a: Ta có: \(142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\)

\(=142-\left[50-80+40\right]\)

=142-10

=132

a. 142 - [50-(2³.10-2³.5)]

= 142 - [ 50 - ( 8 . 10 - 8 . 5 )]

= 142 - [ 50 - ( 80 - 40 )]

= 142 - [ 50 - 40 ]

= 142 - 10 = 132

b. 375 : { 32[ 4+(5.3² - 42 )]} - 14

= 375 : { 32[4+(5.9 - 42 )]} - 14

= 375 : { 32[4 + ( 45 - 42 )]} - 14

= 375 : {32[4+3]} - 14

= 375 : 224 - 14

c.{210 : [ 16+3.(6+3.2^2)]} - 3

= {210 : [ 16 + 3.(6+3.4)]} - 3

= { 210 :[16+3.(6+12)]}-3

= {210 : [ 16+3.18)]} - 3

= { 210 : [ 16 + 54]} - 3

= { 210 : 70 } - 3

= 30 - 3 = 27

d.500-{5.[409-(2^3 . 3-21^2)] - 1724}

= 500-{5.[409-(8 . 63 - 21^2] - 1724}

= 500 - { 5.[409- (504 -  441) - 1724}

= 500 -{ 5.[ 409 - 63 ] - 1724}

= 500 - { 5.346 - 1724}

= 500 - { 1720 - 1724 }

= 500 - 6 

= 494

1) 5x = 120 : 2

5x = 60

x = 60 : 5

x = 12.

2) 3x + 27 = 45

3x = 45 - 27

3x = 18

x = 18 : 3

x = 6.

3) 73 + x - 122 = 234 : 2

73 + x - 122 = 117

73 + x = 117 + 122

73 + x = 239

x = 239 - 73

x = 166.

4) 32 - 2 ( x - 1 ) = 12

2 ( x - 1 ) = 32 - 12

2 ( x - 1 ) = 20

x - 1 = 20 : 2

x - 1 = 10

x = 10 + 1

x = 11.

5) 3 ( x + 1 ) - 15 = 36

3 ( x + 1 ) = 36 + 15

3 ( x + 1 ) = 51

x + 1 = 51 : 3

x + 1 = 17

x = 17 - 1

x = 16.

6) 32 - ( 4x - 16 ) = 21

4x - 16 = 32 - 21

4x - 16 = 11

4x = 11 + 16

4x = 27

x = 27 : 4

x = 6,75.

7) 13x - 3x = 500

( 13 - 3 )x = 500

10x = 500

x = 500 : 10

x = 50.

8) 9x + x - 35 = 68

9x + 1x - 35 = 68

9x + 1x = 68 + 35

9x + 1x = 103

( 9 + 1 )x = 103

10x = 103

x = 103 : 10

x = 10,3.

1) 5x = 120 : 2

5x = 60

x = 60 : 5

x = 12.

2) 3x + 27 = 45

3x = 45 - 27

3x = 18

x = 18 : 3

x = 6.

3) 73 + x - 122 = 234 : 2

73 + x - 122 = 117

73 + x = 117 + 122

73 + x = 239

x = 239 - 73

x = 166.

4) 32 - 2 ( x - 1 ) = 12

2 ( x - 1 ) = 32 - 12

2 ( x - 1 ) = 20

x - 1 = 20 : 2

x - 1 = 10

x = 10 + 1

x = 11.

5) 3 ( x + 1 ) - 15 = 36

3 ( x + 1 ) = 36 + 15

3 ( x + 1 ) = 51

x + 1 = 51 : 3

x + 1 = 17

x = 17 - 1

x = 16.

6) 32 - ( 4x - 16 ) = 21

4x - 16 = 32 - 21

4x - 16 = 11

4x = 11 + 16

4x = 27

x = 27 : 4

x = 6,75.

7) 13x - 3x = 500

( 13 - 3 )x = 500

10x = 500

x = 500 : 10

x = 50.

