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<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
\(\frac{-12}{x}=\frac{4}{13}\)
\(\Rightarrow4x=13.\left(-12\right)=-156\)
\(\Rightarrow x=-156:4=-39\)
\(\frac{-15}{72}=\frac{125}{-3y}\)
\(\Rightarrow\frac{-15}{72}=\frac{-3000}{72y}\)
\(\Rightarrow-15y=-3000\)
\(\Rightarrow y=-3000:-15=200\)
\(\frac{49}{21}=\frac{-14z}{-7}\)
\(\Rightarrow\frac{49}{21}=\frac{42z}{21}\)
\(\Rightarrow49=42z\)
\(\Rightarrow z=\frac{49}{42}\)
a) \(\frac{3x-2}{1\frac{2}{5}}=\frac{2\frac{3}{7}}{2\frac{3}{5}}\Rightarrow3x-2=1.\frac{2}{5}.\frac{2\frac{3}{7}}{2\frac{3}{5}}=\frac{7}{5}.\frac{17}{7}.\frac{5}{13}\)
\(3x-2=\frac{17}{13}\Rightarrow3x=2+\frac{17}{13}=\frac{43}{13}\Rightarrow x=\frac{43}{39}.\)
b) \(\frac{x}{0,16}=\frac{9}{x}\Rightarrow x^2=9.0,16=1,44\Rightarrow x=\pm1,2\)
a, \(\frac{x-6}{x+4}=\frac{2}{7}\Rightarrow7x-42=2x+8\)ĐK : \(x\ne-4\)
\(\Leftrightarrow5x=50\Leftrightarrow x=10\)(tm)
b, \(\left(x+5\right):2\frac{1}{2}=\frac{40}{x+5}\)ĐK : \(x\ne-5\)
\(\Leftrightarrow\frac{5\left(x+5\right)}{2}=\frac{40}{x+5}\Rightarrow5\left(x+5\right)^2=80\Leftrightarrow\left(x+5\right)^2=16\)
TH1 : \(x+5=4\Leftrightarrow x=-1\)
TH2 : \(x+5=-4\Leftrightarrow x=-9\)
a/ x:0,16 = 9:x hay \(\frac{x}{0,16}=\frac{9}{x}\)
=>x.x=0,16.9
=>x2=1,44
=>x2=(1,2)2
=>x=1,2 hoặc x=-1,2
b)\(\frac{72-x}{7}=\frac{x-10}{9}\)
=>(72-x).9=7.(x-10)
648-9x=7x-70
-9x-7x=-70-648
-16x=-718
x=44,875
a) x: 0,16 =9:x hay \(\frac{x}{0,16}\)=\(\frac{9}{x}\)
\(\Rightarrow\)x.x= 0,16.9
\(\Rightarrow\)\(^{x^2}\)=1,44
\(\Rightarrow\)\(^{x^2}\)= \(^{\left(1.2\right)^2}\)
\(\Rightarrow\)x=1,2 hoặc -1,2.
b) \(\frac{72-x}{7}\)= \(\frac{x-10}{9}\)
\(\Rightarrow\)(72-x).9 = 7.(x-10)
648-9x=7x-70
-9x - 7x=-70-648
-16x=-718
\(\Rightarrow\)x= 44,875.
a) - 0,52 : x = - 9,36 : 16,38
- 0,52 : x = \(-\frac{4}{7}\)
\(x=\left(-0,52\right):\left(-\frac{4}{7}\right)=\frac{91}{100}=0,91\)
b) \(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)
\(\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)
\(\frac{x}{1,61}=\frac{34}{23}\)
\(x=\frac{34}{23}.1,61=2,38\)
Bài 2:
a: =>x/4=1/8
hay x=1/2
b: \(\Leftrightarrow\dfrac{x}{-5}=\dfrac{11}{6}\)
hay x=-55/6
c: \(\Leftrightarrow\dfrac{-3.5}{x}=\dfrac{4.25}{8}\)
hay x=-112/17
\(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)
\(\Leftrightarrow\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)
\(\Leftrightarrow\frac{34}{23}=\frac{x}{1,61}\)
\(\Leftrightarrow x=\frac{34}{23}.1,61\)
\(\Leftrightarrow x=\frac{119}{50}\)
\(\begin{array}{l}a)\frac{x}{{ - 3}} = \frac{7}{{0,75}}\\ \Rightarrow x.0,75 = ( - 3).7\\ \Rightarrow x = \frac{{( - 3).7}}{{0,75}} = - 28\end{array}\)
Vậy x = 28
\(\begin{array}{l}b) - 0,52:x = \sqrt {1,96} :( - 1,5)\\ - 0,52:x = 1,4:( - 1,5)\\ x = \dfrac{(-0,52).(-1,5)}{1,4}\\x = \frac{39}{{70}}\end{array}\)
Vậy x = \(\frac{39}{{70}}\)
\(\begin{array}{l}c)x:\sqrt 5 = \sqrt 5 :x\\ \Leftrightarrow \frac{x}{{\sqrt 5 }} = \frac{{\sqrt 5 }}{x}\\ \Rightarrow x.x = \sqrt 5 .\sqrt 5 \\ \Leftrightarrow {x^2} = 5\\ \Leftrightarrow \left[ {_{x = - \sqrt 5 }^{x = \sqrt 5 }} \right.\end{array}\)
Vậy x \( \in \{ \sqrt 5 ; - \sqrt 5 \} \)
Chú ý:
Nếu \({x^2} = a(a > 0)\) thì x = \(\sqrt a \) hoặc x = -\(\sqrt a \)
a: \(\dfrac{x}{-3}=\dfrac{7}{0.75}=\dfrac{28}{3}\)
=>\(x=\dfrac{28\left(-3\right)}{3}=-28\)
b: \(-\dfrac{0.52}{x}=\dfrac{\sqrt{1.96}}{-1.5}=\dfrac{1.4}{-1.5}\)
=>\(x=0.52\cdot\dfrac{1.5}{1.4}=\dfrac{39}{70}\)
c: \(\dfrac{x}{\sqrt{5}}=\dfrac{\sqrt{5}}{x}\)
=>\(x^2=5\)
=>\(x=\pm\sqrt{5}\)
=> (72 - x) . 9 = (x - 40) . 7
=> 648 - 9x = 7x - 280
=> (-280) - 648 = -9x - 7x
-928 = -16x
=> x = 58