(x + 1)⁴ = (x + 1)³

⇒    (x + 1)⁴ - (x + 1)³ = 0

⇒ (x + 1)³ . (x + 1 - 1) = 0

⇒               (x + 1)³ . x = 0

⇒ \(\left[{}\begin{matrix}\left(x+1\right)^3=0\\x=0\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

   Vậy \(x\in\left\{0;-1\right\}\)

`(x+1)^4 =(x+1)^3`

`@TH1: x+1=0 =>x=-1`

    `=>(-1)^4 = (-1)^3`

    `=>1=-1` (Vô lí) 

 `=>x=-1` loại

`@TH2: x+1`\(\ne 0<=>x \ne -1\)

   `=>x+1=1`

   `=>x=0` (t/m)

Vậy `x=0`

\(A=1+2+2^2+...+2^{2017}\)

\(\Rightarrow A=\dfrac{2^{2017+1}-1}{2-1}\)

\(\Rightarrow A=2^{2018}-1\)

mà \(B=2^{2018}\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}\)

\(\Rightarrow A-B=-1\)

\(2A=2+2^2+2^3+...+2^{2018}\)

\(\Rightarrow A=2A-A=2^{2018}-1\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)

a) \(S_{xq}=\left(a+b\right).2.h\)

mà \(\left\{{}\begin{matrix}S_{xq}=120\left(cm^2\right)\\h=60\left(cm\right)\end{matrix}\right.\)

\(\Rightarrow120\left(a+b\right)=120\)

\(\Rightarrow a+b=1\)

\(\Rightarrow\left(a+b\right)^2=1\)

\(\Rightarrow a^2+b^2+2ab=1\)

mà \(a^2+b^2\ge2ab\) (do \(\left(a-b\right)^2=a^2+b^2-2ab\ge0,\forall ab>0\))

\(\Rightarrow4ab\le1\)

\(\Rightarrow ab\le\dfrac{1}{4}\left(1\right)\)

Để thể tích hình hộp chữ nhật có thể tích lớn nhất khi :

\(\left(ab\right)max\left(V=abh;h=60cm\right)\)

\(\left(1\right)\Rightarrow\left(ab\right)max=\dfrac{1}{4}\)

Vậy \(ab=\dfrac{1}{4}\) thỏa mãn đề bài

A = \(\dfrac{2}{5.7}\) + \(\dfrac{5}{7.12}\) + \(\dfrac{7}{12.19}\) + \(\dfrac{9}{19.28}\) + \(\dfrac{11}{28.39}\) + \(\dfrac{1}{30.40}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{19}\)  + \(\dfrac{1}{19}\) - \(\dfrac{1}{28}\) + \(\dfrac{1}{28}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)

A = \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\) 

B = \(\dfrac{1}{20}\) + \(\dfrac{1}{44}\) + \(\dfrac{1}{77}\) + \(\dfrac{1}{119}\) + \(\dfrac{1}{170}\)

B = 2 \(\times\) ( \(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.44}\) + \(\dfrac{1}{2.77}\) + \(\dfrac{1}{2.119}\) + \(\dfrac{1}{2.170}\))

B = 2 \(\times\) ( \(\dfrac{1}{40}\) + \(\dfrac{1}{88}\) + \(\dfrac{1}{154}\) + \(\dfrac{1}{238}\) + \(\dfrac{1}{340}\))

B = 2 \(\times\) ( \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) + \(\dfrac{1}{11.14}\) + \(\dfrac{1}{14.17}\) + \(\dfrac{1}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\)+ \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) + \(\dfrac{3}{17.20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{20}\))

B = \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{20}\)

B = \(\dfrac{1}{10}\)  = \(\dfrac{34}{340}\) < \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)

Vậy A > B 

giúp với

Bạn xem lại đề.

\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)

\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)

mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)

\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)

Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)

\(\Rightarrow A>-\dfrac{11}{21}\)

\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)

\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)

\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)

Dễ thấy A có 9 thừa số, suy ra

\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)

Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)

Vậy \(A< \dfrac{-11}{21}\)

\(3x-2=x+7\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(\dfrac{9}{2}\)nhé

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