a) 3x + 2( 5 - x) = 0

3x + 10 - 2x = 0

x + 10 = 0

x = -10.

b) x(2x - 1)(x + 5) - ( 2x² + 1)(x + 4,5) = 3,5

* x(2x² +10x-x-5) - (2x³ + 9x² + x + 4,5)=3,5.

2x³ + 10x² - x² -5x- 2x³ - 9x² -x -4,5=3,5

-6x - 4,5 =3,5

-6x = 8

x = -8/6.

a)Vì  |x - 3,5 | luôn lớn hơn hoặc = 0

        | 4,5 - x | luôn lớn hơn hoặc =0

Mà  |x - 3,5 | + | 4,5 - x | = 0

=> x-3,5=0 và 4,5-x= 0

=> x= 3,5 và x= 4,5 ( vô lí)

=> x thuộc rỗng

b) Vì lx+3l luôn lớn hơn hoặc = 0 vs mọi x

=> 5-x luôn lớn hơn hoặc = 0

=> x luôn lớn hơn hoặc = 5 

Ta có: | x + 3 | = 5 - x

=> x+3 = 5-x hoặc x+3 = -5+x

<=> x+x= -3+5 hoặc x-x= -3-5

<=> x= 1 hoặc 0= -8(vô lí)

Vậy x= 1

c) Ôi bạn làm tương tự đi nhé, mik đánh mỏi tay ^^

vậy x=1

nhé bn

bài này viết

ra dài dòng lắm

a) \(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow2x^3-x^2+10x^2-5x-2x^3-x-9x^2-4,5=3,5\)

\(\Leftrightarrow-6x=8\Leftrightarrow x=-\frac{4}{3}\)

b) \(\left(2x-5\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\y-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=3\end{cases}}}\)

a) x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5x(2x−1)(x+5)−(2x2+1)(x+4,5)=3,5

\Leftrightarrow2x^3-x^2+10x^2-5x-2x^3-x-9x^2-4,5=3,5⇔2x3−x2+10x2−5x−2x3−x−9x2−4,5=3,5

\Leftrightarrow-6x=8\Leftrightarrow x=-\frac{4}{3}⇔−6x=8⇔x=−34​

b) \left(2x-5\right)^2+\left(y-3\right)^2=0(2x−5)2+(y−3)2=0

\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\y-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=3\end{cases}}}\)

a) x (2x - 1) (x + 5) - (2x² + 1) (x + 4,5) = 3,5

=> 2x³ - x² + 10x² - 5x - 2x³ - x - 9x² - 4,5 = 3,5

 (2x³ - 2x³) + (-x² - 9x²) + 10x² - (5x - x) - 4,5 = 3,5

 -10x² + 10x² - 4x - 4,5 = 3,5

 (-10x² + 10x²) - 4x - 4,5 = 3,5

  -4x = 3,5 + 4,5

 -4x = 8

 x = 8 : (-4) = -2

Vậy x = 2 hoẵ x = -2

b) (2x - 5)² + (y - 3)² = 0

=> (2x - 5)² = 0 hoặc (y - 3)² = 0

* (2x - 5)² = 0 => x = \(\frac{5}{2}\)

* (y - 3)² = 0 => y = 3

(Mình không chắc câu a nha)

a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)

b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)

\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)

\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)

a: =>3x+10-2x=0

hay x=-10

c: \(\Leftrightarrow3x^2-3x^2+6x=36\)

=>6x=36

hay x=6

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)