\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)

Ta có: �=(14⋅9+19⋅14+114⋅19+...+144⋅49)⋅1−3−5−7−...−4989A=(4⋅91​+9⋅141​+14⋅191​+...+44⋅491​)⋅891−3−5−7−...−49​

⇔�=15⋅(54⋅9+59⋅14+514⋅19+...+544⋅49)⋅1−3−5−7−...−4989⇔A=51​⋅(4⋅95​+9⋅145​+14⋅195​+...+44⋅495​)⋅891−3−5−7−...−49​

⇔�=15⋅(14−19+19−114+114−119+...+144−149)⋅1−3−5−7−...−4989⇔A=51​⋅(41​−91​+91​−141​+141​−191​+...+441​−491​)⋅891−3−5−7−...−49​

⇔�=15⋅(14−149)⋅1−3−5−7−...−4989⇔A=51​⋅(41​−491​)⋅891−3−5−7−...−49​

⇔�=15⋅(49−44⋅49)⋅1−3−5−7−...−4989⇔A=51​⋅(4⋅4949−4​)⋅891−3−5−7−...−49​

⇔�=15⋅45196⋅1−3−5−7−...−4989⇔A=51​⋅19645​⋅891−3−5−7−...−49​

⇔�=9196⋅1−3−5−7−...−4989⇔A=1969​⋅891−3−5−7−...−49​

⇔�=9196⋅−62389=−928⇔A=1969​⋅89−623​=−289​
 

\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)

Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5

đề là j v bạn

   \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)

= \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)

= \(\dfrac{-4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)

= \(\dfrac{-4}{13}.\dfrac{17}{17}\)

= \(\dfrac{-4}{13}.1\)

= \(\dfrac{-4}{13}\)

= \(\dfrac{-4.5-12.4}{13.17}\)

=\(\dfrac{-4\left(5+12\right)}{13.17}\)

=\(\dfrac{-4.17}{13.17}\)

=\(\dfrac{-4}{13}\)

                      (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