Tính [x]=?
[2.(6)]3=?
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
6 x 2 =12
3 x 6 = 18
6 x 5 = 30
2 x 6 = 12
6 x 3 = 18
5 x 6 = 30
5 x 6 = 30 2 x 6 = 12 3 x 6 = 18 4 x 6 = 24
6 x 5 = 30 6 x 2 = 12 6 x 3 = 18 6 x 4 = 24
\(\dfrac{2x-6}{x^3-3x^2-x+3}+Q=\dfrac{6}{x-3}+\dfrac{2x^2}{x^2-1}\)
\(\Rightarrow Q=\dfrac{6}{x-3}+\dfrac{2x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x-6}{\left(x-3\right)\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{6x^2-6+2x^3-6x^2-2x+6}{\left(x-3\right)\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x^3-2x}{\left(x-3\right)\left(x+1\right)\left(x-1\right)}=\dfrac{2x\left(x-1\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)\left(x-1\right)}=\dfrac{2x}{x-3}\)
a) 6 x 2 = 12 6 x 3 = 18 6 x 8 = 48 6 x 1 = 6
6 x 4 = 24 6 x 5 = 30 6 x 9 = 54 1 x 6 = 6
6 x 6 = 36 6 x 7 = 42 6 x 10 = 60 0 x 6 = 0
b) 6 x 5 = 30 6 x 4 = 24 3 x 6 = 18 2 x 6 = 12
5 x 6 =30 4 x 6 = 24 6 x 3 = 18 6 x 2 =12
a,C1
(5/6 + 5/8) x 2/3
= 35/24 x 2/3
=35/36
C2
(5/6 + 5/8) x 2/3
= 5/6x2/3 + 5/8x2/3
= 10/18 + 10/24
= 35/36
b,C1
7/5 x 3/4 - 1/2 x 3/4
= 21/20 - 3/8
= 27/40
C2
7/5 x 3/4 - 1/2 x 3/4
=3/4 x (7/5-1/2)
=3/4x9/10
=27/40
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
a, 5/6 x (1/8x2/3)
=5/6 x 1/12
=5/72
b,(7/5 - 1/2) x 3/4
=0,9 x 3/4
=3,06
c,(5/6+5/8) x 2/3
=70/48 x 2/3
=35/36
1/
\(x^2+y^2=\left(x-y\right)^2+2xy=2^2+2.1=6\)
2/
\(x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)=2\left(6+1\right)=14\)
3/
\(x^2-y^2=\left(x-y\right)\left(x+y\right)=2\left(x+y\right)\) (3)
Ta có
\(x^2+y^2=\left(x+y\right)^2-2xy=\left(x+y\right)^2-2=6\)
\(\Rightarrow\left(x+y\right)^2=8\Rightarrow\left(x+y\right)=\pm2\sqrt{2}\) Thay vào (3)
\(\Rightarrow x^2-y^2=2.\pm2\sqrt{2}=\pm4\sqrt{2}\)
4/
\(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)\) (4)
Ta có
\(x^3-y^3=14\) (cmt)
Ta có
\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right).5=\pm2\sqrt{2}.5=\pm10\sqrt{2}\)
\(\Rightarrow x^6-y^6=\pm10\sqrt{2}.14=\pm140\sqrt{2}\)
a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)
\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)
b: x=2 ko thỏa mãn ĐKXĐ
=>Loại
Khi x=3 thì A=-1/(3-2)=-1
c: A=2
=>x-2=-1/2
=>x=3/2