CÂU NÀO CÁC BẠN LÀM ĐC THÌ GIÚP MK NHA !!!! 

a, x=-505

b, x=35/8 hoac -37/8

nhung cau con lai thi tong tu

a. \(\left|4x+2020\right|=0\)

\(\Rightarrow4x+2020=0\)

\(\Rightarrow4x=-2020\)

\(\Rightarrow x=-505\)

b. \(\left|2x+\frac{1}{4}\right|+\left|-5\right|=\left|-14\right|\)

\(\Rightarrow\left|2x+\frac{1}{4}\right|+5=14\)

\(\Rightarrow\left|2x+\frac{1}{4}\right|=9\)

\(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{4}=9\\2x+\frac{1}{4}=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{35}{4}\\2x=-\frac{37}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{35}{8}\\x=-\frac{37}{8}\end{cases}}\)

c. \(\left|2020-5x\right|-\left|3\right|=-\left|-8\right|\)

\(\Rightarrow\left|2020-5x\right|-3=-8\)

\(\Rightarrow\left|2020-5x\right|=-5\left(vl\right)\)

=> x vô nghiệm

d. \(\left|x^2+4x\right|=0\)

\(\Rightarrow x^2+4x=0\)

\(\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

1,\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{13}{5}\end{cases}}\)

2,\(\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\Leftrightarrow\orbr{\begin{cases}5x^2+3x-2=4x^2-3x-2\\5x^2+3x-2=-4x+3x+2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+6x=0\\9x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\left(x+6\right)=0\\\left(3x\right)^2=2^2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=0or-6\\x=-\frac{2}{3}or\frac{2}{3}\end{cases}}\)

3,\(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2+4x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x\left(x+4\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=0or-4\end{cases}}\)

4,\(5x\left(x-2000\right)-x+2000=0\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)

\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}\)

5,\(5x\left(x-2\right)-x+2=0\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\5x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

6,\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\4x-8=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

7,\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(2x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

tí làm nửa kia 

8,\(x^2-6x+8=0\Leftrightarrow x^2-6x+9-1=0\Leftrightarrow\left(x-3\right)^2-1^2=0\)

\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

9,\(9x^2+6x-8=0\Leftrightarrow9x^2+6x+1-9=0\Leftrightarrow\left(3x+1\right)^2-3^2=0\)

\(\Leftrightarrow\left(3x+1-3\right)\left(3x+1+3\right)=0\Leftrightarrow\left(3x-2\right)\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\3x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{4}{3}\end{cases}}\)

10,\(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}x=-1\)

11,\(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

12,\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\Leftrightarrow\left(5-2x\right)\left(2x+7\right)-4x^2+25=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)-\left(5-2x\right)\left(5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7-5-2x\right)=0\Leftrightarrow\left(5-2x\right).2=0\Leftrightarrow5-2x=0\Leftrightarrow x=\frac{5}{2}\)

13,\(x\left(2x-1\right)+\frac{1}{3}.\frac{2}{3}x=0\Leftrightarrow x\left(2x-1\right)+\frac{2}{9}x=0\)

\(\Leftrightarrow x\left(2x-1+\frac{2}{9}\right)=0\Leftrightarrow x\left(2x-\frac{7}{9}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\2x=\frac{7}{9}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{18}\end{cases}}\)

14,\(4\left(2x+7\right)-9\left(x+3\right)^2=0\Leftrightarrow8x+28-9x^2-54x-81=0\)

\(\Leftrightarrow-9x^2+\left(8x-54x\right)+\left(28-81\right)=0\Leftrightarrow-9x^2-46x-53=0\)

\(\Leftrightarrow9x^2+46x+53=0\)Ta có : \(\Delta'=\frac{2116}{4}-477=529-477=52\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-23+\sqrt{52}}{9}\\x=\frac{-23-\sqrt{52}}{9}\end{cases}}\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

* a mũ 2 hay 4 hay 6 ,... ( những số tự nhiên chẵn khác 0 ) đều lớn hơn hoặc bằng 0 với mọi a

Áp dụng :

a) (2x-8)^4 + (3y+45)^2 = 0

Vì : (2x-8)^4 >=0 , (3y+45)^2 >=0 với mọi x,y

=> (2x-8)^4 + (3y+45)^2 >=0

Dấu "=" xảy ra khi : 2x-8=3y+45=0

->(x;y)=(4;-15)

Những câu sau làm tương tự, ta được :

b) ...

