a) = \(12a^2b\left(a^2-b^2\right)\)

= \(12a^4b-12a^2b^3\)

b)nhân ra :

= \(2x^4-16x^3+4x^2-3x^3+24x^2-6x+5x^2-40x+10\)

= \(2x^4-19x^3+33x^2-46x+10\)

Tìm x:

a) \(\frac{1}{4}x^2-\left(\frac{1}{4}x^2-2x\right)=-14\)

= \(\frac{1}{4}x^2-\frac{1}{4}x^2+2x=-14\)

=\(2x=-14=>x=-7\)

b) \(x^3+27-x\left(x^2-1\right)=27\)

= \(x^3+27-x^3+x=27\)

= \(27+x=27=>x=0\)

Lời giải:

Tập A sửa lại thành \(A=\left\{\frac{1}{6};\frac{1}{12};\frac{1}{20}; \frac{1}{30};....;\frac{1}{420}\right\}\)

Ta thấy:

\(\frac{1}{6}=\frac{1}{2.3}\)

\(\frac{1}{12}=\frac{1}{3.4}\)

\(\frac{1}{20}=\frac{1}{4.5}\)

.....

\(\frac{1}{420}=\frac{1}{20.21}\)

Do đó công thức tổng quát của các phần tử thuộc tập A là \(\frac{1}{x(x+1)}|x\in \mathbb{N}; 2\leq x\leq 20\)

Đáp án D.

a) \(\frac{x-3}{3}-1=\frac{x}{-4}\)

\(\Leftrightarrow\frac{x-3}{3}-\frac{3}{3}=\frac{x}{-4}\)

\(\Leftrightarrow\frac{x-6}{3}=\frac{x}{-4}\)

\(\Leftrightarrow-4\left(x-6\right)=3x\)

\(\Leftrightarrow-4x+24=3x\)

\(\Leftrightarrow24=3x+4x\)

\(\Leftrightarrow7x=24\)

\(\Leftrightarrow x=\frac{24}{7}\)

b) \(\frac{5}{8}-\left(x-\frac{1}{2}\right)=\frac{-3}{4}\)

\(\Leftrightarrow\frac{5}{8}-x+\frac{1}{2}=\frac{-3}{4}\)

\(\Leftrightarrow\frac{5}{8}+\frac{4}{8}-x=\frac{-3}{4}\)

\(\Leftrightarrow\frac{9}{8}-x=\frac{-3}{4}\)

\(\Leftrightarrow x=\frac{9}{8}+\frac{3}{4}\)

\(\Leftrightarrow x=\frac{15}{8}\)

c) \(\frac{x}{38}-\frac{1}{19}=\frac{-1}{2}\)

\(\Leftrightarrow\frac{x}{38}-\frac{2}{38}=\frac{-1}{2}\)

\(\Leftrightarrow\frac{x-2}{38}=\frac{-1}{2}\)

\(\Leftrightarrow2\left(x-2\right)=-38\)

\(\Leftrightarrow2x-4=-38\)

\(\Leftrightarrow2x=-38+4\)

\(\Leftrightarrow2x=-34\)

\(\Leftrightarrow x=-34\div2\)

\(\Leftrightarrow x=-17\)

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

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Ta lại có: 

\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)^3=1\)

\(\Leftrightarrow x+y+z=1\)

Làm nốt

\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)

\(\Rightarrow5,5-\left|x-0,4\right|=-\frac{6}{5}\)

\(\Rightarrow-\left|x-0,4\right|=-\frac{6}{5}-5,5=-6,7\)

\(\Rightarrow\left|x-0,4\right|=6,7\)

\(\Rightarrow x-0,4=\pm6,7\)

\(\Rightarrow\orbr{\begin{cases}x-0,4=6,7\\x-0,4=-6,7\end{cases}\Rightarrow\orbr{\begin{cases}x=7,1\\x=-6,3\end{cases}}}\)

\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)

