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1) b) \(\frac{36}{x}=\frac{54}{3}\)
\(\Rightarrow54x=36.3\)
\(\Rightarrow54x=108\)
\(\Rightarrow x=\frac{108}{54}\)
\(\Rightarrow x=2\)
vay \(x=2\)
d) \(1,56:2,88=2,6:x\)
\(2,6:x=\frac{1,56}{2,88}\)
\(2,6:x=\frac{13}{24}\)
\(x=2,6:\frac{13}{24}\)
\(x=\frac{13}{5}.\frac{24}{13}\)
\(x=\frac{24}{5}\) hay \(x=4,8\)
vay \(x=4,8\)
a/|x|-2,5=27,5
=>|x|=27,5+2,5=30
=>x=30 hoặc x=-30
b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)
=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)
c/(x-1)\(^5\)=-32
=>x-1=-2 vì (-2)\(^5\)=-32
=>x=-2+1=-1
d/\(\dfrac{4}{5}.x+0,5=4.5\)
=>\(\dfrac{4}{5}.x+0,5=20\)
=>\(\dfrac{4}{5}.x=20-0,5=19,5\)
=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)
a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-1}{30+60-28}=\dfrac{186}{62}=3\)
\(\dfrac{x}{15}=3\Rightarrow x=45\\ \dfrac{y}{20}=3\Rightarrow y=60\\ \dfrac{z}{28}=3\Rightarrow x=84\)
b) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-20}{-4}=5\)
\(\dfrac{x}{2}=5\Rightarrow x=10\\ \dfrac{y}{3}=5\Rightarrow y=15\\ \dfrac{z}{4}=5\Rightarrow z=20\)
c) x : y :z : t = 3 : 4 : 5 :6\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{t}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{t}{6}=\dfrac{x+y+z+t}{3+4+5+6}=\dfrac{3,6}{18}=\dfrac{1}{5}\)
\(\dfrac{x}{3}=\dfrac{1}{5}\Rightarrow x=\dfrac{3}{5}\\ \dfrac{y}{4}=\dfrac{1}{5}\Rightarrow y=\dfrac{4}{5}\\ \dfrac{z}{5}=\dfrac{1}{5}\Rightarrow z=1\\ \dfrac{t}{6}=\dfrac{1}{5}\Rightarrow t=\dfrac{6}{5}\)
d) \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\)
\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=-\dfrac{49}{7}=-7\)
\(\dfrac{x}{10}=-7\Rightarrow x=-70\\ \dfrac{y}{15}=-7\Rightarrow y=-105\\ \dfrac{z}{12}=-7\Rightarrow z=-84\)
e) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2-y^2+2z^2}{4-9+32}=\dfrac{108}{27}=4\)
\(\dfrac{x}{2}=4\Rightarrow x=8\\ \dfrac{y}{3}=4\Rightarrow y=12\\ \dfrac{z}{4}=4\Rightarrow z=16\)
a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)
a. x : 15 = 8: 24
x : 15 = \(\frac{1}{3}\) \(\Rightarrow x=\frac{1}{3}\times15=5\)
b. \(3\frac{1}{2}\div0,4=x\div1\frac{1}{7}\)
\(\frac{35}{4}=x:\frac{8}{7}\) \(\Rightarrow x=\frac{35}{4}\times\frac{8}{7}=10\)
c. 36 : x = 45 : 3
36: x = 15 \(\Rightarrow x=36:15=\frac{12}{5}\)
d. 1,56 : 2,88 = 2.6 : x
\(\frac{13}{24}\) = 12 : x \(\Rightarrow x=\frac{288}{13}\).
a) (3x + 1)3 = -27
=> (3x + 1)3 = (-3)3
=> 3x + 1 = -3
=> 3x = -3 - 1
=> 3x = -4
=> x = -4/3
b) |2,5 - x| = 1,3
=> \(\orbr{\begin{cases}2,5-x=1,3\\2,5-x=-1,3\end{cases}}\)
=> \(\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)
c) 0,5 - |x - 3,5| = 0
=> |x - 3,5| = 0,5
=> \(\orbr{\begin{cases}x-3,5=0,5\\x-3,5=-0,5\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)
d) Ta có: |x + 2| \(\ge\)0 \(\forall\)x
|x2 - 4| \(\ge\)0 \(\forall\)x
=> |x + 2| + |x2 - 4| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra khi: x + 2 + x2 - 4 = 0
=> x2 + x - 2 = 0
=> x2 + 2x - x - 2 = 0
=> x(x + 2) - (x + 2) = 0
=> (x - 1)(x + 2) = 0
=> \(\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\left(l\right)\\x=-2\end{cases}}\)
\(a,\left(3x+1\right)^3=-27\)
\(\Leftrightarrow3x+1=\sqrt[3]{-27}\)
\(\Leftrightarrow3x+1=-3\)
\(\Leftrightarrow3x=-4\Leftrightarrow x=-\frac{4}{3}\)
b, \(|2,5-x|=1,3\)
\(Th1:2,5-x=1,3\Leftrightarrow x=2,5-1,3\)
\(\Leftrightarrow x=1,2\)
\(Th2:x-2,5=1,3\Leftrightarrow x=1,3+2,5\)
\(\Rightarrow x=3,8\)
c, \(0,5-|x-3,5|=0\)
\(th1:0,5-x+3,5=0\Leftrightarrow4-x=0\)
\(\Rightarrow x=4\)
\(Th2:0,5+x-3,5=0\Leftrightarrow x-3=0\)
\(\Rightarrow x=3\)
d, \(|x+2|+|x^2-4|=0\)
\(x+2=0\Leftrightarrow x=-2\)
9: =>x-3=2
=>x=5
10: =>x+1/2=1/5 hoặc x+1/2=-1/5
=>x=-7/10 hoặc x=-3/10
12:
a: =>x^2=900
=>x=30 hoặc x=-30
b: =>x=1/18*27=3/2
7: =>|x-0,4|=1,1
=>x-0,4=1,1 hoặc x-0,4=-1,1
=>x=1,5 hoặc x=-0,7
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