a: 2x+16=30

=>2x=30-16=14

=>x=14/2=7

b: \(2\left(x+1\right)^3+5=59\)

=>\(2\left(x+1\right)^3=54\)

=>\(\left(x+1\right)^3=27\)

=>x+1=3

=>x=2

c: \(x\inƯ\left(24\right)\)

=>\(x\in\left\{1;2;3;4;6;8;12;24\right\}\)

mà 2<=x<=12

nên \(x\in\left\{2;3;4;6;8;12\right\}\)

d: \(x-2⋮5\)

=>\(x-2\in\left\{0;5;10;15;20;25;30;35;...\right\}\)

=>\(x\in\left\{2;7;12;17;22;27;32;...\right\}\)

mà x<=30

nên \(x\in\left\{2;7;12;17;22;27\right\}\)

a) 2.x + 16 = 30

2x = 30 - 16

2x = 14

x = 14 : 2

x = 7

b) 2.(x + 1)³ + 5 = 59

2.(x + 1)³ = 59 - 5

2.(x + 1)³ = 54

(x + 1)³ = 54 : 2

(x + 1)³ = 27

(x + 1)³ = 3³

x + 1 = 3

x = 3 - 1

x = 2

c) x ∈ Ư(20) = {1; 2; 4; 5; 10; 20}

Mà 2 ≤ x ≤ 12

⇒ x ∈ {2; 4; 5; 10}

d) (x - 2) ⋮ 5 nên x - 2 ∈ B(5) = {0; 5; 10; 15; 20; 25; 30; 35; ...}

x {2; 7; 12; 17; 22; 27; 32; 37; ...}

Mà x ≤ 30

⇒ x ∈ {2; 7; 12; 17; 22; 27}

a)2.x+16=30

2.x=30-16

2.x=14

x=14:2

x=7

vậy x=7

b)\(2.\left(x+1\right)^3+5=59\)

\(2.\left(x+1\right)^3=59-5\)

\(2.\left(x+1\right)^3=54\)

\(\left(x+1\right)^3=54:2\)

\(\left(x+1\right)^3=27\)

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

\(x+1=3\)

\(x=3-1\)

\(x=2\)

vậy x=2

c)x∈U(24) và 2≤x≤12

\(x\in\left\{1;2;3;4;6;8;12;24\right\}\)

mà \(2\le x\le12\)

->\(x\in\left\{2;3;4;6;8;12\right\}\)

vậy \(x\in\left\{2;3;4;6;8;12\right\}\)

d)(x-2):5 và x≤30

x≤30 

->\(x=\left\{1;2;3;4;5;6;7;8;...;29;30\right\}\)

->\(x-2=\left\{-1;0;1;2;3;4;5;6....;28\right\}\)

mà x:5

->\(x⋮5\)

->\(x=\left\{5;10;15;20;25\right\}\)

vậy\(x=\left\{5;10;15;20;25\right\}\)

b. (x + 4)² - (x + 1)(x - 1) = 16

<=> x² + 4x + 16 - (x² - 1) = 16

<=> x² + 4x + 16 - x² + 1 - 16 = 0

<=> x² - x² + 4x = 16 - 16 - 1

<=> 4x = -1

<=> x = \(\dfrac{-1}{4}\)

\(a,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\\ \Leftrightarrow x=\dfrac{23}{24}\\ b,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

2.

a. 3x(12x - 4) - 9x(4x - 3) = 30

<=> 36x² - 12x - 36x² + 27x = 30

<=> 36x² - 36x² - 12x + 27x = 30

<=> 15x = 30

<=> x = 2

b. x(5 - 2x) + 2x(x - 1) = 15

<=> 5x - 2x² + 2x² - 2x = 15

<=> -2x² + 2x² + 5x - 2x = 15

<=> 3x = 15

<=> x = 5

a) x² ( 5x³ - x - 2323x²y=  6969x³y²- 2323x⁴y+ 2323x²y²

c) x² ( 4x^{3 -} 5xy + 2x ) ( -

a) 3x\(\left(x-1\right)^2\)-\(\left(1-x\right)^3\)

