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19 tháng 4 2017

\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)

\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)

\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)

\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)

Vì (2) > (1) => B > A

29 tháng 8 2017

A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)

=>-x^2+2x-1=10x-5x^2-11x-22

=>-x^2+2x-1=-5x^2-x-22

=>4x^2+3x+21=0

=>PTVN

b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)

=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)

=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80

=>20x+16=32x-80

=>-12x=-96

=>x=8

c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)

=>6x-18+7x-35=13x+4

=>-53=4(loại)

d: =>3(2x-1)-5(x-2)=3(x+7)

=>6x-3-5x+10=3x+21

=>3x+21=x+7

=>x=-7

e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1

=>-9x^2+9x-9=-9x^2+1

=>9x=10

=>x=10/9

30 tháng 3 2018

Hỏi đáp Toán

30 tháng 3 2018

Dài quá c ơi :<

4 tháng 12 2018

a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)

( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)

(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)

1/x +1/x+4

2x+4/x(x+4)

4 tháng 12 2018

Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé

8 tháng 6 2017

a ) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)(1)

ĐKXĐ : \(x\ne1;x\ne2\)

(1)\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Leftrightarrow2-x+5x+5=15\)

\(\Leftrightarrow4x+7=15\\\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\left(KTMĐKXĐ\right)\)

Vậy pt vô nghiệm .

b ) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( 2 )

ĐKXĐ : \(x\ne3;x\ne-2\)

(2) \(\Leftrightarrow3x-x^2+6-2x+x^2+2x=3x+6-x^2-2x\)

\(\Leftrightarrow x^2+2x=0\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-2\left(KTMĐKXĐ\right)\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là S={0}.

c ) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) (3)

ĐKXĐ : \(x\ne1;x\ne3\)

\(\left(3\right)\Leftrightarrow\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)

\(\Leftrightarrow6\left(3-x\right)+4\left(x-1\right)=8\)

\(\Leftrightarrow18-6x+4x-4=8\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow x=-3\)

Vậy tập nghiệm của phương trình là S={-3}

d ) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (4)

ĐKXĐ : \(x\ne0;x\ne2\)

\(\left(4\right)\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là S={-1}

8 tháng 6 2017

a) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) ( đk: x ≠ -1; x ≠ 2 )

\(\Leftrightarrow\) \(\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)

\(\Leftrightarrow\) \(2-x+5\left(x+1\right)=15\)

\(\Leftrightarrow\) \(2-x+5x+5=15\)

\(\Leftrightarrow\)\(4x=8\)

\(\Rightarrow\) \(x=2\) ( KTM )

S = ∅

b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( đk: x ≠ - 2 ; x ≠ 3 )

\(\Leftrightarrow\) \(\left(x+2\right)\left(3-x\right)+x\left(x+2\right)=5x+2\left(3-x\right)\)

\(\Leftrightarrow\) \(3x-x^2+6-2x+x^2+2x=5x+6-2x\)

\(\Leftrightarrow\) \(3x+6=3x+6\)

\(\Rightarrow\)\(0x=0\) ( TM )

\(\Rightarrow\) Phương trình vô số nghiệm

S = R

c) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) ( đk: x ≠ 1 ; x ≠ 3 )

\(\Leftrightarrow\) \(\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)

\(\Leftrightarrow\)\(6\left(3-x\right)+4\left(x-1\right)=8\)

\(\Leftrightarrow\) \(18-6x+4x-4=8\)

\(\Leftrightarrow\) \(-2x=-6\)

\(\Rightarrow x=3\) ( KTM )

S = ∅

d) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (đk: x ≠ 2; x ≠ 0 )

\(\Leftrightarrow\) \(x\left(x+2\right)-x+2=2\)

\(\Leftrightarrow\) \(x^2+2x-x+2=2\)

\(\Leftrightarrow\) \(x^2+x=0\)

\(\Leftrightarrow\) \(x\left(x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)

S = \(\left\{2\right\}\)