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a)
x-3=15
x = 15+3=18
b)
x+7-13=13
x+7 =13-13=0
x = 0-7=-7
c)
x+3-16=-4
x+3 =-4+16=12
x = 12-3=9
d)
26-x-9=-13
-x-9= -13-26=-39
-x =-39+9=-30
x=30
a: =>x=-7/6+5/8=-13/24
b: =>x=-14/25-3/4=-131/100
c: \(x=\dfrac{-33}{26}:\dfrac{-9}{13}=\dfrac{33}{26}\cdot\dfrac{13}{9}=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)
d: \(x=\dfrac{4}{9}:\dfrac{5}{3}=\dfrac{4}{9}\cdot\dfrac{3}{5}=\dfrac{12}{45}=\dfrac{4}{15}\)
a, \(\left|x+3\right|=15\)
TH1 : \(x+3=15\Leftrightarrow x=12\)
TH2 : \(x+3=-15\Leftrightarrow x=-18\)
b, \(\left|x-7\right|+13=15\Leftrightarrow\left|x-7\right|=2\)
TH1 : \(x-7=2\Leftrightarrow x=9\)
TH2 : \(x-7=-2\Leftrightarrow x=5\)
c, \(\left|x-3\right|-16=-4\Leftrightarrow\left|x-3\right|=12\)
TH1 : \(x-3=12\Leftrightarrow x=15\)
TH2 : \(x-3=-12\Leftrightarrow x=-9\)
d, \(26-\left|x+9\right|=-13\Leftrightarrow\left|x+9\right|=39\)
TH1 : \(x+9=39\Leftrightarrow x=30\)
TH2 : \(x+9=-39\Leftrightarrow x=-48\)
a)|x-7|+13=25
|x-7| = 25-13
|x-7| = 12
|x| = 12+7
=> x = 19
b)|x-3|-16=-4
|x-3| = -4+16
|x-3| = 12
|x| = 12+3
=> x = 15
c)26-|x+9|=-13
|x+9| = -13-26
|x+9| = -39
|x| = -39 -9
=> x = 48
a, ( x + 1 ) + ( x + 2 ) + ... + ( x + 199 ) = 0
x + 1 + x + 2 + ... + x + 199 = 0
( x + x + ... + x ) + ( 1 + 2 + ... + 199 ) = 0
199x + 19900 = 0
199x = 0 - 19900
199x = -19900
x = -19900 : 199
x = -100
Vậy ...
b, ( x - 30 ) + ( x - 29 ) + ( x - 28 ) = 11
x - 30 + x - 29 + x - 28 = 11
( x + x + x ) - ( 30 + 29 + 28 ) = 11
3x - 87 = 11
3x = 11 + 87
3x = 98
x = \(\frac{98}{3}\)
Vậy ...
a, -19 - x = -20
x = -19 - (-20)
x = -19 + 20
x = 1
b, 5x - 6 = 3x + 12
5x - 6 - 3x = 12
5x - 3x = 12 + 6
(5 - 3)x = 18
2x = 18
x = 18 : 2
x = 9
c, 15 - 3 (x - 1) = 8 - 2x
15 - 3 (x - 1) + 2x = 8
-3x - 3 - 2x = 8 - 15
-3x - 3 - 2x = -7
-3x - 2x - 3 = 7
-3x - 2x = 7 + 3
(-3 - 2) x = 10
-5x = 10
x = 10 : (-5)
x = -2
d, (5x - 6)2 = 16
(5x - 6)2 = 42
=> 5x - 6 = 4
5x = 4 + 6
5x = 10
x = 10 : 5
x = 2
f, 26 - | x + 9 | = 13
| x + 9 | = 26 - 13
=> | x + 9 | = 13
=> x + 9 = +- 13
* Với x + 9 = 13
x = 13 - 9
x = 4
* Với x + 9 = -13
x = -13 - 9
x = -22
Vậy x = {4;-22}
e, | 3 + x | = 19
=> 3 + x = +- 19
* Với 3 + x = 19
x = 19 - 3
x = 16
* Với 3 + x = -19
x = -19 - 3
x = -22
Vậy x = {16;-22}
a, X = -19+20=1
b, (5-3)X = 18
2X = 18
=> X = 9
c, 3X + 3 -2X = 7
X+3 =7
X = 4
f, |X+9| = 13
ta có 2 trường hợp:
TH1: X+9 = 13
=> X= 4
TH2 : X+9 = -13
=> X= -22
e, ta có 2 trường hợp:
TH1: 3+X = 19
=> X= 16
TH2: 3+X = -19
=> X= -22
a) \(\left|x+3\right|=15\)
\(\Rightarrow\orbr{\begin{cases}x=15-3\\x=-15-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=-18\end{cases}}\)
b) Đề sai
c) Đề sai
d) \(26-\left|x+9\right|=-13\)
\(\Rightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Rightarrow\left|x+9\right|=39\)
\(\Rightarrow\orbr{\begin{cases}x=39-9\\x=-39-9\end{cases}\Rightarrow\orbr{\begin{cases}x=30\\x=-48\end{cases}}}\)
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
\(\left|x+3\right|=15\)
\(\Rightarrow\orbr{\begin{cases}x+3=15\\x+3=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=15-3\\x=-15-3\end{cases}\Rightarrow}}\orbr{\begin{cases}x=12\\x=-18\end{cases}}\)
Vậy .........
a, bạn Forevër™ ( 30 - 03 - 2007 ) giải rồi nhé
b, Ta có :\(|x-7|+13=25\)
<=> : \(|x-7|=25-13\)
<=> : \(|x-7|=12\)
=> : \(x-7=12\) :
< => : \(x=19\)
: \(x=-5\)
Vậy\(x=\left\{-5;19\right\}\)
C,Ta có : \(|x-3|-16=-4\)
\(< =>|x-3|=-4+16\)
\(< =>|x-3|=12\)
\(< =>x-3=12\)hoặc \(x-3=-12\)
\(< =>x=15\)hoặc \(x=-9\)
Vậy\(x=\left\{15;-9\right\}\)