c) (4x - 8)[x + (-3)] = 0

=> 4(x - 2)(x - 3) = 0

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

=> x - 2 = 0 hoặc x - 3 = 0

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

+) x - 3 = 0 => x = 3

Vậy x \(\in\){2;3}

11(x - 6) = 4x + 11

=> 11x - 66 = 4x + 11

=> 11x - 4x = 11 + 66

=> 7x = 77

=> x = 77/7

=> x = 11

a)11(x-6)=4x+11

11x - 11*6=4x+11

11x-4x=11+11*6

7x=77

x=77/7

x=11

b)lx-3l+1=x

lx-3l=x-1

\(\Rightarrow x-3\in\left\{-\left(x-1\right);x-1\right\}\)

Ta có:

TH1:x-3=-(x-1)

x-3=-x+1

x-(-x)=1+3

2x=4

x=4/2

x=2

TH2:x-3=x-1

x-x=3-1

0=2

\(x\in rỗng\)

Vậy x=2

`2//(5x-8)-3(4x-5)=4(3x-4)`

`<=>5x-8-12x+15=12x-16`

`<=>-19x=-23`

`<=>x=23/19`     Vậy `x=23/19`

`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`

`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`

`<=>4x=8`

`<=>x=2`     Vậy `x=2`

`A`

a)

<=> 66 - 11x = 4x + 11

<=> -11x - 4x = 11 - 66

<=> -15x        = -55

<=>    x          = -55 : (-15)

<=>   x           = 11/3

b)

x = 15 . 8 : 6 = 20

bạn có thể check lại đề bài câu a được không ạ

là x^2 = x^5 á bạn

\(11\cdot\left(x-6\right)=4x+11\)

\(11x-66=4x+11\)

\(11x-4x=11+66\)

\(7x=77\)

\(\Rightarrow x=11\)

\(11.\left(x-6\right)=4x+11\)
=) \(11x-66=4x+11\)
=) \(11x-4x=11+66\)
=) \(7x=77\)
=) \(x=77:7=11\)

11.(x-6)=4x+11

11.x-66=4x+11

7x=66+11

7x=77

x=11 

vậy x=11

\(11\left(x-6\right)=4x+11\)

\(\Rightarrow11x-66=4x+11\)

\(\Rightarrow11x-4x=66+11\)

\(\Rightarrow7x=77\)

\(\Rightarrow x=11\)

Vậy x = 11 

\(\left(3x+0,8\right)\div x+14,5=15\)

\(\Rightarrow\left(3x+0,8\right)\div x=15-14,5\)

\(\Rightarrow\left(3x+0,8\right)\div x=0,5\)

\(\Rightarrow3x+0,8=0,5x\)

\(\Rightarrow-2,5x=0,8\)

\(\Rightarrow x=0,8\div\left(-2,5\right)=\frac{-8}{25}\)

Vậy x = \(\frac{-8}{25}\)

thanks

\(I=-\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+2021\)

\(=-\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2021\)

\(=-\left[\left(x^2+5x\right)^2-6^2\right]+2021\)

\(=-\left(x^2+5x\right)^2+2057\le2057\)

\(I_{max}=2057\) khi \(x^2+5x=0\)

\(K=-\left(x-2\right)\left(x-7\right)\left(x-5\right)\left(x-4\right)+102\)

\(=-\left(x^2-9x+14\right)\left(x^2-9x+20\right)+102\)

\(=-\left(x^2-9x+14\right)\left(x^2+9x+14+6\right)+102\)

\(=-\left[\left(x^2-9x+14\right)^2+6\left(x^2-9x+14\right)\right]+102\)

\(=-\left[\left(x^2-9x+14\right)+6\left(x^2-9x+14\right)+9-9\right]+102\)

\(=-\left(x^2-9x+17\right)^2+111\le111\)

\(K_{max}=111\) khi \(x^2-9x+17=0\)

\(M=-\left(4x^2+4x+1\right)\left(16x^2+16x+3\right)-11\)

Đặt \(4x^2+4x+1=t\Rightarrow16x^2+16x=4t-4\)

\(\Rightarrow M=-t\left(4t-4+3\right)-11\)

\(M=-4t^2+t-11\)

\(M=-4\left(t-\dfrac{1}{8}\right)^2-\dfrac{175}{16}\le-\dfrac{175}{16}\)

\(M_{max}=-\dfrac{175}{16}\) khi \(t=\dfrac{1}{8}\)