Tìm x:
a) x x 3 = 8,4;
b) 5 x x = 0,25.
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a) `(x-8)(x^3+8)=0`
`<=>(x-8)(x+2)(x^2-2x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\) (Vì `x^2-2x+4 \ne 0 forall x)`
Vậy `A={8;-2}`.
b) `(4x-3)-(x+5)=3(10-x)`
`,=>4x-3-x-5=30-3x`
`<=>3x-8=30-3x`
`<=>6x=38`
`<=>x=19/3`
Vậy `S={19/3}`.
a: \(x\cdot3\cdot5=2,7\)
=>\(x\cdot15=2,7\)
=>\(x=\dfrac{2,7}{15}=0,18\)
b: \(1,2:x\cdot4=20\)
=>\(1,2:x=20:4=5\)
=>\(x=1,2:5=0,24\)
a) x + \(\dfrac{3}{4}\) = \(\dfrac{5}{3}\)
x = \(\dfrac{5}{3}\) - \(\dfrac{3}{4}\)
x = \(\dfrac{20}{12}\) - \(\dfrac{9}{12}\)
x = \(\dfrac{11}{12}\)
b) x - \(\dfrac{2}{3}\) = \(\dfrac{7}{2}\)
x = \(\dfrac{7}{2}\) + \(\dfrac{2}{3}\)
x = \(\dfrac{21}{6}\) + \(\dfrac{4}{6}\)
x = \(\dfrac{25}{6}\)
a) \(\dfrac{x.2}{-15}=\dfrac{-5}{3}\)
\(\dfrac{x.2}{-15}=\dfrac{25}{-15}\)
x.2=25
x=12,5
b) \(\dfrac{x-1}{-12}=\dfrac{-3}{x-1}\)
(x-1)2=-3.(-12)
(x-1)2=36
⇒(x-1)2\(\Rightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
a) \(\Leftrightarrow x^2-36=64\)
\(\Leftrightarrow x^2=100\)
\(\Leftrightarrow x=\pm10\)
Vậy \(x=\pm10\)
b) \(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}\)
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+7x-x^2-x+6=0\)
hay x=-1
b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
b. (x + 2)2 - x2 + 4 = 0
<=> (x + 2 - x)(x + 2 + x) + 4 = 0
<=> 2(2 + 2x) + 4 = 0
<=> 4(1 + x) + 4 = 0
<=> 4(1 + x) = -4
<=> 1 + x = -1
<=> x = -1 - 1
<=> x = -2
a) (5 + x)(x - 5) - x(x + 5) = 10
x² - 25 - x² - 5x = 10
-5x = 10 + 25
-5x = 35
x = 35 : (-5)
x = -7
b) x.(2x + 3) - 2(x² + x) = 2
2x² + 3x - 2x² - 2x = 2
x = 2
a: \(\left(x+5\right)\left(x-5\right)-x\left(x+5\right)=10\)
=>\(x^2-25-x^2-5x=10\)
=>-5x-25=10
=>-5x=35
=>x=-7
b: \(x\left(2x+3\right)-2\left(x^2+x\right)=2\)
=>\(2x^2+3x-2x^2-2x=2\)
=>x=2
a) x x 3 = 8,4
x = 8,4 : 3
x = 2,8
b) 5 x x = 0,25
x = 0,25 : 5
x = 0,05
Tìm x:
a) x x 3 = 8,4;
x=8,4:3
x=2,8
b) 5 x x = 0,25.
x=0,25:5
x=0,05