a) \(5\frac{8}{17}:x+\frac{-1}{17}:x+3\frac{1}{17}:17\frac{1}{3}=\frac{4}{17}\)

\(\frac{93}{17}:x+\frac{-1}{17}:x+\frac{52}{17}:\frac{52}{3}=\frac{4}{17}\)

\(\left(\frac{93}{17}+\frac{-1}{17}\right):x+\frac{52}{17}.\frac{3}{52}=\frac{4}{17}\)

\(\frac{92}{17}:x+\frac{3}{17}=\frac{4}{17}\)

\(\frac{92}{17}:x=\frac{4}{17}-\frac{3}{17}\)

\(\frac{92}{17}:x=\frac{1}{17}\)

\(x=\frac{92}{17}:\frac{1}{17}\)

\(x=92\)

b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)

\(1-\frac{1}{x+3}=\frac{18}{19}\)

\(\frac{1}{x+3}=1-\frac{18}{19}\)

\(\frac{1}{x+3}=\frac{1}{19}\)

\(\Rightarrow x+3=19\)

\(\Rightarrow x=19-3\)

\(\Rightarrow x=16\)

\(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)

\(\Rightarrow\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}-\frac{x+11}{15}-\frac{x+11}{16}=0\)

\(\Rightarrow\left(x+11\right)\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)

Mà \(\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)\ne0\)

\(\Rightarrow x+11=0\Rightarrow x=-11\)

hk bít tính A

a: ĐKXĐ của A là: \(\begin{cases}x+2<>0\\ x^2-4<>0\\ x^2+3x+2<>0\end{cases}\)

=>\(\begin{cases}x<>-2\\ x^2<>4\\ \left(x+1\right)\left(x+2\right)<>0\end{cases}\)

=>x∉{-2;2;-1}

ĐKXĐ cua B là \(x^3-1<>0\)

=>\(x^3<>1\)

=>x<>1

b: \(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\)

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

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

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

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x+2}\cdot\frac{x^2+3x+2}{x^2+x+1}\)

\(=\frac{-16}{\left.\left(x+2\right)^2\right.}\cdot\frac{x+2}{16}\cdot\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}=\frac{-\left(x+1\right)}{x^2+x+1}\)

\(B=\frac{x^2+x-2}{x^3-1}\)

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

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

b: Đặt P=A+B

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

\(=\frac{1}{x^2+x+\frac14+\frac34}=\frac{1}{\left(x+\frac12\right)^2+\frac34}\le1:\frac34=\frac43\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x+1/2=0

=>x=-1/2

\(x+x+x\times2+x\times4+x\times2=2016\)

\(x\times1+x\times1+x\times2+x\times4+x\times2=2016\)

\(x\times\left(1+1+2+4+2\right)=2016\)

\(x\times10=2016\)

\(x=2016\div10\)

\(x=201,6\)

\(x+x+x\times2+x\times4+x\times2=2016\)

\(x\times2+x\times2+x\times4+x\times2=2016\)

                    \(x\times\left(2+2+4+2\right)=2016\)

                                                \(x\times10=2016\)

                                                            \(x=2016:10\)

                                                             \(x=201,6\)

ai k cho sakura trường mình khác mình thì mình k lại

a) <=? |(x-1/4)| = 1/4-x

Th1: x >= 1/4 => x - 1/4 = 1/4 - x

<=> 2x = 2.1/4 <=> x = 1/4(nhân)

Th2: x<1/4 => -x + 1/4 = 1/4-x

<=> 0x = 0

<=> x thuộc R và x <1/4.

Vậy S ={x|x<=1/4}

\(\text{a)}\sqrt{x^2-\frac{1}{2}x+\frac{1}{16}}=\frac{1}{4}-x\)

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

\(\Leftrightarrow\sqrt{\left(x-\frac{1}{4}\right)^2}=\frac{1}{4}-x\)

\(\Leftrightarrow x-\frac{1}{4}=\frac{1}{4}-x\)

\(\Leftrightarrow2x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{4}\)

\(\text{b)}\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)

\(ĐKXĐ:x\ge-2\)

\(\Leftrightarrow\left(\sqrt{x-2\sqrt{x-1}}\right)^2=\left(\sqrt{x-1}-1\right)^2\)

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

\(\Leftrightarrow x-2\sqrt{x-1}=x-1-2\sqrt{x-1}+1\)

\(\Leftrightarrow x-2\sqrt{x-1}-x+2\sqrt{x-1}=-1+1\)

\(\Leftrightarrow0x=0\)

Vậy \(S=\left\{x\inℝ|x\ge-2\right\}\)

A = ( x - 3 )³ - ( x + 1 )³ + 12x( x - 1 )

= x³ - 9x² + 27x - 27 - ( x³ + 3x² + 3x + 1 ) + 12x² - 12x

= x³ - 9x² + 27x - 27 - x³ - 3x² - 3x - 1 + 12x² - 12x

= ( x³ - x³ ) + ( 12x² - 9x² - 3x² ) + ( 27x - 3x - 12x ) + ( -27 - 1 )

= 12x - 28

+)Với x = -2/3 => A = \(12\times\left(-\frac{2}{3}\right)-28=-8-28=-36\)

+) Để A = -16 => 12x - 28 = -16

                      => 12x = 12

                      => x = 1

a) \(A=\left(x-3\right)^3-\left(x+1\right)^3+12x\left(x-1\right)\)

\(=\left(x^3-9x^2+27x-27\right)-\left(x^3+3x^2+3x+1\right)+\left(12x^2-12x\right)\)

\(=12x-28\)

b) Thay \(x=\frac{-2}{3}\)vào biểu thức A ta có:

\(A=12.\left(\frac{-2}{3}\right)-28=-36\)

Vậy giá trị của A là -36 tại x=-2/3

c) \(A=-16\Rightarrow12x-28=-16\)

\(\Leftrightarrow12x=-16+28\Leftrightarrow12x=12\Leftrightarrow x=1\)

Vậy để A=-16 thì x=1