Tìm x biết:
\(|x-3,5|=7,5\) \(|x+\dfrac{4}{5}|\)- \(\dfrac{1}{2}\)= 0 3,6 - \(|x-0,4|=0\)
Bài1: tìm x biết
a) l x - 3,5 l = 7,5
b) l x + \(\dfrac{4}{5}\) l - \(\dfrac{1}{2}\) = 0
c) 3,6 - l x - 0,4 l = 0
d) \(\dfrac{-5}{12}\) : | \(\dfrac{-5}{6}\) : x | = \(\dfrac{-5}{9}\)
e) | x - 3,5|+|4,5- x|=0
bài 2:tính hợp lý
a) (-4,3)=[(-7,5)=(4,3)]
b) 45,3 + [(7,3)+(-22)]
c) [(-11.7)+5.5]+[11.7+(2.5)]
d) [(-6.8)+(-56.9)]+[2.8+5.9]
1)
a) \(|x-3,5|=7,5\)
\(\Rightarrow x-3,5=7,5\)
hay \(x-3,5=-7,5\)
TH1 : \(x-3,5=7,5\Rightarrow x=7,5+3,5=11\)
TH2 : \(x-3,5=-7,5\Rightarrow x=-7,5+3,5=-4\)
b) \(|x+\dfrac{4}{5}|-\dfrac{1}{2}=0\)
\(\Rightarrow\left(x+\dfrac{4}{5}\right)-\dfrac{1}{2}=0\) (chỉ có 1 TH vì số 0 ko phải dương or âm)
\(\left(x+\dfrac{4}{5}\right)=0+\dfrac{1}{2}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{4}{5}=\dfrac{5-8}{10}=\dfrac{-3}{10}\)
c) \(3,6-|x-0,4|=0\)
\(\Rightarrow3,6-\left(x-0,4\right)=0\) ( giải thích giống câu b )
\(\Rightarrow-\left(x-0,4\right)=0-3,6\)
\(\Rightarrow-\left(x-0,4\right)=-3,6\)
\(\Rightarrow-x+0,4=-3,6\) ( Phá dấu )
\(\Rightarrow-x=-3,6-0,4=-3,6+\left(-0,4\right)=-4\)
\(\Rightarrow x=4\)
d) \(-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)
\(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)
hay \(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{5}{9}\)
TH1 : \(-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{-5}{9}\Rightarrow\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\left(-\dfrac{5}{9}\right)\)
\(\Rightarrow\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{12}.\dfrac{9}{5}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{5}{6}:\dfrac{3}{4}=-\dfrac{5.4}{6.3}=-\dfrac{5.2}{3.3}=-\dfrac{10}{9}\)
TH2 : \(\Rightarrow-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{9}\)
\(\Rightarrow\)\(\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\dfrac{5}{9}=-\dfrac{5.9}{12.5}=-\dfrac{9}{12}=-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{5}{6}:\left(-\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{4}{3}=\dfrac{10}{9}\)
Vậy x = ....
e)
Vì \(|x-3,5|\ge0;|4,5-x|\ge0\) với mọi x
Do đó : \(|x-3,5|+|4,5-x|=0\)
\(\Rightarrow|x-3,5|=0;|4,5-x|=0\)
\(\Rightarrow x-3,5=0\) và \(4,5-x=0\)
\(\Rightarrow x=0+3,5=3,5\) và \(-x=0+4,5=4,5\Rightarrow x=-4,5\)
( không đồng thời xảy ra)
\(\Rightarrow\) Không tồn tại x thuộc Q để \(|x-3,5|+|4,5-x|=0\)
2)
a) Đề sai
b) (45,3 + 7,3) + (-22)
= 52,6 + (-22) = 30,6
c) [(-11.7) + (11.7)] + [5.5+10]
= 0 + 15.5 = 15.5
( Câu c bạn cho rối quá )
d) [(-6.8) + 2.8] + [(-56.9) + 5.9 ]
= (-4) + (-51) = 55
tìm x
a) I x - 3,5 I =7,5
b) I x + 4/5 I - 1/2 = 0
c) 3,6 - I x - 0,4 I = 0
d) I x - 3,5 I + I 4,5 - x I =0
a) \(\Leftrightarrow\left[{}\begin{matrix}x-3,5=7,5\\x-3,5=-7,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-4\end{matrix}\right.\)
b) \(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{2}\\x+\dfrac{4}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{10}\\x=-\dfrac{13}{10}\end{matrix}\right.\)
c) \(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
d) \(\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\4,5-x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=4,5\end{matrix}\right.\)(vô lý)
Vậy \(S=\varnothing\)
tim x biet
a)|x-3,5|=7,5
b)3,6-|x-0,4|=0
c)|x-3,5|+|4,5-x|=0
a) |x-3,5| = 7,5
TH1: => x - 3,5 = 7,5
=> x = 7,5 + 3,5 = 11
TH2 : x - 3,5 = -7,5
=> x = -7,5 + 3.5 = -4
b) 3,6 - | x - 0,4| = 0
=> | x - 0,4| = 3,6 - 0 = 3,6
Th1: x - 0,4 = 3,6
=> x = 0,4 + 3,6 = 4
th2: x - 0,4 = -3,6
=> x = 0,4 + (-3,6) = -3,2
c) |x - 3,5| + |4,5 - x | = 0
= a + a = 0 ( loại bỏ vì nếu vậy thì phép tính trên sẽ ko hợp lí)
= -a + a = 0
Ta có: x - 3,5 = -a
4,5 - x = a
=> 3,5 + -a = 4,5 - a = 4,5 + (-a)
Vậy , không có số x nào thỏa mãn đk trên (theo mk là thế!)
