tìm x∈N biết
56-( x - 14 )=22
172-(137-x )=93
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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
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
\(\frac{x+11}{115}+\frac{x+22}{104}=\frac{x+33}{93}+\frac{x+44}{82}\)
\(\Leftrightarrow\frac{1+\left(x+11\right)}{115}+\frac{1+\left(x+22\right)}{104}=\frac{1+\left(x+33\right)}{93}+\frac{1+\left(x+44\right)}{82}\)
\(\Leftrightarrow\frac{x+126}{115}+\frac{x+126}{104}=\frac{x+126}{93}+\frac{x+126}{82}\)
\(\Leftrightarrow\left(x+126\right).\left(\frac{1}{115}+\frac{1}{104}+\frac{1}{93}+\frac{1}{82}\right)=0\)
\(\Leftrightarrow x+126=6\Leftrightarrow x=-126\)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 $2x+1=0\Leftrightarrow x=\frac{-1}{2}$
56 - ( x - 14 ) = 22
x - 14 = 56 - 22
x - 14 = 34
x = 34 + 14
x = 48
172 - ( 137 - x ) = 93
137 - x = 172 - 93
137 - x = 79
x = 137 - 79
x = 58
1/ 56 - (x - 14) = 22 => x - 14 = 56 - 22 => x - 14 = 34 => x = 34 + 14 => x = 48
2/ 172 - (137 - x) = 93 => 137 - x = 172 - 93 => 137 - x = 79 => x = 137 - 79 => x = 58