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\(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}=\frac{\left(0,3^2\right)^5}{\left(0,3\right)^6}=\frac{\left(0,3\right)^6\cdot\left(0,3\right)^4}{\left(0,3\right)^6}=\left(0,3\right)^4\)
a) Ko tìm được gt nào của x vì \(\left|x\right|\ge0\)
d)=> x - 3,5 = 5 hoặc x - 3,5 = -5
=> x = 5 + 3,5 hoặc x = -5 + 3,5
=> x = 8,5 hoặc x = -1.5
e: =>|x+0,75|=1,4
=>x+0,75=1,4 hoặc x+0,75=-1,4
=>x=-2,15 hoặc x=0,65
g: =>|x-2|=1/2
=>x-2=1/2 hoặc x-2=-1/2
=>x=5/2 hoặc x=3/2
a)x/0,9=5/6 =>6x=5.0,9=4,5
=>x=4,5/6=0,75
b)16/x2=x/-4 =>16.(-4)=x3 =>x3=-64=(-4)3 =>x=-4
c)x/2=8/x =>x2=2.8=16 => x=+-4
d)x-1/x+2=x+2/x+3 =>(x+2)(x+2)=(x-1)(x+3)
=>x2+4x+4=x2+2x-3
=>x2-x2+4x-2x+4+3=0
=>2x+7=0
=>2x=-7 =>x=-7/2=-3,5
a) có \(\dfrac{x}{0,9}=\dfrac{5}{6}\)
x.6=5.0,9
x.6=\(\dfrac{9}{2}\)
x=\(\dfrac{9}{2}:6\)
x=\(\dfrac{9}{2}.\dfrac{1}{6}\)
x=\(\dfrac{3}{4}\)
b) \(\dfrac{16}{x^2}=\dfrac{x}{-4}\)
16.(-4)=\(x^3\)
-64=\(x^3\)
Mà 64=(-4)^3
Suy ra x=-4
c)\(\dfrac{x}{2}=\dfrac{8}{x}\)
x.x=8.2
\(x^2\) =16
Suy ra x=4 hoặc x=-4
d) \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+3}\)
(x-1)(x+3)=(x+2)(x+2)
\(x^2+3x-x-3=x^2+2x+2x+4\)
\(x^2-4x-3=x^2+4x+4\)
8x=-7
x=-7/8
a: \(\dfrac{x}{0.9}=\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
b: \(\dfrac{-6}{x}=\dfrac{9}{-15}\)
\(\Leftrightarrow x=10\)
c: \(\dfrac{\dfrac{14}{15}}{\dfrac{9}{10}}=\dfrac{x}{\dfrac{3}{7}}\)
\(\Leftrightarrow x=\dfrac{3}{7}\cdot\dfrac{14}{15}:\dfrac{9}{10}=\dfrac{2}{5}\cdot\dfrac{10}{9}=\dfrac{20}{45}=\dfrac{4}{9}\)
\(7\cdot\left(-28\right)=\left(-49\right)\cdot4\)
từ đó ta suy ra được các tỉ lệ thức sau :
\(\frac{7}{-49}=\frac{4}{-28}\)
\(\frac{-49}{7}=\frac{-28}{4}\)
\(\frac{4}{7}=\frac{-28}{-49}\)
\(\frac{7}{4}=\frac{-49}{-28}\)
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
a)\(\dfrac{0,4}{x}=\dfrac{x}{0,9}\Rightarrow x^2=0,4.0,9=0,36\Rightarrow x=0,6;-0,6\)
\(b)\dfrac{0,2}{1\dfrac{1}{5}}=\dfrac{\dfrac{2}{3}}{6x+7}\Rightarrow6x+7=\dfrac{1\dfrac{1}{5}.\dfrac{2}{3}}{0,2}=4\Rightarrow6x=-3\Rightarrow x=-\dfrac{3}{6}=-\dfrac{1}{2}\)
c)\(\dfrac{13\dfrac{1}{3}}{1\dfrac{1}{3}}=\dfrac{26}{2x+1}\Rightarrow2x+1=\dfrac{1\dfrac{1}{3}.26}{13\dfrac{1}{3}}=2,6\Rightarrow2x=1,6\Rightarrow x=0,8\)
d) mk ko hiểu
e)\(\dfrac{-2,6}{x}=\dfrac{-12}{42}\Rightarrow x=\dfrac{-2,6.42}{-12}=9,1\)
f)\(\dfrac{x^2}{6}=\dfrac{24}{25}\Rightarrow x^2=\dfrac{6.24}{25}=5,76\Rightarrow x=-2,4;2,4\)
n)mk chịu thua
xin lỗi bạn nha
\(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}=\frac{\left(3.0,3\right)^5}{\left(0,3\right)^6}=\frac{3^5.\left(0,3\right)^5}{\left(0,3\right)^6}=\frac{3^5}{0,3}=\frac{243}{0,3}=810\)
