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\(\dfrac{0,3.x+2,5}{x}-3=17\)
<=> \(\dfrac{0,3.x+2,5}{x}=20\)
<=> 0,3.x + 2,5 = 20x
<=> 2,5 = 20x - 0,3.x
<=> 2,5 = 19,7.x
<=> x = \(\dfrac{25}{197}\)
@Khánh Linh
Bài 1 :
Lý luận chung cho cả 2 câu a) và b) :
Vì giá trị tuyệt đối luôn lớn hơn hoặc bằng 0, mà tổng của chúng lại bằng 0
a) \(\Rightarrow\hept{\begin{cases}x-2y=0\\y-1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)
b) \(\Rightarrow\hept{\begin{cases}x-3=0\\x-2y-5=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}\)
a)x∈ B(15); 20<x≤ 65
B(15)={0;15;30;45;.....}
Vì \(x\in B\left(15\right)\)
\(\Rightarrow x\in\left\{0;15;30;45;...\right\}\)
mà 20<x≤ 65
\(\Rightarrow x\in\left\{30;45;60\right\}\)
b)x⋮13;10<x<70
B(13)={0;13;26;39....}
Vì \(x\in B\left(13\right)\)
\(\Rightarrow x\in\left\{0;13;26;39;...\right\}\)
mà 10<x< 70
\(\Rightarrow x\in\left\{13;26;39;52;65\right\}\)
c)x∈Ư(42);x>5
Ư(42)={1;2;3;6;7;14;21;42}
Vì \(x\inƯ\left(42\right)\)
\(\Rightarrow x\in\left\{1;2;3;6;714;21;42\right\}\)
mà \(x>5\)
\(\Rightarrow x\in\left\{6;7;14;21;42\right\}\)
\(x^{20}+81x^{16}=0\)
\(\left\{{}\begin{matrix}x^{20}\ge0\forall x\\x^{16}\ge0\Rightarrow81x^{16}\ge0\forall x\end{matrix}\right.\)
\(\Rightarrow x^{20}+81x^{16}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x^{20}=0\Rightarrow x=0\\81x^{16}=0\Rightarrow x^{16}=0\Rightarrow x=0\end{matrix}\right.\)
\(\Rightarrow x=0\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
\(\dfrac{x-1}{2}=\dfrac{8}{x-1}\)
\(\Rightarrow\left(x-1\right)\left(x-1\right)=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-1=4\Rightarrow x=5\\x-1=-4\Rightarrow x=-3\end{matrix}\right.\)
\(\dfrac{\left(x+1\right)^2}{2}=\dfrac{4}{x+1}\)
\(\Rightarrow\left(x+1\right)^2\left(x+1\right)=8\)
\(\Rightarrow\left(x+1\right)^3=8\)
\(\Rightarrow\left(x+1\right)^3=2^3\)
\(\Rightarrow x+1=2\Rightarrow x=1\)
70%.x + 0,3.x + x : 2 = 20
<=> x(70% + 0,3) + x : 2 = 20
<=> x + x : 2 = 20
<=> x.(\(\dfrac{1}{2}+1\)) = 20
<=> x.\(\dfrac{3}{2}=20\)
<=> x = \(\dfrac{40}{3}\)
@Khánh Linh