anh chị giúp em nhé
\(x = {{2.1.2+1} \over 1.2}+{{2.2.3+1} \over 2.3}+ {{2.3.4+1} \over 3.4}+{{2.4.5+1} \over 4.5}+...+{{2.9.10+1} \over 9.10}\)
Những câu hỏi liên quan
(2.1.2 + 1/1.2) + (2.2.3 + 1/2.3) + (2.3.4 + 1/3.4) + (2.4.5 + 1/4.5)+...+(2.9.10 + 9/10) = ?
(2.1.2+1/1.2) + (2.2.3+1/2.3) + (2.3.4+1/3.4) +...+ (2.9.10/9.10) = ?
Giúp em với ạ !! Hichic
Cho x, y, z là các số lớn hơn hoặc bằng 1. C/mr
\({1\over 1+x^2}+{1\over 1+y^2}+{1\over 1+z^2}>={3\over 1+xyx} \)
Giải các phương trình sau : a, {8 over x-8} + { 11over x-11} {9 over x-9} +{10 over x-10}b, {x over x-3} - {x over x-5} { x over x-4} - { xover x-6}c, { 4over x^2 - 3x + 2 } - { 3 over 2x^2 - 6x +1 } +1 0d, {1over x-1} + {2over x-2} + {3 over x-3} {6 over x-6}e, {2over 2x+1} - {3 over 2x-1} {4over 4x^2 -1}f, { 2xover x +1 } + { 18 over x^2 +2x-3} {2x-5 over x+3}g, {1 over x-1} + { 2x^2 -5 over x^3 -1 } { 4 over x^2 +x+1}
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Giải các phương trình sau :
a, \({8 \over x-8} + { 11\over x-11} = {9 \over x-9} +{10 \over x-10}\)
b, \({x \over x-3} - {x \over x-5} = { x \over x-4} - { x\over x-6}\)
c, \({ 4\over x^2 - 3x + 2 } - { 3 \over 2x^2 - 6x +1 } +1 =0\)
d, \({1\over x-1} + {2\over x-2} + {3 \over x-3} = {6 \over x-6}\)
e, \({2\over 2x+1} - {3 \over 2x-1} = {4\over 4x^2 -1}\)
f, \({ 2x\over x +1 } + { 18 \over x^2 +2x-3} = {2x-5 \over x+3}\)
g, \({1 \over x-1} + { 2x^2 -5 \over x^3 -1 } = { 4 \over x^2 +x+1}\)
a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10
b, x/x-3 - x/x-5 = x/x-4 - x/x-6
c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0
d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6
e, 2/2x+1 - 3/2x-1 = 4/4x^2-1
f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3
g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1
\(x = {{3} \over 1^2.2^2}+ {{5} \over 2^2.3^2}+...+{{39} \over 19^2.20^2}\)
(\({{1} \over 2}\)+\({{1} \over 3}\)+...+\({{1} \over 2014}\))*x=\({{2013} \over 1}\)+\({{2012} \over 2}\)+...+\({{2} \over 2012}\)+\({{1} \over 2013}\)
GIẢI GIÚP EM VỚI MN ƠI
\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(VP=2013+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(VP=1+\left(\dfrac{2012}{2}+1\right)+....+\left(\dfrac{2}{2012}+1\right)+\left(\dfrac{1}{2013}+1\right)\)
\(VP=\dfrac{2014}{2014}+\dfrac{2014}{2}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}\)
\(VP=2014\left(\dfrac{1}{2}+..+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)
\(VP-VT=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)-x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)=0\)
\(\Rightarrow\left(2014-x\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)=0\)
\(\Rightarrow x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\ne0\right)\)
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Bài 1: cho a,b,c khác đôi một{1 over a} + {1 over b} + {1 over c} 0Rút gọn các biểu thứcM {1 over a^2+2bc} + {1 over b^2+2ac} + {1 over c^2+2ab}N {bc over a^2+2bc}+ {ca over b^2+2ac} + {ab over c^2+2ab}Bài 2: Cho {x over a} + {y over b} + {z over c}0 và {a over x} + {b over y} + {c over z} 2Chứng Minh Rằng {a^2 over x^2} + {b^2 over y^2} + {c^2 over z} 4
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Bài 1: cho a,b,c khác đôi một\({1 \over a} + {1 \over b} + {1 \over c}= 0\)
Rút gọn các biểu thức
