24:(3x-2)=-3
1. Tính
a) x3y2+2x3y2+3x3y2+.......+100x3y2
b) x3y24-2x3y24+3x3y24+.....+2009x3y24-2010x3y24
a/10-2(4-3x)=-4
b/-12+3(-x+7)+=-18
c/24:(3x-2)=-3
d/-45:5.(-3-2x)=3
a) \(10-2\left(4-3x\right)=-4\)
\(2\left(4-3x\right)=10-\left(-4\right)\)
\(2\left(4-3x\right)=14\)
\(4-3x=\dfrac{14}{2}\)
\(4-3x=7\)
\(3x=4-7\)
\(3x=-3\)
\(x=-1\)
b) \(-12+3\left(-x+7\right)=-18\)
\(3\left(-x+7\right)=-18+12\)
\(3\left(-x+7\right)=-6\)
\(-x+7=-2\)
\(-x=-2-7\)
\(-x=-9\)
\(x=9\)
c) \(24:\left(3x-2\right)=-3\)
\(3x-2=24:\left(-3\right)\)
\(3x-2=-8\)
\(3x=-8+2\)
\(3x=-6\)
\(x=-2\)
d) \(-45:5\left(-3-2x\right)=3\)
\(5\left(-3-2x\right)=-45:3\)
\(5\left(-3-2x\right)=-15\)
\(-3-2x=-15:5\)
\(-3-2x=-3\)
\(-2x=-3+3\)
\(-2x=0\)
\(x=0\)
(2x^2+3x-1)^2-5*(2x^2+3x+3)+24=0
Tìm x:
C, X^2-9=2×(x+3)^2
b, x^3-3x^2+3x-1=0
d, x^2-8x+3x-24=0
Giúp mk với. Mk cảm ơn
c) \(x^2-9=2\cdot\left(x+3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
d) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
a) \(x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)
\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
c) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
các bạn giải giúp mình nka, mik dg cần gấp
1. Tính
a) x3y2+2x3y2+3x3y2+.......+100x3y2
b) x3y24-2x3y24+3x3y24+.....+2009x3y24-2010x3y24
1. Tính
a) x3y2+2x3y2+3x3y2+.......+100x3y2
b) x3y24-2x3y24+3x3y24+.....+2009x3y24-2010x3y24
các thánh giải giúp em với bí rồi ạ
tìm x : a) (x + 1)^3 + (3 - 2)^3 = 2x^3 + 2(2x - 1)^2 - 9
b) (3x^3+24) : (x+2) + (2x^3−54) : (x^2+3x+9) = 6
a: \(\left(x+1\right)^3+\left(x-2\right)^3=2x^3+2\left(2x-1\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=2x^3+2\left(4x^2-4x+1\right)-9\)
\(\Leftrightarrow2x^3-3x^2+15x-7=2x^3+8x^2-8x-7\)
\(\Leftrightarrow-11x^2+23x=0\)
\(\Leftrightarrow x\left(-11x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{11}\end{matrix}\right.\)
1/ (2x - 5) +17 =6
2/ 10 - 2(4 - 3x) = - 4
3/- 12 +3(- x + 7) = -18
4/ 24 :(3x - 2)= - 3
5/ - 45 :5.(- 3- 2x) = 3
\(\text{1/ (2x - 5) +17 =6}\\ 2x-5=6-17\\ 2x-5=-11\\ 2x=-11+5\\ x=-6:2\\ x=-3\)
\(\text{2/ 10 - 2(4 - 3x) = - 4}\\ 2\left(4-3x\right)=10-\left(-4\right)\\ 2\left(4-3x\right)=14\\ 4-3x=14:2\\ 4-3x=7\\ 3x=4-7\\ x=-3:3\\ x=-1\)
tìm x
12-2(4-3x)=-4
24:(3x-2)=-3
-45:5.(-3-2x)=3
12 - 2(4 - 3x) = -4
2(4 - 3x) =-4+12
2(4 - 3x) = 8
4 - 3x = 8 : 2
4 - 3x = 4
3x = 4 - 4
3x = 0
x = 0 : 3
x = 0
Vậy x = 0
Đợi mk chút xíu tý mk giải nốt
Tk cho mk nha
Mk cảm ơn
Trả lời :
a, = 0
Hok tốt nhé bạn ((:
b)24 : (3x - 2) = -3
3x - 2 = 24 : -3
3x - 2 = -8
3x = -8 + 2
3x = -6
x = -6 : 3
x = -2
Vậy x = -2
Tk cho mk nha
Mk cảm ơn
Giải phương trình:
(2x^2+3x-1)^2-5(2x^2+3x+3)+24=0
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