7574675

Lời giải:

\(\left\{x^2-[8^2-(5^2-8.3)^3-7.9]^3-4.12\right\}^3=1\\ \Rightarrow x^2-[8^2-(5^2-8.3)^3-7.9]^3-4.12=1\\ \Rightarrow x^2-(64-1-63)^3-48=1\\ \Rightarrow x^2-48=1\\ \Rightarrow x^2=49=7^2=(-7)^2\\ \Rightarrow x=\pm 7\)

\(\left\{x^2-\left[8^2-\left(5^2-8.3\right)^3-7.9\right]^3-4.12\right\}^3=8000.\)

\(\Rightarrow\left\{x^2-\left[8^2-\left(25-24\right)^3-7.9\right]^3-4.12\right\}^3=8000\)

\(\Rightarrow\left\{x^2-\left[8^2-\left(25-24\right)^3-63\right]^3-48\right\}^3=8000\)

\(\Rightarrow\left\{x^2-\left[64-1-63\right]-48\right\}^3=8000\)

\(\Rightarrow\left(x^2-48\right)^3=8000\)\(\Rightarrow x^2-48=20\)

\(\Rightarrow x^2=68\)\(\Rightarrow x=2\sqrt{17}\)

1/4.12/13+1/4.1/13-3/25

1/4.(12/13+1/13)-3/25

1/4.1-3/25

1/4-3/25

1/8

\(\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-12\%=\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-\frac{3}{25}=\frac{1}{4}\cdot\left(\frac{12}{13}+\frac{1}{13}\right)-\frac{3}{25}\)

\(=\frac{1}{4}\cdot1-\frac{3}{25}=\frac{1}{4}-\frac{3}{25}=\frac{13}{100}\)

Nhớ bài đây sửa đi sửa lại cũng vì do cái số " % " :(((

a) \(\left|\frac{2}{5}:x\right|=\frac{1}{4}\)

Trường hợp 1 : \(\frac{2}{5}\) : x = \(\frac{1}{4}\)

=> x = \(\frac{2}{5}:\frac{1}{4}=\frac{2}{5}\cdot4=\frac{8}{5}\)

Trường hợp 2 : \(\frac{2}{5}:x=-\frac{1}{4}\)

=> \(x=\frac{2}{5}:\left(-\frac{1}{4}\right)=\frac{2}{5}\cdot\left(-4\right)=-\frac{8}{5}\)

Vậy \(x=\pm\frac{8}{5}\)

b) \(\frac{x}{24}=-\frac{1}{3}-\frac{1}{8}=-\frac{11}{24}\)

=> x = -11

c) \(\frac{3}{x+3}=\frac{-7}{21}\)

=> \(\frac{3}{x+3}=\frac{-1}{3}\)

=> -1(x + 3) = 9

=> -x - 3 = 9

=> -x = 12

=> x = -12

\(\frac{1}{4}.\frac{12}{13}+\frac{1}{4}.\frac{1}{13}-12\%\)

=\(\left(\frac{1}{4}+\frac{1}{4}\right).\left(\frac{12}{13}+\frac{1}{13}\right)-\frac{3}{25}\)

=\(\frac{1}{2}.1-\frac{3}{25}\)

=\(\frac{1}{2}-\frac{3}{25}\)

=\(\frac{25}{50}-\frac{6}{50}\)

=\(\frac{19}{50}\)

Tìm x

a)\(\left|\frac{2}{5}:x\right|=\frac{1}{4}\)

=>\(\hept{\begin{cases}\frac{2}{5}:x=\frac{1}{4}\\\frac{2}{5}:x=\frac{-1}{4}\end{cases}}\)

=>\(\hept{\begin{cases}x=\frac{2}{5}:\frac{1}{4}\\x=\frac{2}{5}:\frac{-1}{4}\end{cases}}\)

=>\(\hept{\begin{cases}x=\frac{8}{5}\\x=\frac{-8}{5}\end{cases}}\)

+)Vậy x \(\varepsilon\left[\frac{8}{5};\frac{-8}{5}\right]\)

\(30-\left[4\left(x-2\right)+15\right]=3\)                                                \(\left(8x-120:4\right).3^3=3^6\)

\(4\left(x-2\right)+15=27\)                                                            \(\left(8x-120:4\right)=27\)

\(4\left(x-2\right)=12\)                                                                               \(8x-30=27\)

\(x-2=3\)                                                                                          \(8x=57\)

\(x=5\)                                                                                                        \(x=\frac{57}{8}\)

Chia 2 vế cho \(27^x\) ta được:

\(3\left(\dfrac{8}{27}\right)^x+4\left(\dfrac{12}{27}\right)^x-\left(\dfrac{18}{27}\right)^x-2=0\)

\(\Leftrightarrow3\left(\dfrac{2}{3}\right)^{3x}+4\left(\dfrac{2}{3}\right)^{2x}-\left(\dfrac{2}{3}\right)^x-2=0\)

Đặt \(\left(\dfrac{2}{3}\right)^x=t>0\)

\(\Rightarrow3t^3+4t^2-t-2=0\)

\(\Leftrightarrow\left(t+1\right)^2\left(3t-2\right)=0\)

\(\Rightarrow t=\dfrac{2}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=\dfrac{2}{3}\)

\(\Rightarrow x=1\)

15: \(\Leftrightarrow\left\{x^2-\left[8^2-\left(25-24\right)^3-63\right]^3-48\right\}=1\)

\(\Leftrightarrow x^2-48=1\)

=>x=7 hoặc x=-7