a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

Giúp mình vs.

\(\left|x-1\right|=2x-1\)

\(\Rightarrow\orbr{\begin{cases}x-1=2x-1\\x-1=-2x+1\end{cases}\left(đk:2x\ge\frac{1}{2}\right)}\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{2}{3}\left(tm\right)\end{cases}}\)

vậy \(x=\frac{2}{3}\)

nhìu dữ

a)3/2

b)-1/3

c)-5/6

d)0

e)-1/2

Bài 2

a=3

b=1/2

c=-1/3

d=0

e=9

f=-2/3

mk ko làm rõ đâu  nhe

Mình trả lời hai câu đầu trước nha!

1)

x= -6+2

x= -4

2)

=-5 x = 13 + (-3)

= -5 .x =10

x = 10 ÷ -5

x = -2

bài 1:

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

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

=2x^2-3x^2-3

=-x^2-3

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

Bn gửi từng câu sẽ có nhều ng trl hơn nhé

tý mk giải câu a cho cần ko

Bài 2 : 

a. A = 2 ( x³ + y³ ) - 3 ( x² + y² ) với x + y = 1

=> A = 2 ( x + y ) ( x² - xy + y² ) - 3 [ ( x + y )² - 2xy ]

=> A = 2 [ ( x + y )² - 3xy ] - 3 ( 1 - 2xy )

=> A = 2 ( 1 - 3xy ) - 3 + 6xy

=> A = 2 - 6xy - 3 + 6xy

=> A = - 1

B = x³ + y³ + 3xy với x + y = 1

=> B = ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² - 3xy )

=> B = ( x + y )³ - 3xy ( x + y - 1 )

=> B = 1³ - 3xy . 0

=> B = 1

Bài 1.

a) ( x - 1 )³ + ( 2 - x )( 4 + 2x + x² ) + 3x( x + 2 ) = 16

<=> x³ - 3x² + 3x - 1 + 8 - x³ + 3x² + 6x = 16

<=> 9x + 7 = 16

<=> 9x = 9

<=> x = 1

b) ( x + 2 )( x² - 2x + 4 ) - x( x² - 2 ) = 15

<=> x³ + 8 - x³ + 2x = 15

<=> 2x + 8 = 15

<=> 2x = 7

<=> x = 7/2

c) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 9( x + 1 )² = 15

<=> ( x - 3 )[ ( x - 3 )² - ( x² + 3x + 9 ) + 9( x² + 2x + 1 ) = 15

<=> ( x - 3 )( x² - 6x + 9 - x² - 3x - 9 ) + 9x² + 18x + 9 = 15

<=> ( x - 3 ).(-9x) + 9x² + 18x + 9 = 15

<=> -9x² + 27x + 9x² + 18x + 9 = 15

<=> 45x + 9 = 15

<=> 45x = 6

<=> x = 6/45 = 2/15

d) x( x - 5 )( x + 5 ) - ( x + 2 )( x² - 2x + 4 ) = 3

<=> x( x² - 25 ) - ( x³ + 8 ) = 3

<=> x³ - 25x - x³ - 8 = 3

<=> -25x - 8 = 3

<=. -25x = 11

<=> x = -11/25

Bài 2.

a) A = 2( x³ + y³ ) - 3( x² + y² )

= 2( x + y )( x² - xy + y² ) - 3x² - 3y²

= 2( x² - xy + y² ) - 3x² - 3y²

= 2x² - 2xy + 2y² - 3x² - 3y²

= -x² - 2xy - y²

= -( x² + 2xy + y² )

= -( x + y )²

= -(1)² = -1

b) B = x³ + y³ + 3xy 

= x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² + 3xy

= ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² - 3xy )

= ( x + y )³ - 3xy( x + y - 1 )

= 1³ - 3xy( 1 - 1 )

= 1 - 3xy.0

= 1

Bài 1:

a,\(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(9x+7=16\)

\(9x=9\)

\(x=1\)

b,\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(x^3+8-x^3+2x=15\)

\(8+2x=15\)

\(2x=7\)

\(x=\frac{7}{2}\)

c,\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)

\(45x+9=15\)

\(45x=6\)

\(x=\frac{2}{15}\)

a.\(\Leftrightarrow\left(x-1\right)^3+8-x^3+3x\left(x+2\right)=17\)

    \(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

   \(\Leftrightarrow9x+7=17\)

   \(\Leftrightarrow9x=10\Leftrightarrow x=\frac{10}{9}\)

`#040911`

`a,`

`15 + 25 \div (2x - 1) = 20`

`\Rightarrow 25 \div (2x - 1) = 20 - 15`

`\Rightarrow 25 \div (2x - 1) = 5`

`\Rightarrow 2x - 1 = 25 \div 5`

`\Rightarrow 2x - 1 = 5`

`\Rightarrow 2x = 6`

`\Rightarrow x = 3`

Vây, `x = 3.`

`b,`

\(3^{x-1}+2\cdot3^x=21\)

`\Rightarrow 3^x \div 3 + 2. 3^x = 21`

`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`

`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`

`\Rightarrow 3^x . \frac{7}{3} = 21`

`\Rightarrow 3^x = 21 \div \frac{7}{3}`

`\Rightarrow 3^x = 9`

`\Rightarrow 3^x = 3^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

`c,`

\(2^{x-3}+2^{x+1}=17\)

`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`

`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`

`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`

`\Rightarrow 2^x . \frac{17}{8} = 17`

`\Rightarrow 2^x = 17 \div \frac{17}{8}`

`\Rightarrow 2^x = 8`

`\Rightarrow 2^x = 2^3`

`\Rightarrow x = 3`

Vậy, `x = 3`

`d,`

\(5^x-5^{x-1}=20\)

`\Rightarrow 5^x - 5^x \div 5 = 20`

`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`

`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`

`\Rightarrow 5^x . \frac{4}{5} = 20`

`\Rightarrow 5^x = 20 \div \frac{4}{5}`

`\Rightarrow 5^x = 25`

`\Rightarrow 5^x = 5^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

\(a.25:\left(2x-1\right)=5\)

\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)

\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)

\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)

\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)

\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)

\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)

\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)

\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)