(x-1) : 0,16= -9 : (1-x)
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\(3^{x+1}=9^x\)
\(\Rightarrow3^{x+1}=3^{2x}\)
\(\Rightarrow x+1=2x\Rightarrow x=1\)
( 1/2 + 1/3 + 1/4 + ... + 1/10 ) . x = 1/9 + 2/8 + ... + 9/1
=> x = ( 1/9 + 2/8 + ... + 9/1 ) : ( 1/2 + 1/3 + ... 1/10 )
=> x = ( 9/1 + 8/2 + ... + 2/8 + 1/9 ) : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=>x = [ ( 9 - 1 - 1 -... - 1 ) +( 8/2 + 1 ) + ( 7/3 + 1 ) + ... + ( 1/9 + 1 ) ] : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = ( 1 + 10/2 + 10/3 + ... + 10/9 ) : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = [10 . ( 1/2 + 1/3 + ... + 1/9 ) ] : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = 10
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#𝗝𝘂𝗻𝗻
1) 41 - 2x + 1 = 9
-2x + 1 = 9 - 41
-2x + 1 = -32
2x + 1 = 32
2x + 1 = 25
x + 1 = 5
x = 5 - 1
x = 4
`7(x-1/2)^2=9`
`(x-1/2)^2=9/7`
\(=>\left[{}\begin{matrix}x-\dfrac{1}{2}=\sqrt{\dfrac{9}{7}}\\x-\dfrac{1}{2}=-\sqrt{\dfrac{9}{7}}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\\x=-\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{6+\sqrt{7}}{2\sqrt{7}}\\x=\dfrac{-6+\sqrt{7}}{2\sqrt{7}}\end{matrix}\right.\)
7.(x-\(\dfrac{1}{2}\))2=9
7.x+\(\dfrac{1}{4}\) =9
7.x=\(\dfrac{37}{4}\)
x=\(\dfrac{37}{28}\)
a/ (x+1)+(x+2)+(x+3)+...+(x+9)+(x+10)
= x+1 + x+2 + x+3 + ... + x+9 + x+10
= (x+x+x +...+ x+x ) + 1+2+3+...+9+10
= 10.x +55
a) \(\left(x-3\right)\left(4-5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\4-5x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
Vậy \(x\in\left\{3;\frac{4}{5}\right\}\)
b) \(\left(9x-1\right):0,16=9:\left(x-1\right)\)
Đề sai chắc chắn lun
\(x:0,16=9:x\)
\(\Rightarrow x^2=0,16.9\)
\(\Rightarrow x^2=1,44\)
\(\Rightarrow x=\pm1,2\)
\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)
\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)
\(\Rightarrow2xy=54+y\)
\(\Rightarrow2xy-y=54\)
\(\Rightarrow xy-\dfrac{y}{2}=27\)
\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)
\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)
\(\left|x+1\right|+\left|x+2\right|+...+\left|x+9\right|\\ =\left|-x-1\right|+\left|-x-2\right|+...+\left|x+5\right|+...+\left|x+9\right|\\ \text{Áp dụng }BDT\text{ }\left|a\right|\ge a:\\ \Rightarrow\left|-x-1\right|+\left|-x-2\right|+...+\left|x+5\right|+...+\left|x+9\right|\\ \ge-x-1-x-2+...+\left|x+5\right|+...+x+9\\ =\left|x+5\right|-\left(1+2+3+4\right)+\left(6+7+8+9\right)\\ \\ =\left|x+5\right|+20\ge20\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}-\left(x+1\right)\ge0\\...\\ -\left(x+4\right)\ge0\\x+5=0\\x+6\ge0\\ ...\\ x+9\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le1\\...x\le-4\\x=-5\\x\ge-6\\ ...\\ x\ge-9\end{matrix}\right. \\ \Leftrightarrow\left\{{}\begin{matrix}x\le-4\\x=-5\\x\ge-6\end{matrix}\right.\Leftrightarrow x=-5\)
Vậy \(GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là:20\) khi \(x=-5\)