Dấu "=" xảy ra khi : 2x-10=0 và x+y-7=0

->x=5 và 5+y-7=0

->(x;y)=(5;2)

c) 5x-15=0 và 2x-y+4=0

->x=3 và 6-y+4=0

->(x;y)=(3;10)

d) Trùng câu a

a)x=4,y=-15

b)x=5,y=2

còn câu c) mik chịu 

Lời giải:

1. $(x+2)-2=0$

$x+2=2$

$x=0$

2.

$(x+3)+1=7$

$x+3=7-1=6$

$x=6-3=3$

3.

$(3x-4)+4=12$

$3x-4+4=12$

$3x=12$

$x=12:3=4$

4.

$(5x+4)-1=13$

$5x+4=13+1=14$

$5x=14-4=10$

$x=10:5=2$

5.

$(4x-8)-3=5$

$4x-8=5+3=8$

$4x=8+8=16$

$x=16:4=4$

6.

$3+(x-5)=7$

$x-5=7-3=4$

$x=4+5=9$

7.

$8-(2x-4)=2$

$2x-4=8-2=6$

$2x=6+4=10$

$x=10:2=5$

8.

$7+(5x+2)=14$

$5x+2=14-7=7$

$5x=7-2=5$

$x=5:5=1$

9.

$5-(3x-11)=1$

$3x-11=5-1=4$

$3x=11+4=15$

$x=15:3=5$

10.

$16-(8x+2)=6$

$8x+2=16-6=10$

$8x=10-2=8$

$x=8:8=1$

`a) x^2 + 5x + 6 = 0`

Ptr có: `\Delta = b^2 - 4ac = 5^2 - 4 . 1 . 6 = 1 > 0`

`=>` Ptr có `2` `n_o` pb

`x_1 = [ -b + \sqrt{\Delta} ] / [ 2a ] = [ -5 + \sqrt{1} ] / 2 = -2`

`x_2 = [ -b - \sqrt{\Delta} ] / [ 2a ] = [ -5 - \sqrt{1} ] / 2 = -3`

Vậy `S = { -2 ; -3 }`

_________________________________________________

`b) x^4 + 7x^2 - 8 = 0`

Đặt `x^2 = t` `(t >= 0)`

  `=> t^2 + 7t - 8 = 0`

Ptr có: `\Delta = b^2 - 4ac = 7^2 - 4 . 1 . (-8) = 81 > 0`

`=>` Ptr có `2` `n_o` pb

`t_1 = [ -b + \sqrt{\Delta} ] / [ 2a ] = [ -7 + \sqrt{81} ] / 2 = 1`  (t/m)

`t_2 = [ -b - \sqrt{\Delta} ] / [ 2a ] = [ -7 - \sqrt{81} ] / 2 = -8`  (ko t/m)

   `@ t = 1 => x^2 = 1 <=> x = +-1`

Vậy `S = { +-1 }`

|5\(x\) - 4| = |\(x+2\)|

\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)

\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}

|2\(x\) - 3| - |3\(x\) + 2| = 0

|2\(x\) - 3| = | 3\(x\) + 2|

\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)

vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}

|\(\dfrac{5}{4}\)\(x\) - \(\dfrac{7}{2}\)| - | \(\dfrac{5}{8}\)\(x\) + \(\dfrac{3}{5}\)| = 0

|\(\dfrac{5}{4}x\) - \(\dfrac{7}{2}\)|    = | \(\dfrac{5}{8}x+\dfrac{3}{5}\)|

\(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)

  \(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{2}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)

 \(\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)

Vậy \(x\in\) { \(\dfrac{116}{75}\); \(\dfrac{164}{25}\)}

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

Mấy cái này chuyển vế đổi dấu là xong í mà :3

1,

16-8x=0

=>16=8x

=>x=16/8=2

2, 

7x+14=0

=>7x=-14

=>x=-2

3,

5-2x=0

=>5=2x

=>x=5/2

Mk làm 3 cau làm mẫu thôi

Lúc đăng đừng đăng như v :>

chi ra khỏi ngt nản

từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại

1, 16 - 8x = 0

<=>-8x = 16

<=> x = -2

Vậy_

2, 7x + 14 = 0

<=> 7x = -14

<=> x = -2

3, 5 - 2x = 0

<=> - 2x = -5

<=> x =\(\frac{5}{2}\)

Vậy_

4, 3x - 5 = 7

<=> 3x = 7 + 5

<=> 3x = 12

<=> x = 4

Vậy...

5, 8 - 3x = 6

<=> - 3x = 6 - 8

<=> -3x = - 2

<=> x =\(\frac{2}{3}\)

Vậy......