bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

1- (5\(\dfrac{4}{9}\) +x+7\(\dfrac{7}{18}\)) : 15\(\dfrac{3}{4}\) = 0 

1- (\(\dfrac{49}{9}+x+\dfrac{133}{18}\)) : \(\dfrac{63}{4}=0\) 

 (\(\dfrac{49}{9}+\dfrac{133}{18}\)+\(x\) ) : \(\dfrac{63}{4}\) = 1 - 0 

       (\(\dfrac{77}{6}\) + \(x\) ) : \(\dfrac{63}{4}\) = 1

         \(\dfrac{77}{6}+x\)         = 1 x \(\dfrac{63}{4}\) 

          \(\dfrac{77}{6}\) + \(x\)          = \(\dfrac{63}{4}\)

                  \(x\)            = \(\dfrac{63}{4}\) - \(\dfrac{77}{6}\)

                  \(x=\) \(\dfrac{35}{12}\)

Trả lời

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\frac{8+4+2+1}{16}=\frac{23}{16}\)

\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)

\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)

\(\Leftrightarrow4x=\frac{8}{16}\)

\(\Leftrightarrow4x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:4\)

\(\Leftrightarrow x=\frac{1}{8}\)

Vậy x=\(\frac{1}{8}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\left(\frac{8+4+2+1}{16}\right)=\frac{23}{16}\)

\(\Leftrightarrow4x+\frac{15}{16}=\frac{23}{16}\)

\(\Leftrightarrow4x=\frac{23}{16}-\frac{15}{16}\)

\(\Leftrightarrow4x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:4\)

\(\Leftrightarrow x=\frac{1}{8}\)

x=1/8

mình nhanh nhất

Nhưng Azure phan bảo linh ơi, mình cần cả bài giải cơ nếu bạn ghi hẳn bài giải ra thì mình sẽ k cho nhé

a x * 4 + [1/2 + 1/4 + 1/8 + 1/6] = 23/16

x * 4 +             25/24             = 23/16

x * 4                                     = 23/16 - 25/24

x * 4                                     =   38/96 = 19/48

x                                          = 19/48 / 4

x                                          = 19/172

b x *         1              = 425

x                              = 425 / 1

x                              =    425

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

a: Ta có: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)

\(\Leftrightarrow x^2+5x-10=x^2+3x-4\)

\(\Leftrightarrow2x=6\)

hay x=3

4².4^x=(4²)^x+1

=>4^x+2=4^2x+2

=>x+2=2x+2

=>x=0

Bài 2: 

a: Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)

\(=\left(x^2-4\right)\left(x^2+4\right)\)

\(=x^4-16\)

b: Ta có:\(\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)

\(=x^3+y^3\)

Bài 1: 

Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x+1\right)\left(x+3\right)+3x^2=0\)

\(\Leftrightarrow x^3+64-x\left(x^2+4x+3\right)+3x^2=0\)

\(\Leftrightarrow x^3+64-x^3-4x^2-3x+3x^2=0\)

\(\Leftrightarrow-x^2-3x+64=0\)

\(\Leftrightarrow x^2+3x-64=0\)

\(\text{Δ}=3^2-4\cdot1\cdot\left(-64\right)=265\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{265}}{2}\\x_2=\dfrac{-3+\sqrt{265}}{2}\end{matrix}\right.\)

4x=1-(1/16+1/4+1/5+1/2)=0,0625=1/16

x=1/16/4=1/64

hỏi google ấy 

x=1/64 nha