Bài 1: tính
\(1^2+2^2+3^2+.....+100^2\)
Bài 2: tìm x
a, x+ (x+1)+ (x+2)+........+(x+2003) = 2004
b, \(2\frac{1}{3}+\left(x-\frac{3}{2}\right)-\left(3-\frac{3}{2}\right)x\)
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Bài 1 : Thực hiện phép tính(1) D 1+frac{1}{2}left(1+2right)+frac{1}{3}left(1+2+3right)+...+frac{1}{16}left(1+2+...+16right)(2) M frac{frac{1}{99}+frac{2}{98}+frac{3}{97}+...+frac{99}{1}}{frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{100}}Bài 2 : Tìm x biết(1) x-left{x-left[x-left(-x+1right)right]right}1(2) left[frac{1}{2}+frac{1}{3}+...+frac{1}{2016}right]cdot xfrac{2015}{1}+frac{2014}{2}+...+frac{1}{2015}(3) frac{x}{left(a+5right)left(4-aright)}frac{1}{a+5}+frac{1}{4-a}(4) frac{x+2}{11}+frac{x+...
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Bài 1 : Thực hiện phép tính
(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)
(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Bài 2 : Tìm x biết
(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)
(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot x=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)
(3) \(\frac{x}{\left(a+5\right)\left(4-a\right)}=\frac{1}{a+5}+\frac{1}{4-a}\)
(4) \(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
(5) \(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}+4=0\)
Bài 3 :
(1) Cho : A =\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\); B =\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
CMR : \(\frac{A}{B}\)Là 1 số nguyên
(2) Cho : D =\(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}\)CMR : \(D< \frac{3}{4}\)
Bài 4 : Ký hiệu [x] là số nguyên lớn nhất không vượt quá x , gọi là phần nguyên của x.
VD : [1.5] =1 ; [3] =3 ; [-3.5] = -4
(1) Tính :\(\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]\)
(2) So sánh : A =\(\left[X\right]+\left[X+\frac{1}{5}\right]+\left[X+\frac{2}{5}\right]+\left[X+\frac{3}{5}\right]+\left[X+\frac{4}{5}\right]\)và B = [5x]. Biết x=3.7
Bài 1:Thực hiện phép tínha,75%-1frac{1}{2}+0,5:frac{5}{12}-left(frac{-1}{2}right)^2 b,left(frac{5}{7}.0,6-5:3frac{1}{2}right).left(40%-1,4right).left(-2right)^3Bài 2: Tính nhanh left(frac{-4}{5}+frac{4}{3}right)+left(frac{-5}{4}+frac{14}{5}right)-frac{7}{3}Bài 3:Tìm x biếta,frac{-2}{3}-frac{1}{3}.left(2x-5right)frac{3}{2} b,frac{2}{5}.x+frac{1}{2}frac{-3}{4} frac{1}{3}.x-8frac{1}{2} x-frac{1}{4}frac{5}{8}.frac{2}{3}Bài 4: Tìm...
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Bài 1:Thực hiện phép tính
\(a,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2\) \(b,\left(\frac{5}{7}.0,6-5:3\frac{1}{2}\right).\left(40\%-1,4\right).\left(-2\right)^3\)
Bài 2: Tính nhanh
\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
Bài 3:Tìm x biết
\(a,\frac{-2}{3}-\frac{1}{3}.\left(2x-5\right)=\frac{3}{2}\) \(b,\frac{2}{5}.x+\frac{1}{2}=\frac{-3}{4}\) \(\frac{1}{3}.x-8=\frac{1}{2}\) \(x-\frac{1}{4}=\frac{5}{8}.\frac{2}{3}\)
Bài 4: Tìm x biết
\(a,\left(\frac{3}{4}.x+2\frac{1}{2}\right).\frac{-2}{3}=\frac{1}{8}\) \(b,\frac{1}{3}.x-0,5=0,75\)
Bài 5: Tìm x biết
\(a,\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=-\frac{53}{10}\) \(b,5,2.x+7\frac{2}{5}=6\frac{3}{4}\) \(2,4:\left(\frac{-1}{2}-x\right)=1\frac{3}{5}\)
Bài 6:Tìm số tự nhiên x,biết: \(\left(x-5\right).\frac{30}{100}=\frac{20.x}{100}+5\)
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1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Bài 1 : Tìm x biết :a, left|frac{3}{2}x+frac{1}{2}right|left|4x-1right|b, left|frac{5}{4}x-frac{7}{2}right|-left|frac{5}{8}x+frac{3}{5}right|0c,left|frac{7}{5}x+frac{2}{3}right|left|frac{4}{3}x-frac{1}{4}right|Bài 2 : Tìm x biết :a, | 2x - 5 | x +1 b, | 3x - 2 | -1 x c, | 3x - 7 | 2x + 1d, | 2x-1 | +1 x
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Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
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a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
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Bài 1: Tính
\(\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{3}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)
Bài 2: Tìm x
\(\frac{-15}{12}.x+\frac{3}{7}=\frac{6}{5}.x-\frac{1}{2}\)
Tìm x :
x + (x + 1 ) + ( x+2 ) + ..+ ( x+2003 ) =2004
\(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)
Bài 1. Tính:
a) (2008.2009.2010.2011).\(\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
Bài 2. Tìm x biết:
a)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
b)\(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
1) Tính :
a) \(\left(2008.2009.2010.2011\right).\left(1+\frac{1}{2}:\frac{2}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(=\left(2008.2009.2010.2011\right).0\)
\(=0\)
2) Tìm x
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2012\)
b) \(\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}.\frac{1}{6}.\left(x-1,010\right)=\frac{1}{360}-\frac{1}{720}\)
\(\Rightarrow\frac{1}{2.3.4.5.6}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow\frac{1}{720}.\left(x-1,01\right)=\frac{1}{720}\)
\(\Rightarrow x-1,01=\frac{1}{720}:\frac{1}{720}\)
\(\Rightarrow x-1,01=1\)
\(\Rightarrow x=1+1,01\)
\(\Rightarrow x=2,01\)
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\(\frac{\left(x-1\right)^3}{3}-\frac{3.\left(2x-3\right)^2}{2}=\frac{\left(2+x\right)^3}{3}-\frac{5.\left(1+3x\right)^2}{2}-\frac{3x}{2}\)Tìm x
(BÀI NÀY ĐÂY)
bài 1: cho x, y thuộc Q. cmr:
|x + y| =< |x| + |y|
bài 2: tính:
\(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
bài 3: cho a + b + c = a^2 + b^2 + c^2 = 1 và x : y : z = a : b : c.
cmr: (x + y + z)^2 = x^2 + y^2 + z^2
1
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
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Bài 3:
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)
Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a2 + b2 + c2 = 1)
Vậy: (x + y + z)2 = x2 + y2 + z2
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