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5x.(-x)2+1=6
5x.x+1=6
5x2+1=6
5x2=6-1
5x2=5
x2=5:5
x2=1
x2=12
=>x=1
(15-x)+(x-12)=7-(-8+x)
15-x+x-12=7+8-x
3-x+x=15-x
3=15-x
x=15-3
x=12
4x3=4x
Để 4x3=4x=>x3=x
=>x=1
Câu cuối cùng mình ko bik.
hoc tốt
a)Ta thấy:
\(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x+a\right)-x}{x\left(x+a\right)}\)
\(=\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrowđpcm\)
b)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)}-\dfrac{x\left(x+1\right)}{x\left(x+1\right)^2\left(x+2\right)}\)
\(=\dfrac{x+2}{x\left(x+1\right)\left(x+2\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)-x}{x\left(x+1\right)\left(x+2\right)}=\dfrac{2}{x\left(x+1\right)\left(x+2\right)}\Rightarrowđpcm\)
c)Ta thấy:
\(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}-\dfrac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}-\dfrac{x\left(x+1\right)\left(x+2\right)}{x\left(x+1\right)^2\left(x+2\right)^2\left(x+3\right)}=\dfrac{x+3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}-\dfrac{x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{x+3-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\dfrac{3}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\Rightarrowđpcm\)
a/ \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\)
Ta có: \(\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{x+a}{x\left(x+a\right)}-\dfrac{x}{x\left(x+a\right)}\)
\(=\dfrac{\left(x-x\right)+a}{x\left(x+a\right)}\) hay \(\dfrac{a}{x\left(x+a\right)}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+a}=\dfrac{a}{x\left(x+a\right)}\left(đpcm\right)\)
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)
\(\left(x\cdot100\right)+\left(1+2+...+100\right)=5750\)
\(\left(x\cdot100\right)+\left(100+1\right)\cdot\frac{100}{2}=5750\)
\(\left(x\cdot100\right)+101\cdot50=5750\)
\(\left(x\cdot100\right)+5050=5750\)
\(x\cdot100=5750-5050\)
\(x\cdot100=700\)
\(x=700\div100\)
\(x=7\)
Ta có: ( x+1)+(x+2)+(x+3)+.....+(x+99)+(x+100)=5750
<=>(x+x+x+....+x+x)+(1+2+3+..+99+100)=5750
<=> 100x+5050=5750
=>100x=5750-5050
=>100x=700
=>x=700:100
=>x=7
Vậy x=7
hoặc mở câu hỏi tương tự tham khảo.
TH1 : x>0
xong phá ngoặc làm nhé
Th2 : x<0
xong cx phá ngoạc làm
|x|, |x+1|, |x+2|, ..., |x+100| đều lớn hơn hoặc bằng 0 với mọi x
=> Tổng của chúng lớn hơn hoặc bằng 0
=> 102x lớn hơn hoặc bằng 0
=> x lớn hơn hoặc bằng 0
=> |x| = x, |x+1| = x+1 , ..., |x+100| = x+100
=> 101x + 1+2+3+...+100 = 102x
=> x = (100+1).100/2 = 5050