ko biet ket qua

\(x\times12,5=15\\ x=15:12,5=1,2\\ x-1,27=3\\ x=3+1,27=4,27\)

\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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}

hãy tự tin vào chính bản thân

tui cũng hỏi giống bạn nhưng không ai giai cho tui cả huhu

a)\(1,5-2\left|x\right|=-0,5\)

\(\Leftrightarrow2\left|x\right|=1,5+0,5\)

\(\Leftrightarrow\left|x\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) Ta có: \(\left|x-\dfrac{2}{3}\right|+2.25=\dfrac{3}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{3}\right|=\dfrac{3}{4}-\dfrac{9}{4}=\dfrac{-3}{2}\)(vô lý)

Vậy: \(x\in\varnothing\)

a) Ta có: \(x+\dfrac{-3}{7}=\dfrac{4}{7}\)

\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{4}{7}\)

hay x=1

Vậy: x=1

b) Ta có: \(\dfrac{1}{2}x-75\%=\dfrac{1}{4}\)

\(\Leftrightarrow x\cdot\dfrac{1}{2}=1\)

hay x=2

Vậy: x=2

a) 32 : (3.x - 2) = 8

3x - 2 = 32 : 8

3x - 2 = 4

3x = 4 + 2

3x = 6

x = 6 : 3

x = 2

b) 75 : (x - 18) = 25

x - 18 = 75 : 25

x - 18 = 3

x = 3 + 18

x = 21

c) (15 - 6.x) . 243 = 729

15 - 6x = 729 : 243

15 - 6x = 3

6x = 15 - 3

6x = 12

x = 12 : 6

x = 2

d) 4.(x - 12) + 9 = 17

4(x - 12) = 17 - 9

4(x - 12) = 8

x - 12 = 8 : 4

x - 12 = 2

x = 2 + 12

x = 14

e) 20 - 2.(x + 4) = 4

2(x + 4) = 20 - 4

2(x + 4) = 16

x + 4 = 16 : 2

x + 4 = 8

x = 8 : 2

x = 4

`32: ( 3xx x -2)=8`

`3xx x-2=32:8`

`3xx x-2=4`

`3 xx x=4+2`

`3xx x=6`

`x=6:3`

`x=2`

__

`75 : (x-18) =25`

`x-18=75:25`

`x-18= 3`

`x=3+18`

`x=21`

__

`(15-6 xx x ) xx 243 =729`

`15-6 xx x = 729 : 243`

`15-6 xx x = 3`

`6 xx x=15-3`

`6 xx x=12`

`x=12:6`

`x=2`

__

`4 xx (x-12)+9=17`

`4 xx (x-12)=17-9`

`4 xx (x-12)= 8`

`x-12=8:4`

`x-12=2`

`x=2+12`

`x=14`

__

`20-2xx(x+4)=4`

`2xx(x+4)=20-4`

`2xx(x+4)=16`

`x+4=16:2`

`x+4=8`

`x=8-4`

`x=4`

\(a,32:\left(3\times X-2\right)=8\\ 3\times X-2=32:8=4\\ 3\times X=4+2=6\\ X=\dfrac{6}{3}=2\\ Vậy:X=2\\ ---\\ b,75:\left(X-18\right)=25\\ X-18=75:25=3\\ X=3+18=21\\ Vậy:X=21\\ ---\\ c,\left(15-6\times X\right)\times243=729\\ 15-6\times X=729:243=3\\ 6\times X=15-3=12\\ X=\dfrac{12}{6}=2\\ Vậy:X=2\\ ---\\ d,4\times\left(X-12\right)+9=17\\ 4\times\left(X-12\right)=17-9=8\\ X-12=8:4=2\\ X=2+12=14\\ Vậy:X=14\\ ---\\ e,20-2\times\left(X+4+4\right)=4\\ 2\times\left(X+4\right)=20-4=16\\ X+4=16:2=8\\ X=8-4=4\\ Vậy:X=4\)

2,5 x 2 + 2,5 x 2 +5 : 0,5 

= 5 + 5 + 10

= 20 

=2,5x2x2,5x 2+5x2

=(2,5+2,5+5)x2

=10x2

=20