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24 tháng 6 2016

a)2.2x=128

=>23+x=128

=>23+x=27

=>3+x=7

=>x=4

b)2x+2x+3=144

=>2x(1+23)=144

=>2x*9=144

=>2x=16

=>2x=24

=>x=4

c)x-287:285=125

=>x-287-85=125

=>x-4=125

=>x=129

d)3x+4-3x=720

=>3x(34-1)=720

=>3x*80=720

=>3x=9

=>3x=32

=>x=2

e)812x:27x=95

=>(34)2x:(33)x=95

=>38x:33x=95

=>38x-3x=(32)5

=>35x=310

=>5x=10

=>x=2

25 tháng 6 2016

a) 2.2x=128

=>23+x = 27

=>3+x=7

=>x=4

24 tháng 7 2023

a) \(2^x\cdot4=128\)

\(\Rightarrow2^x\cdot2^2=2^7\)

\(\Rightarrow2^{x+2}=2^7\)

\(\Rightarrow x+2=7\)

\(\Rightarrow x=5\)

b) \(\left(2x+1\right)^3=125\)

\(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

c) \(2x-2^6=6\)

\(\Rightarrow2x-64=6\)

\(\Rightarrow2x=70\)

\(\Rightarrow x=70:2\)

\(\Rightarrow x=35\)

d) \(64\cdot4^x=45\)

\(\Rightarrow4^3\cdot4^x=45\)

\(\Rightarrow4^{x+3}=45\)

Xem lại đề

e) \(27\cdot3^x=243\)

\(\Rightarrow3^3\cdot3^x=3^5\)

\(\Rightarrow3^{x+3}=3^5\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=2\)

g) \(49\cdot7^x=2401\)

\(\Rightarrow7^2\cdot7^x=7^4\)

\(\Rightarrow7^{x+2}=7^4\)

\(\Rightarrow x+2=4\)

\(\Rightarrow x=2\)

h) \(3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

k) \(3^4\cdot3^x=3^7\)

\(\Rightarrow3^{x+4}=3^7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

n) \(3^x+25=26\cdot2^2+2\cdot3^0\)

\(\Rightarrow3^x+25=104+2\)

\(\Rightarrow3^x+25=106\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(x=4\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`2^x*4 = 128`

`=> 2^x = 128 \div 4`

`=> 2^x = 2^7 \div 2^2`

`=> 2^x = 2^5`

`=> x = 5`

Vậy, `x = 5.`

`b)`

\(\left(2x+1\right)^3=125\)

`=> (2x + 1)^3 = 5^3`

`=> 2x + 1 = 5`

`=> 2x = 5-1`

`=> 2x = 4`

`=> x = 4 \div 2`

`=> x = 2`

Vậy, `x = 2`

`c)`

\(2x-2^6=6\)

`=> 2x = 6+2^6`

`=> 2x = 70`

`=> x = 70 \div 2`

`=> x = 35`

Vậy, `x = 35`

`d)`

\(64\cdot4^x=45\) Bạn xem lại đề

`e)`

`27*3^x = 243`

`=> 3^3 * 3^x = 3^5`

`=> 3^(3 + x) = 3^5`

`=> 3 + x = 5`

`=> x = 5 - 3`

`=> x = 2`

Vậy, `x = 2`

`g)`

`49* 7^x = 2401`

`=> 7^2 * 7^x = 7^4`

`=> 7^(2 + x) = 7^4`

`=> 2 + x = 4`

`=> x = 4 - 2`

`=> x = 2`

Vậy, `x = 2`

`h)`

`3^x = 81`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4`

`k)`

`3^4 * 3^x = 3^7`

`=> 3^(4 + x) = 3^7`

`=> 4 + x = 7`

`=> x = 7 - 4`

`=> x = 3`

Vậy, `x = 3`

`n)`

`3^x + 25 = 26*2^2 + 2*3^0`

`=> 3^x + 25 = 104 + 2`

`=> 3^x + 25 = 106`

`=> 3^x = 106 - 25`

`=> 3^x = 81`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4.`

\(#48Cd\)

25 tháng 9 2021

a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)

b) \(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\Rightarrow x=2\)

c) \(\Rightarrow2^x=32\Rightarrow x=5\)

d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)

e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)

f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)

25 tháng 9 2021

a. 2x . 4 = 128

<=> 2x + 2 = 27

<=> x + 2 = 7

<=> x = 5

b. (2x + 1)3 = 125

<=> (2x + 1)3 - 53 = 0

<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)

