\(\left[{}\begin{matrix}2x-\dfrac{2}{3}+\dfrac{1}{2}x=0\\x^2+5=0\end{matrix}\right.\)

\(x^2+5>0\)

\(\Rightarrow x^2+5=0\) ( vô lý )

x = 4/15

Bài 5 :

a) \(\dfrac{y}{4}=\dfrac{9}{y}\)

\(\Rightarrow y^2=36\left(y\ne0\right)\)

\(\Rightarrow y=\pm6\)

b) \(\dfrac{y+7}{20}=\dfrac{5}{y+7}\left(y\ne-7\right)\)

\(\Rightarrow\left(y+7\right)^2=100=10^2\)

\(\Rightarrow\left[{}\begin{matrix}y+7=10\\y+7=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=3\\y=-17\end{matrix}\right.\)

c) \(\dfrac{4-5y}{3}=\dfrac{y+2}{5}\)

\(\Rightarrow5\left(4-5y\right)=3\left(y+2\right)\)

\(\Rightarrow20-25y=3y+6\)

\(\Rightarrow28y=14\)

\(\Rightarrow y=\dfrac{14}{28}=\dfrac{1}{2}\)

Bài 4 :

\(\dfrac{a}{5}=\dfrac{b}{7}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{2a}{10}=\dfrac{3b}{21}=\dfrac{4c}{40}=\dfrac{2a+3b-4c}{10+21-40}=\dfrac{81}{-9}=-9\)

\(\Rightarrow\left\{{}\begin{matrix}a=-9.5=-45\\b=-9.7=-63\\c=-9.10=-90\end{matrix}\right.\)

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

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

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: $A=\left(\genfrac{}{}{0ex}{}{1}{4\cdot 9}+\genfrac{}{}{0ex}{}{1}{9\cdot 14}+\genfrac{}{}{0ex}{}{1}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{1}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{5}{4\cdot 9}+\genfrac{}{}{0ex}{}{5}{9\cdot 14}+\genfrac{}{}{0ex}{}{5}{14\cdot 19}+...+\genfrac{}{}{0ex}{}{5}{44\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{9}+\genfrac{}{}{0ex}{}{1}{9}-\genfrac{}{}{0ex}{}{1}{14}+\genfrac{}{}{0ex}{}{1}{14}-\genfrac{}{}{0ex}{}{1}{19}+...+\genfrac{}{}{0ex}{}{1}{44}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{49-4}{4\cdot 49}\right)\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{1}{5}\cdot \genfrac{}{}{0ex}{}{45}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{1-3-5-7-...-49}{89}$

$\iff A=\genfrac{}{}{0ex}{}{9}{196}\cdot \genfrac{}{}{0ex}{}{-623}{89}=-\genfrac{}{}{0ex}{}{9}{28}$
 

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

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`