Tìm x thuộc N,biết:
a) x+245=43.11
b) x-382=159:3
c)7x+2x=918
d)3.(x-2)+150=240
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tìm x thuộc N biết:
X+245=43×11
=>x+245=473
=>x=228
X-382=159:3
=>x-382=53
=>x=53+382=438
7×X+2×X=918
=>(7+2).x=918
=>9x=918
=>x=918:9=102
3×(X-2)+150=240
=>3.(x-2)=240-150
=>3.(x-2)=90
=>x-2=90:3=30
=>x=30+2=32
(X+13):5=12
=>(x+13)=12.5
=>x+13=60
=>x=60-13
=>x=47
Bấm **** nha
360:(X-7)=90
X+245=43*11
X+245=473
x=473-245
X=228
X-382=159:3
X-382=53
X=382+53
X=435
7*X+2*X=918
(7+2)*X=918
9*X=918
X=918:9
X=102
3*(X-2)+150=240
3*(X-2)=240-150
3*(X-2)=90
X-2=90:3
X-2=30
X=30-2
X=28
(X+13):5=12
X+13=12*5
X+13=60
X=60-13
X=47
360:(X-7)=90
X-7=360:90
X-7=4
X=7+4
X=11
a,(2x-5^2)-4x(x-3)=0
=> 2x-25-4x2+12x=0
=>-4x2+14x-25=0
đề bài ý a sai nha
b, 6x2-7x=0
=>x(6x-7)=0
=>x=0 và 6x-7=0
=>x=0 và x=7/6
vậy x=0 và x=7/6
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
7x + 2x = 918
(7+2)x=918
9x=918
x = 102
3.(x-2)+150=240
3.(x-2) = 240 - 150 = 90
x - 2 = 90 : 3 = 30
x= 30 + 2
x=32
7x + 2x = 918
(7+2)x=918
9x=918
x = 102
3.(x-2)+150=240
3.(x-2) = 240 - 150 = 90
x - 2 = 90 : 3 = 30
x= 30 + 2
x=32
`a)|2x-15|=13`
`**2x-15=13`
`<=>2x=28`
`<=>x=14.`
`**2x-15=-13`
`<=>2x=-2`
`<=>x=-1.`
`b)|7x+3|=66`
`**7x+3=66`
`<=>7x=63`
`<=>x9`
`**7x+3=-66`
`<=>7x=-69`
`<=>x=-69/7`
`c)|5x-2|=0`
`<=>5x-2=0`
`<=>5x=2`
`<=>x=2/5`
\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
Vậy ...
\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)
Vậy ...
\(c,\Leftrightarrow5x-2=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
a)x+245=43.11
x+245=473
x=473-245
x=228
b) x-382=159:3
x-382=53
x=53+382
x=435
k mk nhe