tìm x
a)x-1,23=5,6
b)x+3=9,87
c)x-2,3=5,6+3,1
d)x+3,4=15,6+1,3
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\) \(x\times5,6+x\times3,4+x=123,4\)
\(x\times5,6+x\times3,4+x\times1=123,4\)
\(x\times\left(5,6+3,4+1\right)=123,4\)
\(x\times10=123,4\)
\(x=123,4\div10\)
\(x=12,34\)
b) \(x\div3,1+x\times5+x\div6,7=52,1\)
\(x\div3,1+x\div0,2+x\div6,7=52,1\)
\(x\div\left(3,1+0,2+6,7\right)=52,1\)
\(x\div10=52,1\)
\(x=52,1\times10\)
\(x=521\)
\(a,2,3:0,25+7,7.4\\ =2,3.4+7,7.4\\ =\left(2,3+7,7\right).4=10.4=40\\ b,4,2.8+5,8:0,125\\ =4,2.8+5,8.8\\ =\left(4,2+5,8\right).8=10.8=80\\ c,3,4.4+5,6:0,25+4\\ =3,4.4+5,6.4+4\\ =\left(3,4+5,6+1\right).4=10.4=40\\ d,4:0,2+12:0,5+1,2:0,25\\ =4.5+12.2+1,2.4\\ 4.5+6.4+1,2.4\\ =4\left(5+6+1,2\right)\\ =4.12,2=48,8\)
Lời giải:
$(2,41+3,12)\times 2,2=5,53\times 2,2=12,166$
$(5,6-3,21)\times 2,3=2,39\times 2,3=5,497$
$3,276:1,4+7,66=2,34+7,66=10$
$28,886:1,3+77,78=22,22+77,78=100$
a) \(210:x=2,1:0,25=8,4\)
\(x=210:8,4=25\)
b) \(x.2+x.0,5=6,25\)
\(2,5x=6,25\)
\(x=6,25:2,5=2,5\)
c) \(x:0,25-x=15,6\)
\(3x=15,6\)
\(x=5,2\)
d) \(x:12,5=3,75.4=15\)
\(x=15.12,5=187,5\)
e) \(x.12,6-x.5,6=42\)
\(7x=42\)
\(x=6\)
g) \(x:0,1-x:4-x.0,75=2,25\)
\(10x-\dfrac{x}{4}-\dfrac{3x}{4}=2,25\)
\(9x=2,25\)
\(x=0,25\)
a) 136 - x : 48 = 27,8 +92
136 - x : 48 = 119,8
x : 48 = 136 - 119,8
x : 48 = 16,2
x = 16,2 * 48
x = 777,6
b) 1,23 + x : 4 = 5,6
x : 4 = 5,6 * 1,23
x : 4 = 6,888
x = 6,888 * 4
x = 27,552
k cho mình nha.
a) 136 - x : 48 = 27,8 +92
136 - x : 48 = 119,8
x : 48 = 136 - 119,8
x : 48 = 16,2
x = 16,2 x48
x = 777,6
Vậy x = 777,6
b) 1,23 + x : 4 = 5,6
x : 4 = 5,6 - 1,23
x : 4 = 4,37
x = 4,37 x 4
x = 17,48
Vậy x = 17,48
Cấp 2 rồi thì phải có VẬY nha
Giải:
a) \(\left|x\right|=1,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,3\\x=-1,3\end{matrix}\right.\)
Vậy ...
b) \(\left|x+1,3\right|=3,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1,3=3,3\\x+1,3=-3,3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4,6\end{matrix}\right.\)
Vậy ...
c) \(\left|5,6-x\right|=4,6\)
\(\Leftrightarrow\left[{}\begin{matrix}5,6-x=4,6\\5,6-x=-4,6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10,2\end{matrix}\right.\)
Vậy ...
d) \(\left|3x-5\right|+\left|x-2\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\x=2\end{matrix}\right.\)
=> Phương trình vô nghiệm
Vậy ...
* Trả lời:
\(a.\left|x\right|=1,3\)
\(\Rightarrow x=1,3\) hoặc \(x=-1,3\)
Vậy \(x=\pm1,3\)
\(b.\left|x+1,3\right|=3,3\)
\(\Rightarrow x+1,3=3,3\) hoặc \(x+1,3=-3,3\)
\(\Rightarrow x=3,3-1,3\) | \(x=-3,3-1,3\)
\(\Rightarrow x=2\) | \(x=-4,6\)
Vậy \(x=2\) ; \(x=-4,6\)
\(c.\left|5,6-x\right|=4,6\)
\(\Rightarrow5,6-x=4,6\) hoặc \(5,6-x=-4,6\)
\(\Rightarrow-x=4,6-5,6\) | \(-x=-4,6-5,6\)
\(\Rightarrow-x=-1\) | \(-x=-10,2\)
\(\Rightarrow x=1\) | \(x=10,2\)
Vậy \(x=1\) ; \(x=10,2\)
\(d.\left|3x-5\right|+\left|x-2\right|=0\)
Lý luận: giá trị tuyệt đối luôn lớn hơn hoặc bằng 0
Nên \(\left|3x-5\right|+\left|x-2\right|\ne0\)
\(\Rightarrow\) Không có giá trị \(x\)
a)x-1,23=5,6
x=5,6+1,23
x=6,83
b)x+3=9,87
x=9,87-3
x=6,87
c)x-2,3=5,6+3,1
x-2,3=8,7
x=8,7+2,3
x=11
d)x+3,4=15,6+1,3
x+3,4=16,9
x=16,9-3,4
x=13,5
`a)x-1,23=5,6`
`x=5,6+1,23`
`x=6,83`
_______________________________
`b)x+3=9,87`
`x=9,87-3`
`x=6,87`
_______________________________
`c)x-2,3=5,6+3,1`
`x-2,3=8,7`
`x=8,7+2,3`
`x=11`
_______________________________
`d)x+3,4=15,6+1,3`
`x+3,4=16,9`
`x=16,9-3,4`
`x=13,5`