2$

$=\sqrt{(3a^2{)}^2}+3a^2$

$=|3a^2|+3a^2$ 

$=3a^2+3a^2=6a^2$ (do $a^2\ge 0$ với mọi a nên $|3a^2|=3a^2$)

d) $5\sqrt{4a^6}-3a^3$

$=5\sqrt{(2a^3{)}^2}-3a^3$

$=5|2a^3|-3a^3$

$=5.(-2a^3)-3a^3=-10a^3-3a^3=-13a^3$ (do $a<0$ $\Rightarrow$ $2a^3<0$ nên $|2a^3|=–2a^3$)

Answer 11

Answer 12

a) = -2a - 5a = -7a

b) = -5a + 3a = -2a

c) 3a² + 3a²= 6a²

d) -5.2a³ - 3a³ = -13a³

Answer 13

a) $2\sqrt{a^2}-5a$

$=2|a|-5a$

$=-2a-5a=-7a$ (do $a<0$ nên $|a|=-a$).

b) $\sqrt{25a^2}+3a$

$=5|a|+3a$

$=5a+3a=8a$ (do $a\ge 0$ nên $|a|=a$).

c) $\sqrt{9a^4}+3a^2$

$=\sqrt{(3a^2{)}^2}+3a^2$

$=|3a^2|+3a^2$ 

$=3a^2+3a^2=6a^2$ (do $a^2\ge 0$ với mọi a nên $|3a^2|=3a^2$).

d) $5\sqrt{4a^6}-3a^3$

$=5\sqrt{(2a^3{)}^2}-3a^3$

$=5|2a^3|-3a^3$

$=5.(-2a^3)-3a^3=-10a^3-3a^3=-13a^3$ (do $a<0$ $\Rightarrow$ $2a^3<0$ nên $|2a^3|=–2a^3$).

Answer 14

a) $2\sqrt{a^2}-5a$

$=2|a|-5a$

$=-2a-5a=-7a$ (do $a<0$ nên $|a|=-a$).

b) $\sqrt{25a^2}+3a$

$=5|a|+3a$

$=5a+3a=8a$ (do $a\ge 0$ nên $|a|=a$).

c) $\sqrt{9a^4}+3a^2$

$=\sqrt{(3a^2{)}^2}+3a^2$

$=|3a^2|+3a^2$ 

$=3a^2+3a^2=6a^2$ (do $a^2\ge 0$ với mọi a nên $|3a^2|=3a^2$).

d) $5\sqrt{4a^6}-3a^3$

$=5\sqrt{(2a^3{)}^2}-3a^3$

$=5|2a^3|-3a^3$

$=5.(-2a^3)-3a^3=-10a^3-3a^3=-13a^3$ (do $a<0$ $\Rightarrow$ $2a^3<0$ nên $|2a^3|=–2a^3$).

Answer 15

Answer 16

Answer 17

Answer 18

a) Ta có: $2\sqrt{a^2}-5a=2|a|-5a$

$=2.(-a)-5a$ (vì $a<0$ nên $\left|a\right|=-a$)

$=-2a-5a$

$=(-2-5)a$

$=-7a$

Vậy $2\sqrt{a^2}-5a=-7a$.

b) Ta có:  $\sqrt{25a^2}+3a=\sqrt{5^2.a^2}+3a$

$=\sqrt{(5a{)}^2}+3a$

$=\left|5a\right|+3a$ 

$=5a+3a$

$=(5+3)a$

$=8a$.

(vì $a\ge 0\Rightarrow |5a|=5a$ ) 

c) Ta có: $\sqrt{9a^4}+3a^2=\sqrt{3^2.(a^2{)}^2}+3a^2$

$=\sqrt{(3a^2{)}^2}+3a^2$

$=\left|3a^2\right|+3a^2$

$=3a^2+3a^2$

$=(3+3)a^2$

$=6a^2$.

