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a, Có \(\dfrac{3x-2y}{7}=\dfrac{4x+3y}{5}\)
=> 5(3x-2y)=7(4x+3y)
=> 15x-10y=28x+21y
=> 15x-28x=21y+10y
=> -13x=31y
=> \(\dfrac{x}{y}=\dfrac{31}{-13}=\dfrac{-31}{13}\)
b,\(\dfrac{5x-2y}{3x+4y}=\dfrac{-3}{4}\)
=> 4(5x-2y)=-3(3x+4y)
=> 20x-8y= -9x-12y
=> 20x+9x=-12y+8y
=> 29x=-4y
=> \(\dfrac{x}{y}=\dfrac{-4}{29}\)
5x2 - 7 = 38 => x2 = 9 => x = \(\pm\)3
Từ đây thay x vào \(\dfrac{3x-2}{4}\) để tìm y,z
\(A=7+7^2+7^3+........+7^{2016}\)
\(A=7\left(1+7+7^2+7^3+........+7^{2012}+7^{2013}+7^{2014}+7^{2015}\right)\)
\(A=7\left[\left(1+7+7^2+7^3\right)+........+\left(7^{2012}+7^{2013}+7^{2014}+7^{2015}\right)\right]\)
\(A=7\left[\left(1+7+7^2+7^3\right)+........+7^{2012}\left(1+7+7^2+7^3\right)\right]\)
\(A=7\left[400+........+7^{2012}.400\right]\)
\(A=7.400\left(1+7^4+7^8+7^{12}+......+7^{2012}\right)⋮400\)
Vì \(20^2=400\) nên \(A⋮20^2\left(dpcm\right)\)
a)Ta có :
\(\dfrac{x}{2}=\dfrac{y}{5}=k\)
Mà x.y=3,6 => 2k+5k=3,6=>7k=3,6
Vậy k = \(\dfrac{18}{35}\)
\(x=2k\Rightarrow x=\dfrac{36}{35}\)
\(y=5k\Rightarrow y=\dfrac{18}{7}\)
\(a,\dfrac{x}{2}=\dfrac{y}{5}\)
\(\rightarrow\)\(x.5=y.2\)
\(x.x.5=y.x.2\)
\(x^2.5=3,6.2\)
\(x^2.5=7,2\)
\(x^2=1,44\)
\(\rightarrow x=1,2\) hoặc \(x=-1,2\)
Ý b bạn làm tường tự nha
a/ \(\left|2x-1,6\right|-2,3=1,4\)
\(\Leftrightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5,3\\2x=-2,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
Vậy ....
b/ \(5,4-\left|3x-1,2\right|=5,5\)
\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)
Mà \(\left|3x-1,2\right|\ge0\)
\(\Leftrightarrow x\in\varnothing\)
c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow3,7=4x-2x\)
\(\Leftrightarrow2x=3,7\)
\(\Leftrightarrow x=1,85\)
Vậy ....
d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\) (loại)
Vậy ..
a, \(\left|2x-1,6\right|-2,3=1,4\)
\(\Rightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
b,\(5,4-\left|3x-1,2\right|=5,5\)
\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)
Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)
Vậy, không có giá trị của x thỏa mãn.
c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow x+x+1,3+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow2x-4x=-3,7\)
\(\Leftrightarrow-2x=-3,7\)
\(\Leftrightarrow x=\dfrac{3,7}{2}\)
d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\)
a) a-b=2(a+b)
a-b=2a+2b
a-2a=2b+b
-a=3b
Ta có -a=3b=> a=-3b =>a:b=-3b:b=-3
a-b=2(a+b)=-3
=>a-b=-3 ; 2(a+b)=-\(\dfrac{3}{2}\).Quay lại tìm hai số khi biết tổng và hiệu => a=-\(\dfrac{9}{4}\) b=\(\dfrac{3}{4}\)
b)
Từ a+b = ab => a = ab-b = b(a-1) => a:b = a-1 ( do b khác 0 )
Mặt khác, theo đề bài, a:b = a+b
Suy ra a-1 = a+b => b = -1
Thay b = -1 vào a+b = ab được a-1 = -a => 2a = 1 => a = 1/2
Vậy a = 1/2 và b = -1. Tick cho mình nhé !!! ^3^ Chúc bạn học giỏi
\(\left\{{}\begin{matrix}x\left(x+y+z\right)=13\\y\left(x+y+z\right)=7\\z\left(x+y+z\right)=-4\end{matrix}\right.\) \(\Leftrightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=13+7-4\)
\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=16\)
\(\Rightarrow\left(x+y+z\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}x+y+z=4\\x+y+z=-4\end{matrix}\right.\)
Với \(x+y+z=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=\dfrac{7}{4}\\z=-1\end{matrix}\right.\)
Với \(x+y+z=-4\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{13}{4}\\y=-\dfrac{7}{4}\\z=1\end{matrix}\right.\)
=>x(6x-5)=0
=>x=0 hoặc x=5/6