a) 19 + (x - 23) = 50

x - 23 = 50 - 19

x - 23 = 31

=> x  = 31 + 23

     x = 54

b) (273 - x) - 140 = 34

(273 - x) = 34 + 140

(273 - x) = 174

x = 273 - 174

x = 99

c) 56 - (x - 14) = 22

x - 14 = 56 - 22

x - 14 = 34

x = 34 + 14

x = 48

d) 172 - (137 - x) = 93

137 - x = 172 - 93

137 - x = 79

x = 137 - 79

x = 58

e) 13. (x - 14) = 91

x - 14 = 91 : 13

x - 14 = 7

x = 14 + 7

x = 21

g) (13 - x).17 = 51

13 - x = 51 : 17

13 - x = 3

x = 13 - 3

x = 10

chúc bn học giỏi ^^ !

ok mk nha!! ! 5654645645767676576558578779769675634636457567687689978978978978766876765657886567567

1.19+(x-23)=50

<=>x-23=50-19

<=>x-23=31

<=>x=31+23

<=.x=54

vậy x=54

2.(273-x)-140=34

<=>273-x=34+140

<=>273-x=174

<=>x=273-174

<=>x=99(thoả mãn)

vậy x=99

a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)

   \(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

    \(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

     \(\dfrac{72}{17x}\)        = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)

      \(\dfrac{72}{17x}\)      = - \(\dfrac{1}{14}\)

      17\(x\)       = 72.(-14)

       17\(x\)     = - 1008

         \(x\)       = - 1008 : 17

         \(x\)       = - \(\dfrac{1008}{17}\)

Vậy \(x\) \(=-\dfrac{1008}{17}\)

b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))³ = - \(\dfrac{24}{27}\)

          - \(\dfrac{32}{27}\)  + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))³ 

          (3\(x-\dfrac{7}{9}\))³ = - \(\dfrac{8}{27}\)

         (3\(x-\dfrac{7}{9}\))³ = (- \(\dfrac{2}{3}\))³

           3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)

           3\(x\)        = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
           3\(x\)        = \(\dfrac{1}{9}\)

             \(x\)        = \(\dfrac{1}{9}\) : 3

             \(x\)       = \(\dfrac{1}{27}\)

Vậy \(x=\dfrac{1}{27}\)

a) 137 x 7 + 137 x 93
= 137 x ( 93 + 7 )
= 137 x 100 = 13700

b) 428 x 102 - 428 x 2
= 428 x ( 102 - 2 )
= 428 x 100
= 42800

c) 124 x 25 - 25 x 24
= 25 x ( 124 - 24 )
= 25 x 100 
= 2500

_HỌC TỐT_
 

a) 137 x 7 + 137 x 93

\(=137\times\left(7+93\right)\\ =137\times100\\ =13700\)

b) 428 x 102 - 428 x 2

\(=428\times\left(102-2\right)\\ =428\times100\\ =42800\)
c) 124 x 25 - 25 x 24

\(=25\times\left(124-24\right)=\\ 25\times100=2500\)

#Châu's ngốc

\(a,137.7+137.93\)

\(=137.\left(7+93\right)\)

\(=137.100=13700\)

\(b,428.102-428.2\)

\(=428.\left(102-2\right)\)

\(=428.100=42800\)

\(c,124.25-25.24\)

\(=25.\left(124-24\right)\)

\(=25.100=2500\)

Cau 1)

21+(x+23)=50

X+23=50-21

X=29-23

X=6

Cau 2)

125+(x-100)=150

X-100=150-125

X=25+100

X=125

Cau 3)

24+(10-x)=26

10-x=26-24

X=10-2

X=8

\(\frac{x+11}{115}+\frac{x+22}{104}=\frac{x+33}{93}+\frac{x+44}{82}\)

\(\Leftrightarrow\frac{1+\left(x+11\right)}{115}+\frac{1+\left(x+22\right)}{104}=\frac{1+\left(x+33\right)}{93}+\frac{1+\left(x+44\right)}{82}\)

\(\Leftrightarrow\frac{x+126}{115}+\frac{x+126}{104}=\frac{x+126}{93}+\frac{x+126}{82}\)

\(\Leftrightarrow\left(x+126\right).\left(\frac{1}{115}+\frac{1}{104}+\frac{1}{93}+\frac{1}{82}\right)=0\)

\(\Leftrightarrow x+126=6\Leftrightarrow x=-126\)vì\(\frac{1}{115}+\frac{1}{104}+\frac{1}{93}+\frac{1}{82}\ne0\)

vậy x=-126

1. 

$A=x^2+8x+17=(x^2+8x+16)+1=(x+4)^2+1$
Vì $(x+4)^2\geq 0$ với mọi $x$

$\Rightarrow A\geq 0+1=1$

Vậy $A_{\min}=1$. Giá trị này đạt tại $x+4=0\Leftrightarrow x=-4$

--------------------

2.

$B=x^2-4x+7=(x^2-4x+4)+3=(x-2)^2+3$
Vì $(x-2)^2\geq 0$ với mọi $x$

$\Rightarrow B\geq 0+3=3$. Vậy $B_{\min}=3$. Giá trị này đạt được khi $x-2=0\Leftrightarrow x=2$

3.

$C=3x^2+6x+1=3(x^2+2x+1)-2=3(x+1)^2-2$

Vì $(x+1)^2\geq 0$ với mọi $x$

$\Rightarrow C\geq 3.0-2=-2$.

Vậy $C_{\min}=-2$. Giá trị này đạt được khi $x+1=0\Leftrightarrow x=-1$
4.

$D=-4x^2-4x$

$-D=4x^2+4x=(4x^2+4x+1)-1=(2x+1)^2-1$

Vì $(2x+1)^2\geq 0$ với mọi $x$

$\Rightarrow -D\geq 0-1=-1$

$\Rightarrow D\leq 1$

Vậy $D_{\max}=1$. Giá trị này đạt tại $2x+1=0\Leftrightarrow x=\frac{-1}{2}$