\(M = {1 \over a^2+2bc} + {1 \over b^2+2ac} + {1 \over c^2+2ab}\)
\(N = {bc \over a^2+2bc}+ {ca \over b^2+2ac} + {ab \over c^2+2ab}\)
Bài 2: Cho \({x \over a} + {y \over b} + {z \over c}=0 \) và \({a \over x} + {b \over y} + {c \over z}= 2\)
Chứng Minh Rằng \({a^2 \over x^2} + {b^2 \over y^2} + {c^2 \over z}= 4 \)
cho a,b,c > 0 . cm:
\(x = {1\over 4a}+{1\over 4b}+{1\over 4c} >= {1\over 2a+b+c}+{1\over 2b+c+a}+{1\over 2c+b+a}\)
Bài 1: Rút gọn
1) x^2-y^2 over 6x^2y^2
÷ x+y over 12xy
2) 5x over 2x+1
÷ 3x(x-1) over 4x^2-1
3)( 2x-1over 2x+1
-2x-1over 2x+1
) ÷ 4x over 10x-5
4) 2over 9x^2+6x+1
- 3x over 9x^2-1
5) (5over x^2+2x+1 +2x over x^2-1
) ÷ 2x^2+7x-5 over 3x-3
6) (3over x-3 + 2x over x^2-9
+ xover x+3
) ÷ 2xover x+3
7) (3over x^2-9
+1over x^2+3x
-1over x^2-3x
) ÷ x-2over 2x^2+6x
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Bài 1: Rút gọn
1) \(x^2-y^2 \over 6x^2y^2 \)÷ \(x+y \over 12xy\)
2) \(5x \over 2x+1 \) ÷ \(3x(x-1) \over 4x^2-1\)
3)( \(2x-1\over 2x+1 \)-\(2x-1\over 2x+1 \)) ÷ \(4x \over 10x-5 \)
4) \(2\over 9x^2+6x+1 \)- \(3x \over 9x^2-1 \)
5) (\(5\over x^2+2x+1 \)+\(2x \over x^2-1 \)) ÷ \(2x^2+7x-5 \over 3x-3\)
6) (\(3\over x-3 \)+ \(2x \over x^2-9 \) + \(x\over x+3 \)) ÷ \(2x\over x+3\)
7) (\(3\over x^2-9 \)+\(1\over x^2+3x \)-\(1\over x^2-3x \)) ÷ \(x-2\over 2x^2+6x\)
1)
ĐK: \(x,y\neq 0\); \(x+y\neq 0\)
\(\frac{x^2-y^2}{6x^2y^2}: \frac{x+y}{12xy}\)
\(=\frac{x^2-y^2}{6x^2y^2}. \frac{12xy}{x+y}=\frac{(x-y)(x+y).12xy}{6x^2y^2(x+y)}=\frac{2(x-y)}{xy}\)
2) ĐK: \(x\neq \frac{\pm 1}{2}; 0; 1\)
\(\frac{5x}{2x+1}: \frac{3x(x-1)}{4x^2-1}=\frac{5x}{2x+1}.\frac{4x^2-1}{3x(x-1)}\)
\(=\frac{5x(2x-1)(2x+1)}{(2x+1).3x(x-1)}=\frac{5(2x-1)}{3(x-1)}\)
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3) ĐK: \(x\neq \frac{\pm 1}{2}; 0\)
\(\left(\frac{2x-1}{2x+1}-\frac{2x-1}{2x+1}\right): \frac{4x}{10x-5}=0: \frac{4x}{10x-5}=0\)
4) ĐK: \(x\neq \frac{\pm 1}{3}\)
\(\frac{2}{9x^2+6x+1}-\frac{3x}{9x^2-1}=\frac{2}{(3x+1)^2}-\frac{3x}{(3x-1)(3x+1)}\)
\(=\frac{2(3x-1)}{(3x+1)^2(3x-1)}-\frac{3x(3x+1)}{(3x-1)(3x+1)^2}\)
\(=\frac{6x-2-9x^2-3x}{(3x+1)^2(3x-1)}=\frac{-9x^2+3x-2}{(3x-1)(3x+1)^2}\)
5) ĐK: \(x\neq \pm 1; \frac{-7\pm \sqrt{89}}{4}\)
\(\left(\frac{5}{x^2+2x+1}+\frac{2x}{x^2-1}\right): \frac{2x^2+7x-5}{3x-3}\)
\(=\left(\frac{5}{(x+1)^2}+\frac{2x}{(x-1)(x+1)}\right). \frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{5(x-1)+2x(x+1)}{(x-1)(x+1)^2}. \frac{3(x-1)}{2x^2+7x-5}=\frac{2x^2+7x-5}{(x+1)^2(x-1)}.\frac{3(x-1)}{2x^2+7x-5}\)
\(=\frac{3}{(x+1)^2}\)
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6) ĐK: \(x\neq \pm 3\); 0
\(\left(\frac{3}{x-3}+\frac{2x}{x^2-9}+\frac{x}{x+3}\right): \frac{2x}{x+3}\)
\(=\left(\frac{3(x+3)}{(x-3)(x+3)}+\frac{2x}{(x-3)(x+3)}+\frac{x(x-3)}{(x+3)(x-3)}\right). \frac{x+3}{2x}\)
\(=\frac{3(x+3)+2x+x(x-3)}{(x-3)(x+3)}.\frac{x+3}{2x}\)
\(\frac{(x^2+2x+9)(x+3)}{(x-3)(x+3).2x}=\frac{x^2+2x+9}{2x(x-3)}\)
7) ĐK: \(x\neq 2; \pm 3;0\)
\(\left(\frac{3}{x^2-9}+\frac{1}{x^2+3x}-\frac{1}{x^2-3x}\right): \frac{x-2}{2x^2+6x}\)
\(=\left(\frac{3x}{x(x-3)(x+3)}+\frac{x-3}{x(x-3)(x+3)}-\frac{x+3}{(x+3)x(x-3)}\right).\frac{2x(x+3)}{x-2}\)
\(=\frac{3x+x-3-(x+3)}{x(x-3)(x+3)}.\frac{2x(x+3)}{x-2}\)
\(=\frac{3x-6}{x(x-3)(x+3)}.\frac{2x(x+3)}{x-2}=\frac{3(x-2).2x(x+3)}{x(x-3)(x+3)(x-2)}=\frac{6}{x-3}\)
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