=> \(\left|x-0,4\right|=5,5-\left[-\frac{6}{5}\right]=5,5+1,2=6,7\)

=> \(\left|x-0,4\right|=\pm6,7\)

Xét hai trường hợp :

TH1 : x - 0,4 = 6,7

=> x  = 6,7 + 0,4 = 7,1

TH2 : x - 0,4 = -6,7

=> x = -6,7 + 0,4 =-6,3

\(b,\left[1-\frac{3}{4}\left|x\right|\right]^2=\frac{16}{25}\)

=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\sqrt{\frac{16}{25}}\)

=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\frac{4}{5}\)

=> \(\orbr{\begin{cases}1-\frac{3}{4}\left|x\right|=\frac{4}{5}\\1-\frac{3}{4}\left|x\right|=-\frac{4}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\pm\frac{4}{15}\\x=\pm\frac{12}{5}\end{cases}}\)

\(c,\left[0,1\left|x\right|-\frac{1}{2}\right]\left[0,5-\left|x\right|\right]=0\)

=> \(\orbr{\begin{cases}0,1\left|x\right|-\frac{1}{2}=0\\0,5-\left|x\right|=0\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{1}{10}\left|x\right|=\frac{1}{2}\\\left|x\right|=0,5\end{cases}}\)

=> \(\orbr{\begin{cases}\left|x\right|=5\\\left|x\right|=0,5\end{cases}}\)=> \(\orbr{\begin{cases}x\in\left\{5;-5\right\}\\x\in\left\{0,5;-0,5\right\}\end{cases}}\)

d, Xét hai trường hợp rồi ra kết quả thôi

a/ - Với \(x\ge1\):

\(\Leftrightarrow x^2-3x+2+x-1=0\)

\(\Leftrightarrow x^2-2x+1=0\Rightarrow x=1\)

- Với \(x< 1\)

\(\Leftrightarrow x^2-3x+2+1-x=0\)

\(\Leftrightarrow x^2-4x+3=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=3\left(l\right)\end{matrix}\right.\)

Vậy pt có nghiệm duy nhất \(x=1\)

b/ ĐKXĐ: ...

\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}\right)+16+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)

\(\Leftrightarrow\left(x+4\right)^2=16\Rightarrow\left[{}\begin{matrix}x+4=4\\x+4=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-8\end{matrix}\right.\)

\(a,5\frac{4}{7}:x=13\Leftrightarrow x=\frac{39}{7}:13\Leftrightarrow x=\frac{39}{7}.\frac{1}{13}=\frac{3}{7}\)

\(b,\left(2,8x-32\right):\frac{2}{3}=-90\)

\(\Leftrightarrow2,8x-32=-90.\frac{2}{3}=-60\)

\(\Leftrightarrow2,8x=-60+32=-28\)

\(\Leftrightarrow x=\frac{-28}{2,8}=-10\)

d, \(7x=3,2+3x\Leftrightarrow7x-3x=3,2\Leftrightarrow4x=3,2\Leftrightarrow x=3,2:4=3,2.\frac{1}{4}=\frac{4}{5}\)

Câu c bị sai đề :\(\frac{19}{10}-1-\frac{2}{5}=\frac{1}{2}\ne1\)bạn nha.

mình lộn \(\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=\frac{13}{10}\ne1\)ms đúng nha

a) \(\frac{39}{7}:x=13\)

     \(x=\frac{39}{7}:13\)

      \(x=\frac{3}{7}\)

b) \(2,8x-32=-90.\frac{2}{3}\)

  \(2,8x-32=-60\)  

    \(2,8x=-60+32\)

     \(2,8x=-28\)

     \(x=-28:2,8\)

     \(x=-10\)

c)  mik nghĩ b nên xem lại đề có đúng là có 2 dấu = ko

d) \(7x-3x=3,2\)( quy tắc chuyển vế )

    \(4x=3,2\)

    \(x=3,2:4\)

    \(x=0,8\)

tk nha