a)128-3(x+4)=23 

    125.(x+4)=23

             ( x +4)=23 : 125

             x+4 = 0,184

                x = 0,184 -4

                  x = -3 .816

theo mềnh nhóe ( làm từng câu )

c: =>12x-64=32

=>12x=96

hay x=8

g: =>3x-16=14

hay x=10

e: =>5(x+4)=85

=>x+4=17

hay x=13

\(2x+x+x+83=-2\\\Rightarrow3x+x+83=-2\\\Rightarrow4x+83=-2\\\Rightarrow4x=-2-83\\\Rightarrow4x=-85\\\Rightarrow x=\dfrac{-85}{4}\)

2x + x + x + 83 = -2

(2x + 2x) + 83 = -2

4x + 83 = -2

4x = (-2) - 83

4x = -85

x = -21,25

a) Ta có: 2|x + 2| \(\ge\)0 \(\forall\)x

=> 2|x + 2| + 15 \(\ge\)15 \(\forall\)x

Hay A \(\ge\)15 \(\forall\)x

Dấu "=" xảy ra <=>x + 2 = 0 <=> x = -2

Vậy Min A = 15 tại x = -2

b) Ta có: 2(x + 5)⁴ \(\ge\)0 \(\forall\)x

         3|x + y + 2| \(\ge\)0 \(\forall\)x;y

=> 20 - 2(x + 5)⁴ - 3|x + y + 2| \(\le\)20 \(\forall\)x;y

Hay B \(\le\)20 \(\forall\)x;y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+5=0\\x+y+2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-5\\y=-2-x\end{cases}}\) <=> \(\hept{\begin{cases}x=-5\\y=-2-\left(-5\right)=3\end{cases}}\)

Vậy Max B = 20 tại x = -5 và y = 3

a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)

\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)

\(\Leftrightarrow19x-21=-27\)

=>19x=-6

hay x=-6/19

b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2=\dfrac{3}{2}\)

\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)

=>x=1/2 hoặc x=-1/2

c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)

\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)

\(\Leftrightarrow3x^2+22x+16=0\)

\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)

\(\Leftrightarrow10x^2-19x=0\)

=>x(10x-19)=0

=>x=0 hoặc x=19/10

mik chưa học dến số nguyên

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(\Rightarrow100x+5050=5750\Rightarrow x=5750-5050\)

\(100x=700\Rightarrow x=700:100=7\)

\(a,\text{ }720\text{ : }\left[41-\left(2x-5\right)\right]=2^3\cdot5\)

\(720\text{ : }\left[42-2x+5\right]=8\cdot5\)

\(720\text{ : }\left[42+5-2x\right]=40\)

\(720\text{ : }\left[47-2x\right]=40\)

\(47-2x=720\text{ : }40\)

\(47-2x=18\)

\(2x=47-18\)

\(2x=29\)

\(x=\frac{29}{2}\)

\(b,\text{ }\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(100x+\left[\left(100-1\right)\text{ : }1+1\right]\text{ x }\left(100+1\right)\text{ }:\text{ }2=5750\)

\(100x+100\text{ x }101\text{ : }2=5750\)

\(100x+10100\text{ : }2=5750\)

\(100x+5050=5750\)

\(100x=5750-5050\)

\(100x=700\)

\(x=700\text{ : }100\)

\(x=7\)

a: =>41-(2x-5)=720:40=18

=>2x-5=23

=>2x=28

=>x=14

b: =>100x+5050=5750

=>100x=700

=>x=7

a: =>41-(2x-5)=720:40=18

=>2x-5=23

=>2x=28

=>x=14

b: =>100x+5050=5750

=>100x=700

=>x=7