Giải

a) 24 + (15-x) = 16

<=> 15-x=16-24

<=> 15-x=(-8)

<=> x = 15-(-8) => x = 23

b) 11 - (24-x) = 

<=> 24-x= 11 - (-27)

<=> 24 - x = 38

<=> x = 24 - 38 => x = -14

c) |x-16| = 42

<=> x-16=42 và  x-16=-42

<=> x = 42 + 16 ; x =-42+16

<=> x = 58 ; x = -26

d) 10-2|x| = (-2) . (-3)

<=> 10 - (-2) . (-3) = 2|x|

<=> 10 - 6 = 2|x|

<=> 2|x| = 4

<=> |x| = 4 : 2 = 2

=> x = 2 hoặc x = (-2)

A/ 24+(15-x)=16                                                  

          (15-x)= 16- 24

          (15-x)= - 18

               x = 5 - (-18)

               x = 23

B/  11-(24-x)=(-3)³

          (24-x)=11- 27

          (24-x)= -16

               x = 24 -(- 16)

               x= 40

C/ 

\(\left|x-16\right|=42\hept{\begin{cases}x-16=42\\x-16=-42\end{cases}}\hept{\begin{cases}x=42+16\\x=-42+16\end{cases}}\hept{\begin{cases}x=58\\x=-26\end{cases}}\)

  D/ 10-2|x|=(-2)x(-3)

       10-2|x|=6

          2|x|=10-6

         2|x|=4

           x=4:2

           x=2

Theo bài ra ta cs 

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{8}=\frac{y}{12}\left(1\right)\)

\(\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\left(2\right)\)

Từ (1) ; (2) => \(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

ADTC dãy tỉ số bằng nhau ta cs 

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{12}=2\\\frac{z}{15}=2\end{cases}\Rightarrow\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}}\)

Như vậy  ta chọn : A

\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)

Ta có: \(x=\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)

\(=\sqrt{49-2\cdot7\cdot4\sqrt{3}+48}+\sqrt{48+2\cdot4\sqrt{3}\cdot2+4}\)

\(=\sqrt{\left(7-4\sqrt{3}\right)^2}+\sqrt{\left(4\sqrt{3}+2\right)^2}\)

\(=\left|7-4\sqrt{3}\right|+\left|4\sqrt{3}+2\right|\)

\(=7-4\sqrt{3}+4\sqrt{3}+2\)

\(=9\)

Làm luôn phần y :D

y = \(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)

y = \(\sqrt{33+2.10\sqrt{2}}+\sqrt{24-2.8\sqrt{2}}\)

y = \(\sqrt{33+2.5.2\sqrt{2}}+\sqrt{24-2.4.2\sqrt{2}}\)

y = \(\sqrt{25+2.5.\sqrt{8}+8}+\sqrt{16-2.4.\sqrt{8}+8}\)

y = \(\sqrt{\left(5+\sqrt{8}\right)^2}+\sqrt{\left(4-\sqrt{8}\right)^2}\)

y = |5 + \(\sqrt{8}\)| + |4 - \(\sqrt{8}\)| 

y = 5 + \(\sqrt{8}\) + 4 - \(\sqrt{8}\)   (Vì 4 > \(\sqrt{8}\) nên 4 - \(\sqrt{8}\) > 0)

y = 9

Vậy y = 9

Chúc bn học tốt!

x = 0000

 phát biểu cảm nghĩ đi

\(3-\left(\frac{5}{\frac{3}{8}}+x-\frac{7}{\frac{5}{24}}\right):\frac{16}{\frac{2}{3}}=2\)

\(\Rightarrow3-\left(\frac{40}{3}+x-\frac{168}{5}\right):24=2\)

\(\Rightarrow\left(-\frac{304}{15}+x\right):24=1\)

\(\Rightarrow\frac{-304}{15}+x=24\)

\(\Rightarrow x=\frac{664}{15}\)

\(3-\left(\frac{5}{\frac{3}{8}}+x-\frac{7}{\frac{5}{24}}\right):\frac{16}{\frac{2}{3}}=2\)

\(\left(\frac{40}{3}+x-\frac{28}{5}\right):24=3-2\)

\(\left(\frac{40}{3}+x-\frac{28}{5}\right):24=1\)

\(\frac{40}{3}+x-\frac{28}{5}=1\cdot24\)

\(\frac{116}{15}+x=24\)

\(x=24-\frac{116}{15}\)

\(x=\frac{244}{15}\)

TH1:                                              TH2:

-8(x-3)=24-16:2                              -8(-x+3)=24-16:2

-8x+24=16                                      8x-24=16

-8x=16-24                                       8x=16+24

-8x=-8                                             8x=40

x=1                                                  x=5

Vậy  \(x\in1;5\)

-8/x-3/=24-16:2                                                          Suy ra x-3=-8;8

-8/x-3/=24-8                                                                             x=11;-5

-8/x-3/=16

   /x-3/=16+(-8)

    /x-3/=8
 

\(\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{4}\right)^3\)

\(=\dfrac{x^3}{8}-\dfrac{y^3}{64}\)

\(=\dfrac{8x^3-y^3}{64}\)

\(=\dfrac{\left(2x-y\right)^3+3\cdot2x\cdot y\left(2x-y\right)}{64}\)

\(=\dfrac{16^3+6\cdot24\cdot16}{64}=100\)