(Vì $a^2\ge 0$ với mọi $a\in R\Rightarrow |3a^2|=3a^2$).

d) Ta có: 

$5\sqrt{4a^6}-3a^3=5\sqrt{2^2.(a^3{)}^2}-3a^3$

$=5.\sqrt{(2a^3{)}^2}-3a^3$

$=5.\left|2a^3\right|-3a^3$

$=\mathrm{5.2.}(-a^3)-3a^3$  (vì $a<0$ nên $|2a^3|=-2a^3$ )

$=10.(-a^3)-3a^3$

$=-10a^3-3a^3$

$=(-10-3)a^3$

$=-13a^3$.

 

Answer 19

Answer 20

a) Ta có: $2\sqrt{a^2}-5a=2|a|-5a$

$=2.(-a)-5a$ (vì $a<0$ nên $\left|a\right|=-a$)

$=-2a-5a$

$=(-2-5)a$

$=-7a$

Vậy $2\sqrt{a^2}-5a=-7a$.

b) Ta có:  $\sqrt{25a^2}+3a=\sqrt{5^2.a^2}+3a$

$=\sqrt{(5a{)}^2}+3a$

$=\left|5a\right|+3a$ 

$=5a+3a$

$=(5+3)a$

$=8a$.

(vì $a\ge 0\Rightarrow |5a|=5a$ ) 

c) Ta có: $\sqrt{9a^4}+3a^2=\sqrt{3^2.(a^2{)}^2}+3a^2$

$=\sqrt{(3a^2{)}^2}+3a^2$

$=\left|3a^2\right|+3a^2$

$=3a^2+3a^2$

$=(3+3)a^2$

$=6a^2$.

(Vì $a^2\ge 0$ với mọi $a\in R\Rightarrow |3a^2|=3a^2$).

d) Ta có: 

$5\sqrt{4a^6}-3a^3=5\sqrt{2^2.(a^3{)}^2}-3a^3$

$=5.\sqrt{(2a^3{)}^2}-3a^3$

$=5.\left|2a^3\right|-3a^3$

$=\mathrm{5.2.}(-a^3)-3a^3$  (vì $a<0$ nên $|2a^3|=-2a^3$ )

$=10.(-a^3)-3a^3$

$=-10a^3-3a^3$

$=(-10-3)a^3$

$=-13a^3$.

 

Answer 21

Answer 22

Answer 23

Answer 24

a) 2\(\sqrt{a^2}\) - 5a = 2\(\left|-a\right|\) -5a = 2(-a) - 5a ( vì a < 0 ) = -7a

b) \(\sqrt{25a^2}\) + 3a = \(\sqrt{5^2.a^2}\) + 3a = \(\sqrt{\left(5a\right)^2}\) + 3a = \(\left|5a\right|\) + 3a = 5a + 3a ( vì a ≤ 0 ) 

= 8a

c) \(\sqrt{9a^4}\) + 3a² = \(\sqrt{\left(3a^2\right)^2}\) + 3a² = \(\left|3a^2\right|\) + 3a² = 3a² + 3a² ( vì a² ≥ 0 ) = 6a²

d) 5\(\sqrt{4a^6}\) - 3a³ = 5\(\sqrt{\left(2a^3\right)^2}\) - 3a³ = 5\(\left|2a^3\right|\) - 3a³ = 5.( -2a³ ) - 3a³ ( vì a < 0 )

= -10a³ - 3a³ = -13a³

Answer 25

a) 2\(\sqrt{a^2}\) - 5a

= 2\(\left|a\right|\) - 5a 

= - 2a - 5a = -7a

b) \(\sqrt{25a^2}\) +3a

= 5\(\left|a\right|\) + 3a

=5a + 3a = 8a

c) \(\sqrt{9a^4}\) +3a\(^2\)

= \(\sqrt{\left(3a^2\right)^2}\) +3a\(^2\)

= \(\left|3a^2\right|\) + 3a\(^2\)

