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tính nhanh :
a) ( 3 - 1/4 + 2/3 ) - ( 5 + 1/3 - 6/5 ) - ( 6 - 7/4 + 3/2 )
giúp mk vs nha mai nộp bài r
a) \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=3-\frac{1}{4}+\frac{7}{4}-\frac{3}{2}+\frac{2}{3}-\frac{1}{3}-5+\frac{6}{5}-6\)
\(=3+\frac{3}{2}-\frac{3}{2}+\frac{1}{3}-11+\frac{6}{5}\)
\(=3+0+\frac{23}{15}-11\)
\(=\frac{68}{15}-\frac{165}{15}=\frac{-97}{15}.\)
\(D=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
\(E=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-1+\dfrac{1}{99}=\dfrac{2}{99}-1=-\dfrac{97}{99}\)
=1.4.2.5.....98.101/2.3.3.4.....99.100
=(1.2.3.....97.98)(4.5.....100.101)/(2.3.....99)(3.4.....100)
=1.101/99.3
=101/297
Bạn tuấn anh có thể giải thích rõ cho mik vì sao bạn có thể ra dược bước 1ko?
Ta có: \(A=1+2+2^2+2^3+...+2^{99}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow2A-A=A=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+...+2^{99}\right)=2^{100}-1\)
Vậy \(A=2^{100}-1\)
ta có: f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)
\(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)
\(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)
\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)
Chúc bn học tốt !!!!!!
Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............
a) 1/7 - 3/5x = 3/5
3/5x= 1/7 - 3/5
3/5x = -16/35
x= -16/35 : 3/5 = -16/21
b) 3/7 - 1/2x = 5/3
1/2x = 3/7 - 5/3 = -26/21
x= -26/21 : 1/2 = -52/21
Đặt A=1.2+2.3+...+99.100
3A=1.2.3+2.3.3+...+99.100.3
=1.2.(3-0)+2.3(4-1)+....+99.100(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+...+99.100.101-98.99.100
=99.100.101-0.1.2
=99.100.101
=>\(A=\frac{99.100.101}{3}=333300\)
Đặt \(A=1.2+2.3+3.4+4.5+...+99.100\)\(\Rightarrow3.A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3\)\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+99.100.\left(101-98\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.-2.3.4+4.5.6+3.4.5+...+\)\(99.100.101-98.99.100\)
\(=99.100.101\)
\(=999900\Rightarrow B=999900\div3=333300\)
Chưa chắc lắm đâu nha !