(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)

#Học tốt!!!

\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)

\(a\cdot b\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{a\cdot b}\)

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)

Bài 1 :

Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)

Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)

Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)

Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)

Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)

Bài 2 :

Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)

Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)

Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)

Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)

Bài 3 :

Câu a : \(\sqrt{\dfrac{61^2-60^2}{81}}=\sqrt{\dfrac{\left(61-60\right)\left(61+60\right)}{81}}=\sqrt{\dfrac{121}{81}}=\dfrac{\sqrt{121}}{81}=\dfrac{11}{9}\)

Câu b : \(\sqrt{\dfrac{74^2-24^2}{121}}=\sqrt{\dfrac{\left(74+24\right)\left(74-24\right)}{121}}=\sqrt{\dfrac{98.50}{121}}=\dfrac{\sqrt{4900}}{\sqrt{121}}=\dfrac{70}{11}\)

Bài 4 :

Câu a : \(9-4\sqrt{x}=1\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Câu b : \(\sqrt{\dfrac{x}{5}}=4\Leftrightarrow\dfrac{x}{5}=16\Leftrightarrow x=80\)

Câu c : \(\sqrt{7x}< 9\Leftrightarrow7x< 81\Leftrightarrow x< \dfrac{81}{7}\)

a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)

b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)

=10+3/64

=643/64

c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)

\(B=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{36}{x-9}\)

\(=\dfrac{\left(\sqrt{x}+3\right)^2+\left(\sqrt{x}-3\right)^2-36}{x-9}\)

\(=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{x-9}\)

\(=\dfrac{2x-18}{x-9}=\dfrac{2\left(x-9\right)}{x-9}=2\)

ĐKXĐ : \(x\ge0;x\ne9\)

Ta có : \(B=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=2\)

\(B=\dfrac{x+6\sqrt{x}+9-x+6\sqrt{x}-9-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{12\sqrt{x}-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{12\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{12}{\sqrt{x}+3}\)

a: \(A=2^{\dfrac{1}{3}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{3}+\dfrac{2}{3}}=2^{\dfrac{3}{3}}=2^1=2\)

b: \(B=36^{\dfrac{3}{2}}=\left(6^2\right)^{\dfrac{3}{2}}=6^{2\cdot\dfrac{3}{2}}=6^3=216\)

c: \(C=36^{\dfrac{3}{2}}\cdot\left(\dfrac{1}{6}\right)^2=\left(6^2\right)^{\dfrac{3}{2}}\cdot\dfrac{1}{6^2}=\dfrac{6^{2\cdot\dfrac{3}{2}}}{6^2}=\dfrac{6^3}{6^2}=6\)

d: \(D=\sqrt{81}\cdot\left(\dfrac{1}{3}\right)^2=9\cdot\dfrac{1}{3^2}=9\cdot\dfrac{1}{9}=1\)

e: \(E=\left(3+2\sqrt{2}\right)^{50}\cdot\left(3-2\sqrt{2}\right)^{50}\)

\(=\left[\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)\right]^{50}\)

\(=\left(9-8\right)^{50}=1^{50}=1\)

f: \(F=120^{\sqrt{5}+1}\cdot120^{3-\sqrt{5}}\)

\(=120^{\sqrt{5}+1+3-\sqrt{5}}=120^4\)

g: \(G=\left(3+2\sqrt{2}\right)^{2019}\cdot\left(3\sqrt{2}-4\right)^{2018}\)

\(=\left(3+2\sqrt{2}\right)^{2018}\cdot\left(3\sqrt{2}-4\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)

\(=\left[\left(3+2\sqrt{2}\right)\left(3\sqrt{2}-4\right)\right]^{2018}\left(3+2\sqrt{2}\right)\)

\(=\left(9\sqrt{2}-12+12-8\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)

\(=\left(\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)=2^{\dfrac{1}{2}\cdot2018}\cdot\left(3+2\sqrt{2}\right)\)

\(=2^{1009}\cdot\left(3+2\sqrt{2}\right)\)

b: =3/4x(-1)=-3/4

B1

a. = 7/3. ( 37/5 - 32/5)

= 7/3 . 1

= 7/3

Phần b có gì đó sai sao lại có 3:+

c. = 4 + 6 - 3 + 5

= 12

d. = -5/21 : -19/21 : 4/5

= 25/76

B2

a. 1/4 : x =1/2 - 3/4

x = -1/4

x = 1/4 : -1/4

x = -1

b. 2 . | 2x - 3 | = 4 - (-8)

2 . | 2x - 3| = 12

| 2x - 3 | = 12:2

| 2x - 3 | = 6

| x - 3 | = 6:2

| x - 3 | = 3

=> x - 3 = +- 3

* x - 3 = 3

x = 6

* x - 3 = -3

x = 0

Chúc bạn vui vẻ

b. = 3 : 9/4 + 1/9 .6

= 4/3 + 2/3

= 2

sr mik nhầm phải là

3:\(\left(-\dfrac{3}{2}\right)^2\)+\(\dfrac{1}{9}\).\(\sqrt{36}\)