Tìm x biết:
A)|x-1|=2x
B)|x+1,2|+|y-3,8|=0
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a)
` x - 1,44 : 1,2 = 5,9`
` x - 1,2 = 5,9`
`x = 5,9 + 1,2`
`x = 7,1`
b)
` 86,5 - x : 8,6 = 23,7`
` x : 8,6 = 86,5 - 23,7`
` x : 8,6 = 62,8`
` x = 62,8 xx 8,6`
`x = 540,08`
c)
`( 25,34 - x ) : 2,8 - 1,2 = 3,7`
`( 25,34 - x ) : 2,8 = 3,7 + 1,2`
`( 25,34 - x ) : 2,8 = 4,9`
` 25,34 - x = 4,9 xx 2,8`
` 25,34 -x = 13,72`
` x = 25,34 - 13,72`
`x = 11,62`
e)
` ( x + 1 ) + ( x + 5 ) + ( x + 9 ) + ... + ( x + 29 ) = 960`
Có tất cả số hạng là:
\(\left(29-1\right):4+1=8\left(số\right)\)
Tổng dãy số :
\(\left(29+1\right)\times8:2=120\)
` => ( x + x + x+....+ x) + ( 1 + 5 + 9 +....+ 29) = 960`
` 8x + 120 = 960`
` 8x = 960 - 120`
`8x = 840`
` x = 840 : 8`
`x = 105`
a,x - 1,44 : 1,2 = 5,9
x-1,2=5,9
x=5,9+1,2
x=7,1
b,86,5 - x : 8,6 = 23,7
x:8,6=86,5-23,7
x:8,6=62,8
x=62,8 x 8,6
x=540,08
c,( 25,34 - x ) : 2,8 - 1,2 = 3,7
( 25,34 - x ) : 2,8=3,7+1,2
( 25,34 - x ) : 2,8=4,9
( 25,34 - x ) =4,9 x 2,8
( 25,34 - x ) =13,72
x = 25,34-13,72
x=11,62
d,x + 75% x - 0,25 x = 22,5
x + 0,75 x-0,25x=22,5
x (1+0,75-0,25)=22,5
x * 1,5 = 22,5
x= 22,5:1,5
x=15
e,( x + 1 ) + ( x + 5 ) + ( x + 9 ) + ... + ( x + 29 ) = 960
(x+x+x+x+x+x+x+x)+(1+5+9+13+17+21+25+29)=960
8*x+120=960
8*x=960-120
8*x=840
x=840:8
x=105
a)
\(\begin{array}{l}{(1,2)^3}.x = {(1,2)^5}\\x = {(1,2)^5}:{(1,2)^3}\\x = {(1,2)^2}\\x = 1,44\end{array}\)
Vậy \(x = 1,44\).
b)
\(\begin{array}{l}{\left( {\frac{2}{3}} \right)^7}:x = {\left( {\frac{2}{3}} \right)^6}\\x = {\left( {\frac{2}{3}} \right)^7}:{\left( {\frac{2}{3}} \right)^6}\\x = \frac{2}{3}\end{array}\)
Vậy \(x = \frac{2}{3}\).
\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)( vô lý)
Vậy \(S=\varnothing\)
b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a: x/2=-5/y
=>xy=-10
=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)
b: =>xy=12
mà x>y>0
nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
c: =>(x-1)(y+1)=3
=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)
d: =>y(x+2)=5
=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)