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\(\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\\ \dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}=0\\ x=\dfrac{2}{3}\)
a: (x-1)(x+2)(-x-3)=0
=>(x-1)(x+2)(x+3)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
b: (x-7)(x+3)<0
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
Ta có: x + (x + 1) + (x + 2) + (x + 3) + 35 = 0
4x + 41 = 0
4x = -41
=> x = -41/4
-(1/3)-1/15-1/35-...-1/x.(x+2)=20/-21
=>1/3+1/15+1/35+...+1/x(x+2)=20/21
=>1/2 * (2/3 + 2/15 + 2/35 + ... + 2/x(x+2) )=20/21
=>\(\dfrac{2}{1\cdot3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{20}{21}:\dfrac{1}{2}\)
=>\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{40}{41}\)
=>\(1-\dfrac{1}{x+2}=\dfrac{40}{41}\)
=>\(\dfrac{1}{x+2}=1-\dfrac{40}{41}=\dfrac{1}{41}\)
=>x+2=41=>x=39
(1/3)-1/15-1/35-...-1/x.(x+2)=20/-21
=>1/3+1/15+1/35+...+1/x(x+2)=20/21
=>1/2 * (2/3 + 2/15 + 2/35 + ... + 2/x(x+2) )=20/21
=>21⋅3+23.5+25.7+...+2x(x+2)=2021:1221⋅3+23.5+25.7+...+2x(x+2)=2021:12
=>1−13+13−15+15−17+...+1x−1x+2=40411−13+13−15+15−17+...+1x−1x+2=4041
=>1−1x+2=4041
Ý là đề vầy chứ gì:
\(5^x.5^{x+1}.5^{x+2}=10^{18}:2^{18}\)
⇔\(5^{3x+3}=5^{18}\)
⇔\(3x+3=18\)
⇔\(x=5\)
Vậy x=5
1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
đề bài vô lý vcl , x thế cộng đến số tự nhiên lớn nhất r cộng 35 cx được (._.'')
hình như sai đề rồi