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x:0,16 = 9:x
x:0,16.x=9
x2:0,16=9
x2 =9.0,16
x2 =1,44
x2 = (6/5)2
=> x = 6/5
Vậy x =6/5
\(x:0,16=9:x\)
\(\Rightarrow x^2=0,16.9\)
\(\Rightarrow x^2=1,44\)
\(\Rightarrow x=\pm1,2\)
\(\Leftrightarrow x^2=1.44\)
hay \(x\in\left\{1.2;-1.2\right\}\)
\(\frac{x}{0,16}=\frac{9}{x}\)
\(\Rightarrow x^2=0,16.9=\frac{36}{25}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{9}{5}\\x=-\frac{9}{5}\end{array}\right.\)
\(\frac{x}{0,16}=\frac{9}{x}\)
\(\Leftrightarrow x.x=0,16.9\)
\(\Leftrightarrow x^2=1,44=\left(\pm1,2\right)^2\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1,2\\x=-1,2\end{array}\right.\)
Vậy \(x\in\left\{1,2;-1,2\right\}\)
Chúc bạn học tốt!
Câu a bạn Mạnhlàm đúng rồi vậy mik làm câu b nhé
<br class="Apple-interchange-newline"><div id="inner-editor"></div>72−x7=x−709
<=>(72−x).963=(x−70).763
=>648−9x−7x+49063=0
<=>.−16x+113863=0
<=>-16x+1138=0
<=>x=71,125
a) theo đề bài ta có: x.x=0,16.9
=> x2=1,44
=> x2=a) theo đề bài ta có: x.x=0,16.9
=> x2=1,44
=> x2=(1,2)2
vậy x = 1,2
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
x:0,16 = 9:x
=>\(\frac{x}{0,16}\) = \(\frac{9}{x}\)
=> x2 = 9.0,16
x2 =1,44
=>x2 = (\(\pm\)1,2)
=>\(\left[\begin{array}{nghiempt}x=1,2\\x=-1,2\end{array}\right.\)
Vậy x=1,2 hoặc x=-1,2