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

\(a,=x^4+6x^3+8x^2\\ b,=x^2+3x-28\\ c,=x^2-3x-x^2+6x-9+9=3x\)

a: \(\frac{4\left(x+3\right)}{3x^2-x}:\frac{x^2+3x}{1-3x}\)

\(=\frac{4\left(x+3\right)}{x\left(3x-1\right)}\cdot\frac{-\left(3x-1\right)}{x\left(x+3\right)}=\frac{-4}{x^2}\)

b: \(\frac{x+1}{x^2-2x-8}\cdot\frac{4-x}{x^2+x}\)

\(=\frac{x+1}{\left(x-4\right)\left(x+2\right)}\cdot\frac{-\left(x-4\right)}{x\left(x+1\right)}=\frac{-1}{x\left(x+2\right)}\)

c: \(\frac{9x+5}{2\left(x-1\right)\left(x+3\right)^2}-\frac{5x-7}{2\left(x-1\right)\left(x+3\right)^2}\)

\(=\frac{9x+5-5x+7}{2\left(x-1\right)\left(x+3\right)^2}=\frac{4x+12}{2\left(x-1\right)\left(x+3\right)^2}\)

\(=\frac{4\left(x+3\right)}{2\left(x-1\right)\left(x+3\right)^2}=\frac{2}{\left(x-1\right)\left(x+3\right)}\)

d: \(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x}{x^2-9}\)

\(=\frac{18}{\left(x+3\right)\left(x-3\right)^2}-\frac{3}{\left(x-3\right)^2}-\frac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{18-3x-9-x^2+3x}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{-x^2+9}{\left(x-3\right)^2\left(x+3\right)}=\frac{-1}{x-3}\)

e: \(\frac{1}{x^2-x+1}+\frac{1}{1-x^2}+\frac{2}{x^3+1}\)

\(=\frac{1}{x^2-x+1}-\frac{1}{\left(x+1\right)\left(x-1\right)}+\frac{2}{\left(x+1\right)\cdot\left(x^2-x+1\right)}\)

\(=\frac{\left(x+1\right)\left(x-1\right)-x^2+x-1+2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)}=\frac{x^2-1-x^2+x-1+2x-2}{\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)}\)

\(=\frac{3x-4}{\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)}\)

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

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7574675

Lời giải:

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