(2x - 1/3)= 2

2x= 2+1/3

2x= 7/3

x= 7/3 : 2

x= 1,66666

a: =>2x-1/3=2

=>2x=7/3

=>x=7/6

b: Sửa đề: 1/11-5/22

\(\dfrac{8}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{8}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:\dfrac{-9}{15}\)

\(=\dfrac{8}{9}\cdot\dfrac{-22}{3}+\dfrac{5}{9}\cdot\dfrac{-5}{3}\)

\(=\dfrac{-66-25}{27}=\dfrac{-91}{27}\)

a,        2x - 15 = 17

           2x       = 17 +15

          2x        = 32

            x          =16

  b,          (7x - 11)³  =2⁵ . 5² +88      

                (7x - 11)³ =888                                                   

                 7x         =888+ 11³                            7x    =2219                  x=   317

\(2^{x-2}-3.2^x=-88\)

\(2^x:2^2-2^x:\frac{1}{3}=-88\)

\(2^x:\left(2^2-\frac{1}{3}\right)=-88\)

\(2^x:\frac{11}{3}=-88\)

\(2^x=-88.\frac{11}{3}\)

\(2^x=\frac{-968}{3}\)

\(2^{x-2}-\)\(3.2^x=-88\)

\(2^x:2^2-2^x.3=-88\)

\(2^x.\frac{1}{4}-\)\(2^x.3=-88\)

\(2^x.\left(\frac{1}{4}-3\right)=-88\)

\(2^x.\frac{-11}{4}=-88\)

\(2^x=-88:\frac{-11}{4}\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

a)\(2^{x-2}\)-\(3\times2^x\)=(-88)

=>\(2^x\times\frac{1}{2^2}-3\times2^x=\left(-88\right)\)

\(\Rightarrow2^x\times\left(\frac{1}{4}-3\right)=\left(-88\right)\)

\(\Rightarrow2^x\times\left(\frac{-11}{4}\right)=\left(-88\right)\)

\(\Rightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

nhé

a)(2x-1)⁶=(2x-1)⁸

=> (2x-1)⁸-(2x-1)⁶=0

=> (2x-1)⁶.((2x-1)²-1)=0  

+)th1(2x-1)⁶=0

+)th2((2x-1)²-1)=0

a) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)\in\left\{\pm1;0\right\}\)

TH1 : \(2x-1=0\)                       TH2 : \(2x-1=-1\)                      TH3 : \(2x-1=1\)

                   \(2x=1\)                                          \(2x=0\)                                               \(2x=2\)

                      \(x=\frac{1}{2}\)                                          \(x=0\)                                                  \(x=1\)

Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)

b) Tương tự

Áp dụng  : \(\frac{1}{\sqrt{1}.2}< 2.\left(1-\frac{1}{\sqrt{2}}\right)\)

\(\frac{1}{\sqrt{2}.3}< 2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)

...................................

\(\frac{1}{\sqrt{2015}.2016}< 2.\left(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)\)

Cộng các BĐT trên với nhau được : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}}< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)=2\left(1-\frac{1}{\sqrt{2016}}\right)< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)

Từ đó suy ra đpcm

Cái ............... là gì vậy bn

....................... là còn nữa đấy bạn :))

**** ko z, t đag cần 

a/\(\left(x+3\right)\left(\frac{40}{x}+3\right)=88\)(đk: x\(\ne0\))

\(\Leftrightarrow40+\frac{120}{x}+3x+9=88\)

\(\Leftrightarrow3x+\frac{120}{x}=39\) \(\Leftrightarrow\frac{3x^2+120}{x}=39\)

\(\Leftrightarrow x^2+40-13x=0\Leftrightarrow\left(x-8\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)(tm)

vậy x=8 hoặc x=5 là nghiệm của phương trình

b/ \(\frac{x-1}{2x+1}< \frac{1}{2}\left(x\ne-\frac{1}{2}\right)\)

\(\Leftrightarrow\frac{x-1}{2x+1}-\frac{1}{2}< 0\) \(\Leftrightarrow\frac{2x-2}{4x+2}-\frac{2x+1}{4x+2}< 0\)

\(\Leftrightarrow\frac{-3}{4x+2}< 0\) \(\Leftrightarrow4x+2>0\)(vì -3<0)

<=> x>-1/2(tm)

vậy x>-1/2 là nghiệm của phương trình

c/ \(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=2y-2x\left(1\right)\\x^3+1=2y\left(2\right)\end{matrix}\right.\)

(1)\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\)

\(\Leftrightarrow x=y\)(vì x^2 +xy+y^2 +2>0)

thay x=y vào (2) ta được:

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{\sqrt{5}-1}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{matrix}\right.\)

vậy...

a, A = 27x³ + 54x² + 36x + 4

=> A = (3x)³ + 2 . 3 . 9x² + 3 . 3x . 4 + 8 - 4

=> A = [(3x)³ + 2 . 3 . (3x)² + 3 . 3x . 2² + 2³] - 4

=> A = (3x + 2)³ - 4

Thay x = -2002 vào A ta có: A = (3x + 2)³ - 4 

=> A = [3 x (-2002) + 2]³ - 4 = [6006 + 2]³ - 4 = 6008³ - 4

b, B = 2x² + 4x + xy + 2y

=> B = 2x(x + 2) + y(x + 2)

=> B = (x + 2)(2x + y)

Thay x = 88; y = -76 vào B

=> B = (88 + 2)[2 . 88 + (-76)]

=> B = 90 . [176 + (-76)] = 90 . 100 = 9000