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

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

= \(3x^3\)-\(6x^2\)+3x-\(\left(1^3-3\cdot1^2\cdot x+3\cdot1\cdot x^2-x^3\right)\)

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

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

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

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

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b) \(3x\left(x+2\right)+5\left(-x-2\right)\)

= \(3x^2+6x+\left[5\left(-x\right)-10\right]\)

= \(3x^2+6x+5\left(-x\right)-10\)

không biết mình làm thế này có quá gọn không nhỉ :|

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c) \(x\left(x-y\right)+\left(x-y\right)\)

= \(x^2-xy+\left(x-y\right)\)

= \(x^2-xy+x-y\)

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d) \(12a^2b-18ab^2-30b^3\)

Theo mình ở câu này bạn ghi thiếu. Ở câu này hằng dẳng thức ta sử dụng là \(\left(A-B\right)^3=A^3-3A^2B+3AB^2-B^3\), nếu bạn cung cấp dữ kiện \(A^3\) thì mình mới làm được nếu không thì câu này gọn sẵn rồi :))

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e) \(2x\left(x-2\right)-\left(2-x\right)^2\)

= \(2x^2-4x-\left(2^2-2\cdot2\cdot x+x^2\right)\)

= \(2x^2-4x-4+4x-x^2\)

= \(2x^2-x^2-4x+4x-4\)

= \(x^2-4\)

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Chúc bạn học tốt !

Bài 1: 

a) \(a^2-6a+9=\left(a-3\right)^2\)

b) \(\dfrac{1}{4}x^2+2xy^2+4y^4=\left(\dfrac{1}{2}x+2y^2\right)^2\)

Bài 2:

a)  \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)

\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)

b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`

`=> 3x (12x-4) - 3*3x (4x - 3) = 30`

`=> 3x [12x - 4 - 3(4x-3)] = 30`

`=> 3x (12x - 4 - 12x + 9) = 30`

`=> 3x (-4+9)=30`

`=> 3x*5=30`

`=> 3x=6`

`=> x=2`

Vậy, `x=2`

`b)`

`x( 5 - 2x) + 2x( x - 1)`

`=> x(5-2x) + 2x^2 - 2x=15`

`=> 5x - 2x^2 + 2x^2 - 2x =15`

`=> 3x = 15`

`=> x=5`

Vậy, `x=5.`

a: =>36x^2-12x-36x^2+27x=30

=>15x=30

=>x=2

b: =>5x-2x^2+2x^2-2x=15

=>3x=15

=>x=5

Bài 4: 

a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)

\(\Leftrightarrow3x-40=2\)

=>3x=42

hay x=14

b: \(\Leftrightarrow x^3+8-x^3-2x=0\)

=>-2x+8=0

=>-2x=-8

hay x=4

c: \(x\left(x-2\right)+\left(x-2\right)=0\)

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

=>x=2 hoặc x=-1

d: \(5x\left(x-3\right)-x+3=0\)

=>5x(x-3)-(x-3)=0

=>(x-3)(5x-1)=0

=>x=3 hoặc x=1/5

e: \(3x\left(x-5\right)-\left(x-1\right)\left(3x+2\right)=30\)

\(\Leftrightarrow3x^2-15x-3x^2-2x+3x+2=30\)

=>-14x=28

hay x=-2

f: \(\Leftrightarrow\left(x+2\right)\left(x+30-x-5\right)=0\)

=>x+2=0

hay x=-2

a, \(x^2+y^2-2x+6y-30\)

\(=x^2-2x+1+y^2+6y+9-40\)

\(=\left(x-1\right)^2+\left(y+3\right)^2-40\ge-40\)

\(min=-40\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

a)x^2+y^2-2x+6y-30=(x-1)^2+(y+3)^2-40\(\ge\) -40

dấu = xảy ra khi x=1,y=-3

b, \(4-2x^2\le4\)

\(max=4\Leftrightarrow x=0\)

c, \(-x^2+10x-5=-\left(x^2-10x+25\right)+20=-\left(x-5\right)^2+20\le2\text{​​}0\)

\(max=20\Leftrightarrow x=5\)