a, đk x khác 0

<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)

b, <=> 36x +252 = -360 <=> x = -17 

c. đk x khác -1 

<=> (x+1)^2 = 16 

TH1 : x + 1 = 4 <=> x = 3 (tm)

TH2 : x + 1 = -4 <=> x = -5 (tm) 

d, đk x khác 1/2 

<=> (2x-1)^2 = 81 

TH1 : 2x - 1 = 9 <=> x = 5 (tm) 

TH2 : 2x - 1 = -9 <=> x = -4 (tm) 

a: \(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

b: =>x+7/15=-2/3

=>x+7=-10

hay x=-17

c: \(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)

hay \(x\in\left\{3;-5\right\}\)

a) \(\dfrac{x}{4}=\dfrac{4}{x}\)=>x²=4.4=16 =>x²=4²

=>x=2 hay x=-2.

b) \(\dfrac{x+7}{15}=-\dfrac{24}{36}\)=>\(\dfrac{x+7}{15}=-\dfrac{2}{3}\)=>x+7=-\(\dfrac{2}{3}.15\)=-10 =>x=-17

c)\(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)=>(x+1)²=2.8=16=4²

=>x+1=4 hay x+1=-4

=>x=3 hay x=-5.

d) \(\dfrac{2x-1}{\left(-3\right)^2}=\dfrac{\left(-3\right)^2}{2x-1}\)=>\(\dfrac{2x-1}{9}=\dfrac{9}{2x-1}\)=>(2x-1)²=9²

=>2x-1=9 hay 2x-1=-9

=>x=5 hay x=-4.

Cách 1:

a) \(12:6 = 2\)

b) \(24:\left( { - 8} \right)=-(24:8)=-3\)

c) \(\left( { - 36} \right):9=-(36:9)=-4\)

d) \(\left( { - 14} \right):\left( { - 7} \right)=14:7=2\)

Cách 2:

a) Ta có \(12 = 6.2\) nên \(12:6 = 2\).

b) Ta có \(24 = \left( { - 8} \right).\left( { - 3} \right)\)\( \Rightarrow 24:\left( { - 8} \right) = \left( { - 3} \right)\).

c) Ta có \(\left( { - 36} \right) = 9.\left( { - 4} \right)\) nên \(\left( { - 36} \right):9 = \left( { - 4} \right)\).

d) Ta có \(\left( { - 14} \right) = \left( { - 7} \right).2\) nên \(\left( { - 14} \right):\left( { - 7} \right) = 2\)

a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)

\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)

\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)

\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)

`(4^{-2}:(1/3)^2xx1/2)/((-1/6)^2)`
`=((1/16:1/9)xx1/2)/(1/36)`
`=36xx1/2xx(1/16xx9)`
`=18xx9/16`
`=81/8`

Đề sai rồi bạn

\(\dfrac{1}{2}\) là để giống m vuông í

a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)

= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)

= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)

=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)

=\(\frac{1}{2}-1-1\)

=\(\frac{-3}{2}\)

b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)

= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)

= \(-12:\left\{\frac{-1}{12}\right\}^2\)

= \(-12:\frac{1}{144}\)

= \(-12.144\)

= -1728

c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)

= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)

= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)

= \(\frac{-7}{6}\)

d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)

= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)

= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)

= \(10.\frac{7}{5}\)

= 14

e) (1+23−14).(0,8−34)2

= (1+23−14).(\(\frac{4}{5}\)−34)2

= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)

= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)

= \(\frac{17}{20}.\frac{1}{400}\)

= \(\frac{17}{8000}\)

a)Bạn xem lại đề được không

b)Đặt x^2 ra ngoài

c)Đặt x^3=t rồi quy đồng

d)Bt = -17(x^2-1), còn ẩn phụ gì nữa?

tại thấy thầy ghi đề đặt ẩn phụ nên như vậy,tui cũng nghĩ ra như vậy rùi mà

a) Thử bấm nghiệm đi (Dùng solve)

a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)

\(=1+\left(-1\right)+0,5\)

\(=0,5\)

b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=-12:\left(\dfrac{-1}{12}\right)^2\)

\(=-12:\dfrac{1}{144}\)

\(=-1728\)

c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)

\(=\dfrac{7}{23}.\dfrac{-23}{6}\)

\(=\dfrac{-7}{6}\)

d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)

\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)

\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)

\(=10.\dfrac{7}{5}\)

\(=14\)

e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)

\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)

Bài 1:

a.\(12+\left\{45-\left[36:\left(12-9\right)^2\right]\right\}\)

\(=12+\left\{45-\left[36:3^2\right]\right\}\)

\(=12+\left\{45-\left[36:9\right]\right\}\)

\(=12+\left\{45-4\right\}\)

\(=12+41\)

\(=53\)

b.\(24:\left[48-\left(42:7\right)^2\right]\)

\(=24:\left[48-6^2\right]\)

\(=24:\left[48-36\right]\)

\(=24:12=2\)

Bài 2 : 

C1 :  { 0;1;2;3;4;5;6;7;8;9;10}

C2 : \(\left\{x\in N|x\le10\right\}\)

pha ngoac ra la dc

con bai 2 C1 {1:2:3:4:5:6:7:8:9:0}

C2 {nt huoc N /n<10}

a) 12 + {45 - [36 : (12 - 9)² ]}

= 12 + [45 - (36 : 3²)]

= 12 + [45 - (36 : 9)]

= 12 + (45 - 4)

= 12 + 41

= 53