= 3a\(^2\) + 3a\(^2\) = 6a\(^2\)

d) 5\(\sqrt{4a^6}\) - 3a\(^3\)

= 5\(\sqrt{\left(2a^3\right)^2}\) - 3a\(^3\)

= 5\(\left|2a^3\right|\) - 3a\(^3\)

= 5.( -2a\(^3\) ) - 3a\(^3\)

= - 10a\(^3\) - 3a\(^3\) = -13\(^3\)

Answer 26

a) \(2|a|-5a\) = -2a - 5a = -7a

b) \(\sqrt{\left(5a\right)^2}\) + 3a = \(|5a|+3a\) = -5a + 3a = -2a

c) \(\sqrt{\left(3a^2\right)}+3a^2\) = \(|3a|+3a^3\) = 3a + 3a² = 3a(1+a)

d) \(5\sqrt{\left(2a^4\right)}-3a^3\) = \(5|2a^4|-3a^3\) = 10a⁴-3a³ = a³(10a - 3)

Answer 27

a) -7a

b) 8a

c)6a^2

d)-13a^3

Answer 28

a,=2|a|-5a

=-2a-5a=-7a

b,=5|a|+3a

=5a+3a=8a

 

Answer 29

a) $2\sqrt{a^2}-5a$

$=2|a|-5a$

$=-2a-5a=-7a$ (do $a<0$ nên $|a|=-a$).

b) $\sqrt{25a^2}+3a$

$=5|a|+3a$

$=5a+3a=8a$ (do $a\ge 0$ nên $|a|=a$).

c) $\sqrt{9a^4}+3a^2$

$=\sqrt{(3a^2{)}^2}+3a^2$

$=|3a^2|+3a^2$ 

$=3a^2+3a^2=6a^2$ (do $a^2\ge 0$ với mọi a nên $|3a^2|=3a^2$).

d) $5\sqrt{4a^6}-3a^3$

$=5\sqrt{(2a^3{)}^2}-3a^3$

$=5|2a^3|-3a^3$

$=5.(-2a^3)-3a^3=-10a^3-3a^3=-13a^3$ (do $a<0$ $\Rightarrow$ $2a^3<0$ nên $|2a^3|=–2a^3$

Answer 30

\(a) 2\sqrt{a^2} - 5a2 a 2 ​ −5a = 2|a|-5a=2∣a∣−5a = -2a - 5a = -7a=−2a−5a=−7a (do a < 0a<0 nên |a| = -a∣a∣=−a). b) \sqrt{25a^2} + 3a 25a 2 ​ +3a = 5|a| + 3a=5∣a∣+3a = 5a + 3a = 8a=5a+3a=8a (do a \ge 0a≥ 0 nên |a| = a∣a∣=a). c) \sqrt{9a^4} + 3a^2 9a 4 ​ +3a 2 = \sqrt{(3a^2)^2} + 3a^2= (3a 2 ) 2 ​ +3a 2 = |3a^2| + 3a^2=∣3a 2 ∣+3a 2 = 3a^2 + 3a^2 = 6a^2=3a 2 +3a 2 =6a 2 (do a^2 \ge 0a 2 ≥ 0 với mọi a nên |3a^2| = 3a^2∣3a 2 ∣=3a 2 ). d) 5\sqrt{4a^6} - 3a^35 4a 6 ​ −3a 3 = 5\sqrt{(2a^3)^2} - 3a^3=5 (2a 3 ) 2 ​ −3a 3 = 5|2a^3| - 3a^3=5∣2a 3 ∣−3a 3 =5.(-2a^3)-3a^3=-10a^3-3a^3=-13a^3=5.(−2a 3 )−3a 3 =−10a 3 −3a 3 =−13a 3 (do a < 0a<0 \Rightarrow⇒ 2a^3<02a 3 <0 nên |2a^3| = – 2a^3∣2a 3 ∣=–2a 3 ). \)