1 x 1 x 6
a.A=1 1/2 x 1 1/3 x 1 1/4 x 1 1/5 x...x1 1/2020 x 1 1/2021
b.B=1 1/2 + 1 1/6 + 1 1/12 +1 1/20+ ... + 1 1/380 + 1 1/420
c.C=1 6/8 x 1 6/18 x 1 6/30 x 1 6/44 x ... x 1 6/10700
a.A=1 1/2 x 1 1/3 x 1 1/4 x 1 1/5 x...x1 1/2020 x 1 1/2021
b.B=1 1/2 + 1 1/6 + 1 1/12 +1 1/20+ ... + 1 1/380 + 1 1/420
c.C=1 6/8 x 1 6/18 x 1 6/30 x 1 6/44 x ... x 1 6/10700
a.A=1 1/2 x 1 1/3 x 1 1/4 x 1 1/5 x...x1 1/2020 x 1 1/2021
b.B=1 1/2 + 1 1/6 + 1 1/12 +1 1/20+ ... + 1 1/380 + 1 1/420
c.C=1 6/8 x 1 6/18 x 1 6/30 x 1 6/44 x ... x 1 6/10700
1. (x-6)^2 = 2(x-6)
2. 2(x-3)^2 = (x-3)(x+5)
3. 4(x-3)=2x-5(2x+3)
4. x2 +4 -2 (x-1) = (x-2)^2
5. x-3/5 = 6 - 1-2x/3
6. x+2 = 6-5x/2
7. x+2/5 - x+3 = x-2/2
8. 2x-5/x-4 = 2x+1/x+2
9. X+3/x-3 - x-1/x+3 = x2 + 4x + 6/x2 -9
10. 3x-3/x2-9 -1/x-3 = x+1/x+3
11. X+1/x-1 - x-1/x+1 = 4/x2 -1
Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!
1/x(x+3) + 1/(x+3)(x+6) +1/(x+6)(x+9) + 1/(x+9)(x+12) =1/16
\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}+\dfrac{3}{\left(x+3\right)\left(x+6\right)}+...+\dfrac{3}{\left(x+9\right)\left(x+12\right)}=\dfrac{3}{16}\)
=>\(\dfrac{1}{x}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+6}+...+\dfrac{1}{x+9}-\dfrac{1}{x+12}=\dfrac{3}{16}\)=>\(\dfrac{1}{x}-\dfrac{1}{x+12}=\dfrac{3}{16}\)
=>\(\dfrac{x+12-x}{x\left(x+12\right)}=\dfrac{3}{16}\)
=>12/x(x+12)=3/16
=>4/x(x+12)=1/16
=>x(x+12)=64
=>x^2+12x-64=0
=>x^2+16x-4x-64=0
=>(x+16)(x-4)=0
=>x=4 hoặc x=-16
1/x-1-x^3-x/x^2+1(x/x^2-2x+1-1/x^2-1)
[2/(x+1)^3.(1/x+1)+1/x^2+2x+1(1/x^2+1)]:x-1/x^3=x/x-1
(x/x^2-36-x-6/x^2+6x):2x-6/x^2+6x+x/6-x
giúp mik với ;-; mik cần gấp
1, Tìm x, biết \(x^2\) – 36 = 0
A. x = 6. B. x = -6.
C. x = 6; x = -6. D. x = 36 hoặc x = - 36.
2, Tìm x, biết \(x^3\) – 3\(x^2\) + 3x - 1 = 0
A. x = 1. B. x = -1. C. x = 0. D. x = 2.
rút gọn BT
A= (x+1/x)^6 - ( x^6 +1/x^6) -2 : ( x+1/x)^3 +x^3 +1/x^3
x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
1 Tìm x:
a) 5.(1-x)-6.(1+x)=7.(3-x)
b) 2.(x+5)-3.(3x+7)=6.(1-x)+8x
c) x-2+3.(x-4)=5.(x-6)+7
d) x+2+3.(1-x)-5.(2-x)=6.(1-x)+7.(3x)
giờ làm vẫn đc đúng ko bạn