<=> (2x - 4)(4x2 + 4x + 1 + 10x + 5 + 25) = 0

<=> (2x - 4)(4x2 + 14x + 31) = 0

<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\VôNghiệm\end{matrix}\right.\)

c. 2x - 26 = 6

<=> 2x = 32

<=> x = 5

d. 64 . 4x = 45

<=> 43 . 4x = 45

<=> 43 + x = 45

<=> 3 + x = 5

<=> x = 2

e. 27 . 3x = 243

<=> 33 . 3x = 35

<=> 33 + x = 35

<=> 3 + x = 5

<=> x = 2

g. 49 . 7x = 2401 (Bn xem lại đề câu này)

<=> 72 . 7x = 74

<=> 72 + x = 74

<=> 2 + x = 4

<=> x = 2

a: x^3-7x-6

=x^3-x-6x-6

=x(x-1)(x+1)-6(x+1)

=(x+1)(x^2-x-6)

=(x-3)(x+2)(x+1)

b: =2x^3+x^2-2x^2-x+6x+3

=x^2(2x+1)-x(2x+1)+3(2x+1)

=(2x+1)(x^2-x+3)

c: =2x^3-3x^2-2x^2+3x+2x-3

=x^2(2x-3)-x(2x-3)+(2x-3)

=(2x-3)(x^2-x+1)

d: =2x^3+x^2+2x^2+x+2x+1

=(2x+1)(x^2+x+1)

e: =3x^3+x^2-3x^2-x+6x+2

=(3x+1)(x^2-x+2)

f: =27x^3-9x^2-18x^2+6x+12x-4

=(3x-1)(9x^2-6x+4)

29 tháng 8 2023

a) \(x^3-7x-6\)

\(=x^3-x-6x-6\)

\(=\left(x^3-x\right)-\left(6x+6\right)\)

\(=x\left(x^2-1\right)-6\left(x+1\right)\)

\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)\)

b) \(2x^3-x^2+5x+3\)

\(=2x^3+x^2-2x^2-x+6x+3\)

\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)

\(=\left(x^2-x+3\right)\left(2x+1\right)\)

c) \(2x^3-5x^2+5x+1\)

\(=2x^3-3x^2-2x^2+3x+2x-3\)

\(=\left(2x^3-3x^2\right)-\left(2x^2-3x\right)+\left(2x-3\right)\)

\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)\)

\(=\left(x^2-x+1\right)\left(2x-3\right)\)

d) \(2x^3+3x^2+3x+1\)

\(=2x^3+x^2+2x^2+x+2x+1\)

\(=\left(2x^3+x^2\right)+\left(2x^2+x\right)+\left(2x+1\right)\)

\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)\)

\(=\left(2x+1\right)\left(x^2+x+1\right)\)

e) \(3x^3-2x^2+5x+2\)

\(=3x^3+x^2-3x^2-x+6x+2\)

\(=\left(3x^3+x^2\right)-\left(3x^2+x\right)+\left(6x+2\right)\)

\(=x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)\)

\(=\left(3x-1\right)\left(x^2-x+2\right)\)

f) \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^3-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

2: Tìm x

a) Ta có: x+25=40

nên x=40-25=15

Vậy: x=15

b) Ta có: 198-(x+4)=120

\(\Leftrightarrow x+4=198-120=78\)

hay x=78-4=74

Vậy: x=74

c) Ta có: \(\left(2x-7\right)\cdot3=125\)

\(\Leftrightarrow2x-7=\dfrac{125}{3}\)

\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)

\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)

Vậy: \(x=\dfrac{73}{3}\)

d) Ta có: \(x+16⋮x+1\)

\(\Leftrightarrow x+1+15⋮x+1\)

mà \(x+1⋮x+1\)

nên \(15⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(15\right)\)

\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

17 tháng 1 2021

\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)

Tự kết luận nhé bạn

30 tháng 7 2021

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

⇔  \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

30 tháng 7 2021

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)2 =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

<=> (x-1)(3x+1)=0

<=> x-1=0  hay  3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

<=> 2(x+3)- x(x+3)=0

<=> (x+3)(2-x)=0

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

<=> x(x-2)+3(x-2)=0

<=> (x-2)(x+3)=0

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)2=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2  = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x2-x+3x-3=0

<=> x(x-1)+3(x-1)=0

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

 

10 tháng 8 2023

Câu a) -3 phần 1/2 

10 tháng 8 2023

Câu a) 2 mũ 2

 

18 tháng 12 2021

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

30 tháng 6 2021

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

30 tháng 6 2021

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x2-8x+6x-4x2=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)