8) 9x + x - 35 = 68

9x + 1x - 35 = 68

9x + 1x = 68 + 35

9x + 1x = 103

( 9 + 1 )x = 103

10x = 103

x = 103 : 10

x = 10,3.

a) 5x = 120 : 2

5x = 60

x = 60 : 5

x = 12

b) 3x + 27 = 45

3x = 45 - 27

3x = 18

x = 18 : 3

x = 6

c) 73 + x - 122 = 234 : 2

73 + x - 122 = 117

73 + x = 117 + 122

73 + x = 239

x = 239 - 73

x = 166

d) 32 - 2(x - 1) = 12

2(x - 1) = 32 - 12

2(x - 1) = 20

x - 1 = 20 : 2

x - 1 = 10

x = 10 + 1

x = 11

e) 3(x + 1) - 15 = 36

3(x + 1) = 36 + 15

3(x + 1) = 51

x + 1 = 51 : 3

x + 1 = 17

x = 17 - 1

x = 16

f) 21 - (4x - 16) = 21

4x - 16 = 21 - 21

4x - 16 = 0

4x = 0 + 16

4x = 16

x = 16 : 4

x = 4

g) 13x - 3x = 500

10x = 500

x = 500 : 10

x = 50

h) 9x + x - 35 = 65

10x - 35 = 65

10x = 65 + 35

10x = 100

x = 100 : 10

x = 10

a: =>|5x-2|=|2x-3|

=>5x-2=2x-3 hoặc 5x-2=-2x+3

=>3x=-1 hoặc 7x=5

=>x=5/7 hoặc x=-1/3

b: =>|5x-2|-|2x+2|=3x+5

TH1 x<-1

PT sẽ là 2-5x+2x+2=3x+5

=>-3x+4=3x+5

=>-6x=1

=>x=-1/6(loại)

TH2: -1<=x<2/5

Pt sẽ là 2-5x-2x-2=3x+5

=>-7x=3x+5

=>-4x=5

=>x=-5/4(loại)

Th3: x>=2/5

PT sẽ là 5x-2-2x-2=3x+5

=>3x-4=3x+5

=>0x=9(loại)

Tìm min:

$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$

Tìm min

$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$

$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)

Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$

$\Leftrightarrow x=\frac{-1}{4}$

Tìm min

$H=5x^2-x+1=5(x^2-\frac{x}{5})+1$

$=5[x^2-\frac{x}{5}+(\frac{1}{10})^2]+\frac{19}{20}$

$=5(x-\frac{1}{10})^2+\frac{19}{20}\geq \frac{19}{20}$
Vậy $H_{\min}=\frac{19}{20}$. Giá trị này đạt tại $x-\frac{1}{10}=0$

$\Leftrightarrow x=\frac{1}{10}$

a,  3 x - 2 3 = 2 . 32

3 x - 2 3 = 64

3 x - 2 3 = 4 3

3 x - 2 = 4 ⇔ x = 2

b,  5 x + 1 - 5 x = 500

5 x . 5 - 5 x = 500

5 x . ( 5 - 1 ) = 500

5 x . 4 = 500

5 x = 125 = 5 3

x = 3

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(TH_1:3x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(TH_2:-2x-7=0\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)

b) \(2x^3-5x^2+3x=0\)

\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)

\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)

\(TH_1:x=0\)

\(TH_2:x-1=0\)

\(\Leftrightarrow x=1\)

\(TH_3:2x-3=0\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)

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

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

\(TH_1:3x+4=0\)

\(\Leftrightarrow x=-\dfrac{4}{3}\)

\(TH_2:2x-4=0\)

\(\Leftrightarrow x=2\)

Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)

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

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Rightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x-9x=-6-16+12\)

\(\Leftrightarrow11x=-10\)

\(\Leftrightarrow x=-\dfrac{10}{11}\)

Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow3x+1=5x+8\)

\(\Leftrightarrow3x-5x=8-1\)

\(\Leftrightarrow-2x=7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(X=\dfrac{-7}{2}\)

b) Ta có: \(2x^3-5x^2+3x=0\)

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

\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)

c) \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow9x^2-16-3x^2-4x=0\)

\(\Leftrightarrow6x^2-4x-16=0\)

\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)

\(\Leftrightarrow3x^2-6x+4x-8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)

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

Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)

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

\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)

\(\Leftrightarrow20x+16-12=9x-6\)

\(\Leftrightarrow20x+16-12-9x+6=0\)

\(\Leftrightarrow11x+10=0\)

\(\Leftrightarrow x=\dfrac{-10}{11}\)

Vậy \(x=\dfrac{-10}{11}\)

a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)