Tíc nhá!
Tìm x
a /x-3,5/=7,5
b /x+4phan5/-1phan2=0
c 3,6-/x-0,4/=0
a, |x - 3,5| = 7,5
=> x - 3,5 = 7,5 hoặc x - 3,5 = -7,5
=> x = 11 hoặc x = -4
vậy_
b, |x + 4/5| - 1/2 = 0
=> |x + 4/5| = 1/2
=> x + 4/5 = 1/2 hoặc x + 4/5 = -1/2
=> x = -7/10 hoặc x = -9/10
vậy_
c, 3,6 - |x - 0,4| = 0
=> |x - 0,4| = 3,6
=> x - 0,4 = 3,6 hoặc x - 0,4 = -3,6
=> x = 4 hoặc x = -3,2
vậy_
a)/x-3,5/=7,5
TH1---------x-3,5=7,5 x=7,5+3,5=11
TH2--------x-3,5=-7,5 x=-7,5+3,5=-4
vậy\(x\in\left\{11;-4\right\}\)
a) | x | = 3/7 b) | x | = 0 c) | x | = -8,7 d) | x - 2/5 | = 1/4 e) | x + 0,5 | - 3,9 = 0 f) 3,6 - |x - 0,4 | = 0.
g) | x - 3,5 | = 7,5 h) | x - 3,5 | + | 4,5 - x | = 0
a) |x| = 3/7
=> x = -3/7 hoặc x = 3/7
b) |x| = 0
=> x = 0
c) |x| = -8,7
=> x rỗng
d) |x - 2/5| = 1/4
<=> x - 2/5 = 1/4 hoặc x - 2/5 = -1/4
x = 1/4 + 2/5 x = -1/4 + 2/5
x = 13/20 x = 3/20
=> x = 13/20 hoặc x = 3/20
e) |x + 0,5| - 3,9 = 0
<=> |x + 0,5| = 0 + 3,9
<=> |x + 0,5| = 3,9
<=> x + 0,5 = -3,9; 3,9
<=> x + 0,5 = 39 hoặc x + 0,5 = -3,9
x = 3,9 - 0,5 x = -3,9 - 0,5
x = 3,4 x = -4,4
=> x = 3,4 hoặc x = -4,4
f) 3,6 - |x - 0,4| = 0
<=> -|x - 0,4| = 0 - 3,6
<=> -|x - 0,4| = -3,6
<=> |x - 0,4| = 3,6
<=> x - 0,4 = -3,6; 3,6
x - 0,4 = -3,6 hoặc x - 0,4 = 3,6
x = -3,6 + 0,4 x = 3,6 + 0,4
x = -3,2 x = 4
=> x = -3,2 hoặc x = 4
g) |x - 3,5| = 7,5
<=> x - 3,5 = -7,5; 7,5
x - 3,5 = -7,5 hoặc x - 3,5 = 7,5
x = -7,5 + 3,5 x = 7,5 + 3,5
x = -4 x = 11
=> x = -4 hoặc x = 11
bài 1 :tìm x,y biết
a) (5x+1)=\(\dfrac{36}{49}\) b) (x-2/9) = (2/3) c)(8x-1) 2x+1= 5^2 x+1
d) (x-3,5)^x+(y - 1/10)^4=0
`(5x+1)=36/49`
`<=> 5x = 36/49-1`
`<=> 5x = -13/49`.
`<=> x = -13/245.`
Vậy `x = -13/245`.
`b, x-2/9 = 2/3`.
`<=> x = 2/3 + 2/9`
`<=> x = 8/9`.
Vậy `x = 8/9`.
c: (8x-1)^(2x+1)=5^(2x+1)
=>8x-1=5
=>8x=6
=>x=3/4
d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0
=>x-3,5=0 và y-0,1=0
=>x=3,5 và y=0,1
tìm x \(\in\) Q biết rằng
\(\dfrac{11}{12}\) - ( \(\dfrac{2}{5}\) + x ) = \(\dfrac{2}{3}\)
2x \(\times\) ( x - \(\dfrac{1}{7}\) ) = 0
\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) : x = \(\dfrac{2}{5}\)
1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)
\(\Leftrightarrow x=-\dfrac{3}{20}\)
2) \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)
\(\Leftrightarrow4x=-\dfrac{20}{7}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
Tìm x, biết:
\(\dfrac{1}{2}x+\dfrac{4}{5}=2x-\dfrac{8}{5}\)
\(\sqrt{x}=5\) (x ≥ 0)
x2 = 3
`#3107.101107`
`1/2x + 4/5 = 2x - 8/5`
`=> 1/2x - 2x = -4/5 - 8/5`
`=> -3/2x = -12/5`
`=> x = -12/5 \div (-3/2)`
`=> x = 8/5`
Vậy, `x = 8/5`
_____
`\sqrt{x} = 5`
`=> x = 5^2`
`=> x = 25`
Vậy, `x = 25`
___
`x^2 = 3`
`=> x^2 = (+-\sqrt{3})^2`
`=> x = +- \sqrt{3}`
Vậy, `x \in {-\sqrt{3}; \sqrt{3}}.`
Tìm x biết:
\(a,3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\)
\(b,\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(c,\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(d,\left(2x-3\right)\left(6-2x\right)=0\)
\(e,x:\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{2}{3}\)
\(f,\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(g,2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(h,\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(i,\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(j,\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(k,\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(l,\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(m,3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(n,60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(p,-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(q,3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\)
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3