tim x biet 2x+5/3x-1=1/13
tim x biet: 2x+5/3x-1=1/13
bai 1)tim x biet
a )24-(36+5)=x
b)14-21=(13-x)-(15+8)
bai 2) tim x biet
a)17-x=-25+(-16+9)
b)3x-21=-19-(-2x)
Bài 1: Tìm x, biết
a )24-(36+5)=x b)14-21=(13-x)-(15+8)
24-41=x (13-x)-23=-7
x=-17 13-x=(-7)+23
Vậy x=-17 13-x=16
x=13-16
x=-3 Vậy x=-3
Bài 2:Tìm x, biết
a)17-x=-25+(-16+9) b)3x-21=-19-(-2x)
17-x=-25+(-7) 3x-21=-19+2x
17-x=-32 3x-2x=-19+21
x=17-(-32) x=4
x=49 Vậy x=4
Vậy x=49
Bài 1:
a. 24 - (36+5) = x
=> 24 - 41 = x
=> -17 = x
=> x = -17
b. 14 - 21 = (13 - x) - (15 + 8)
=> -7 = 13 - x - 23
=> -7 - 13 + 23 = -x
=> 3 = -x
=> x = -3
Bài 2:
a. 17 - x = -25 + (-16 + 9)
=> 17 - x = -25 + (-7)
=> 17 - x = -32
=> 17 + 32 = x
=> x = 49
b. 3x - 21 = -19 - (-2x)
=> 3x - 21 = -19 + 2x
=> 3x - 2x = -19 + 21
=> x = 2
Bài1
a)24-(36+5)=x\(\Rightarrow\)x=24-41=-17
b)14-21=(13-x)-(15+8)\(\Rightarrow\)(13-x)-(15+8)=-7\(\Rightarrow\)13-x=-7+23=16\(\Rightarrow\)x=13-16=-3
Bài2
a)17-x=-25+(-16+9)\(\Rightarrow\)17-x=-25+-7\(\Rightarrow\)17-x=-32\(\Rightarrow\)x=17-(-32)=17+32=49
b)3x-21=-19-(-2x)\(\Rightarrow\)3x-(-2x)=-19-21\(\Rightarrow\)3x+2x=-40\(\Rightarrow\)5x=-40\(\Rightarrow\)x=-8
tim x biet;(2x-1)(3x+1)+(3x-4)(3-2x)=5
ta co (2x-1)(3x+1)+(3x+4)(3-2x)=5
(=)6x2-3x+2x-1+6x-6x2+12-8x=5
(=)-4x+11=5
(=)-4x=-6
(=)x=3/2
(2x-1)(3x+1)+(3x-4)(3-2x)=5
<=> 6x2+2x-3x-1+9x-6x2-12+8x=5
<=> 16x-13=5
<=> 16x = 18
<=> x=9/8
tim x biet
45-(3x+9)/13=42
(x+1)+(2x+2)+.....+(20x+20)=1050
a) \(\frac{45-\left(3x+9\right)}{13}=42\)
\(\Leftrightarrow45-\left(3x+9\right)=546\)
\(\Leftrightarrow3x+9=-501\)
\(\Leftrightarrow3x=-510\)
\(\Leftrightarrow x=-170\)
b) \(\left(x+1\right)+\left(2x+2\right)+...+\left(20x+20\right)=1050\)
\(\Leftrightarrow x+1+2x+2+...+20x+20=1050\)
\(\Leftrightarrow\left(x+1\right)+2\left(x+1\right)+....+20\left(x+1\right)=1050\)
\(\Leftrightarrow\left(x+1\right)\left(1+2+3+...+20\right)=1050\)
\(\Leftrightarrow\left(x+1\right)\left[\frac{\left(20+1\right).20}{2}\right]=1050\)
\(\Leftrightarrow\left(x+1\right).210=1050\)
\(\Leftrightarrow x+1=5\)
\(\Leftrightarrow x=4\)
tim x biet a) |3x-2|+5x=4x-10
b)3+ |2x+5|>13
tim x biet :
a) |x+1| +|x-5| = 7
b)|2x + 5|=3x-8
B, => 2x+5=3x-8
2x-3x=-8-5
-x=-13
=>x=13
hoặc 2x+5=-3x+8
2x+3x=8-5
5x=3
x=\(\frac{3}{5}\)
\(\left|x+1\right|+\left|x-5\right|=7.\)
\(Th1:x+1< 0;x-5< 0\)
\(x< -1;x< 5\Rightarrow x< -1\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)-\left(x-5\right)=7\)
\(-x-1-x+5=7\)
\(-2x+4=7\)
\(-2x=3\)
\(x=-1,5\left(tm\right)\)
\(Th2:x+1>0;x-5>0\)
\(x>-1;x>5\Rightarrow x>5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(x+1+x-5=7\)
\(2x-4=7\)
\(2x=11\)
\(x=5,5\)\(\left(tm\right)\)
\(Th3:x+1\le0;x-5>0\)
\(x\le-1;x>5\)(không xảy ra)
\(Th4:x+1>0;x-5\le0\)
\(x>-1;x\le5\Rightarrow-1< x\le5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)+x-5=7\)
\(-x-1+x-5=7\)( không xảy ra)
Vậy x = -1,5 hoặc x = 5,5
\(\)
\(\left|2x+5\right|=3x-8.\)
\(Th1:\)
\(2x+5=-\left(3x-8\right)\)
\(2x+5=-3x+8\)
\(2x+3x=8-5\)
\(5x=3\)
\(x=\frac{3}{5}\)
\(Th2:\)
\(2x+5=3x-8\)
\(2x-3x=-8-5\)
\(-x=-13\)
\(x=13\)
Vậy x = 13 hoặc x = \(\frac{3}{5}\)
Tim x biet: x+1/1+2x+3/3+3x+5/5+...............+20x+39/9=22+4/3+6/5+................+40/39
tim x biet : \(\left|2x-5\right|+1=3x\)
Ta có:
|2x-5|+1=3x
=> |2x-5|=3x-1
=> 2x-5=3x-1 ; 2x-5=1-3x
=> -5+1=3x-2x ; 2x+3x=1+5
=>x=-4 ; 5x =6
; x=\(\dfrac{5}{6}\)
tim x thuoc z
1)26-|x+9|=-13
2)|x+7|-13=25
tim x biet
1)123-3.(x+4)=23
2)720:[41-(2x-5)]=23.5
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự