Trả lời

568-{5.[143-3.(14-1)]+10}:10

=568-{5.[143-3.13]+10}:10

=568-{5.[143-39]+10}:10

=568-{5.104+10}:10

=568-530:10

=568-53

=515

568-{5x[143-3x13+10}:10

=568-{5x[143-39+10}:10

=568-(5x114):10

=568-570:10

=568-57

=511

496

    \(568-\left\{5.\left[143-\left(4-1\right)^2+10\right]\right\}:10\)

\(=568-\left\{5.\left[143-3^2+10\right]\right\}:10\)

\(=568-\left\{5.\left[143-9+10\right]\right\}:10\)

\(=568-\left\{5.144\right\}:10=568-720:10=568-72=496\)

\(568-\left\{5.\left[143-\left(4-1\right)^2\right]+10\right\}:10\)

\(=568-\left[5.\left(143-3^2\right)+10\right]:10\)

\(=568-\left[5.\left(143-9\right)+10\right]:10\)

\(=568-\left(5.134+10\right):10\)

\(=568-68\)

\(=500\)

  568 - { 5 [ 143 - ( 4 - 1 )² ] + 10 } : 10 

=568 - { 5 . ( 143 - 3² ) + 10 } :10

=568 - { 5 . ( 143 - 9 ) + 10 ) : 10

=568 - ( 5 . 134 + 10 ) : 10

=568 - 68

=500

568 - { 5 [ 143 - ( 4 - 1 )² ] + 10 } : 10

= 568 - { 5 [ 143 - 3² ] + 10 } : 10

= 568 - { 5 [ 143 - 9 ] + 10 } :10

= 568 - { 5 . 134 + 10 } : 10

= 568 - { 670 + 10 } : 10

= 568 - 680 : 10

= 568 - 68

= 500

=))

568 - { 5 [ 143-(4-1)]² +10} :10

= 568 - (5 * 140^2 +10) :10

= 568 - 9801

= 9233

thành âm rùi

=568 - { 5 [143-3]²+10}:10=568 - 5.140²:10-10:10=568-5.14².10²:10-1=567-5.14².10=567-9800=-9233

`|2x+1|-3=x+4`

`<=>|2x+1|=x+4+3=x+7(x>=-7)`

`**2x+1=x+7`

`<=>x=7-1=6(tm)`

`**2x+1=-x-7`

`<=>3x=-6`

`<=>x=-2(tm)`

`|3x-5|=1-3x(x<=1/3)`

`**3x-5=1-3x`

`<=>6x=6`

`<=>x=1(l)`

`**3x-5=3x-1`

`<=>-5=-1` vô lý

`|2x+2|+|x-1|=10`

Nếu `x>=1`

`pt<=>2x+2+x-1=10`

`<=>3x+1=10`

`<=>3x=9`

`<=>x=3(tm)`

Nếu `x<=-1`

`pt<=>-2x-2+1-x=10`

`<=>-1-3x=10`

`<=>-11=3x`

`<=>x=-11/3(tm)`

Nếu `-1<=x<=1`

`pt<=>2x+2+1-x=10`

`<=>x+3=10`

`<=>x=7(l)`

Vậy `S={3,-11/3}`

d) 

+) Với \(x< -4\), PT \(\Rightarrow3-x-x-4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{17}{4}\)  (Nhận)

+) Với \(-4\le x\le-3\), PT \(\Rightarrow3-x+x+4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{9}{2}\) (Loại)

+) Với \(-3< x\le3\), PT \(\Rightarrow3-x+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{2}\) (Nhận)

+) Với \(x>3\), PT \(\Rightarrow x+3+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{4}\) (Loại)

  Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{17}{4}\right\}\) 

b cần bài nào thế

Chụp rõ hơn tí đi cậu, mờ quá!