a) x * 2 + x * 3 = 60

x * ( 2 + 3 ) = 60

x = 60 : 5

x = 12

b) x * 15 - x * 9 = 78

x * ( 15 - 9 ) = 78

x = 78 : 6

x = 13

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: \(1,2:x\cdot4=20\)

=>\(1,2:x=20:4=5\)

=>\(x=1,2:5=0,24\)

x=1/6

x=1/4

a) x = 1/6

b) x = 1/4

HT

a) 2/3 + x = 5/6

X = 5/6 - 2/3

X = 1/6

b)  x : 5/6 = 3/10

X = 5/6 x 3/10

X = 17/15

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 = \(\dfrac{25}{6}\)

nhanh lên nhé 

sos

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, 1=(2x+0,5)⁶⁰⁰

=>1⁶⁰⁰=(2x+0,5)⁶⁰⁰

=>1=2x+0,5

=>2x+0,5=1

=> 2x= 1+0,5

=>2x= 1,5

=:>x= 1,5:2

=>x=0,75

b, (x-0,125)²=0,25

=>(x-0,125)²=(0,5)²

=>x-0,125=0,5

=>x=0,5 +0,125

=>x= 0,625

c, (x-3)¹¹=(x-3)⁴¹

(x-3)¹¹ -(x-3)⁴¹=0

(x-3)¹¹-(x-3)¹¹ .(x-3)³⁰=0

(x-3)¹¹.[1-(x-3)^{30 ]}=0

(x-3)¹¹.(4-x)³⁰=0

(x-3)¹¹=0 hoặc (4-x)³⁰=0

(x-3)¹¹=0          ( 4-x)³⁰=0

x-3=0                4-x=0

x=0+3                  x=4-0

x=3                     x=4

vậy xϵ{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

\(a,\) \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)

    \(Vậy...\)

\(b,\) \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)

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