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
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!
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 /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\(x\in\left\{11;-4\right\}\)
Bài 1: tìm x
c) \(|\) x \(|\) = 3,5
d) \(|\) x \(|\) = -2,7
e) \(|\) 1 - x \(|\) + 0,73 = 3
f) 52 . 73 . 11x + 52 . 72 .11 = 0
g) (3 . 5 + 5 . 7)x + ( 3 . 5 + 5 . 7) + (3 . 5 + 5 . 7) = 0
h) 52 . 72 . 112x - 52 . 72 . 114 = 0
l) \(|\) x + \(\dfrac{3}{4}\) \(|\) - 5 = -2
c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
c) x = 3,5 hoặc -3,5
d) x = -2,7
e) x = -1,27
f) x = 0
g) x = -2
h) x = 0
l) x = 9/4
x = -15/4
Tìm x biết:
\(|x-3,5|=7,5\) \(|x+\dfrac{4}{5}|\)- \(\dfrac{1}{2}\)= 0 3,6 - \(|x-0,4|=0\)
\(\left|x-3,5\right|=7,5\)
\(\Leftrightarrow x-3,5=7,5\Rightarrow x=11\)
\(\Leftrightarrow x-3,5=-7,5\Rightarrow x=-4\)
\(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{4}{5}=\dfrac{1}{2}\Rightarrow x=\dfrac{-3}{10}\)
\(\Leftrightarrow x+\dfrac{4}{5}=-\dfrac{1}{2}\Rightarrow x=\dfrac{-13}{10}\)
\(3,6-\left|x-0,4\right|=0\)
\(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow x-0,4=3,6\Rightarrow x=4\)
\(\Leftrightarrow x-0,4=-3,6\Rightarrow x=-3,2\)
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 \(S=\varnothing\)
B2: Tìm x
a) | x | = 3,5
b) |1-x |+ 0,73 = 3
c) -5/8 + x = 4/9
d) x.(1/5 + 1/4) - (1/7 + 1/8) =0
e) | x | + -2,7
f) |x + 3/4 | -5 = -2
a) x ( x + 0 ) = 0
b) ( x - 1 ) ( 7 - x ) = 0
c) ( -x + 5 ) ( 3 - x ) = 0
d) ( x + 5 ) + ( x - 9 ) = 13
e) ( 4 + x ) + ( x - 7 ) = x + 2
f) ( 3x + 5 ) - ( 2x - 7) = 4 - x
g) ( x - 1 )2 = 36
h) ( 3 - x ) 3 = -27
b) ( x - 1 ) ( 7 - x ) = 0
=> x - 1 =0 hoặc 7 - x = 0
TH1 : x - 1 = 0
x = 0 + 1
x = 1
TH2 : 7 - x = 0
x = 7 - 0
x = 7
Vậy x = 1 hoặc x = 7
a: =>x^2=0
=>x=0
b:=>x+1=0 hoặc 7-x=0
=>x=-1 hoặc x=7
c: =>(x-5)(x-3)=0
=>x=5 hoặc x=3
d: =>x+5+x-9=13
=>2x-4=13
=>2x=17
=>x=17/2
e: =>x+4+x-7=x+2
=>x-3=2
=>x=5
f: =>3x+5-2x+7=4-x
=>x+12=4-x
=>2x=-8
=>x=-4
g: =>x-1=6 hoặc x-1=-6
=>x=7 hoặc x=-5
h:=>3-x=-3
=>x=6
A(2,3x-6,5)(0,1x+2)=0
B(2x+7)(x-5)(5x+1)=0
C(x-1)(2x+7)(x2+2)=0
D(4x-10)(24+5x)=0
E(3,5-7x)(0,1x+2,3)=0
F (5x+2)(x+7)=0
G15 (x+9)(x-3)(x+21)=0
H (x2+1)(×2-4x+4)=0
I(3x-2)(2 (x+3)/9-4x-3/5)=0
\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)