Tính B = \(\left(28^2+26^2+24^2+...+4^2+2^2\right)-\left(27^2+25^2+23^2+...+3^2+1^2\right).\)
Những câu hỏi liên quan
Phân tích đa thức thành nhân tử:
23). left(2a-bright)^2-4left(a-bright)^2
24). 9left(a+bright)^2-4left(a-2bright)^2
25). 4left(2a-bright)^2-16left(a-bright)^2
26). 36left(x-yright)^2-25left(2x-1right)^2
27). left(x^2+1right)^2-4x^2
28). 16x^2-left(x^2+4right)
29). left(4x^2+dfrac{1}{4}right)-4x^2
30). left(4x^2+1right)^2-16x^2
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Phân tích đa thức thành nhân tử:
23). \(\left(2a-b\right)^2-4\left(a-b\right)^2\)
24). \(9\left(a+b\right)^2-4\left(a-2b\right)^2\)
25). \(4\left(2a-b\right)^2-16\left(a-b\right)^2\)
26). \(36\left(x-y\right)^2-25\left(2x-1\right)^2\)
27). \(\left(x^2+1\right)^2-4x^2\)
28). \(16x^2-\left(x^2+4\right)\)
29). \(\left(4x^2+\dfrac{1}{4}\right)-4x^2\)
30). \(\left(4x^2+1\right)^2-16x^2\)
23: \(=\left(2a-b\right)^2-\left(2a-2b\right)^2\)
\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)
\(=b\left(4a-3b\right)\)
24: \(=\left(3a+3b\right)^2-\left(2a-4b\right)^2\)
\(=\left(3a+3b-2a+4b\right)\left(3a+3b+2a-4b\right)\)
\(=\left(a+7b\right)\left(5a-b\right)\)
25: \(=\left(4a-2b\right)^2-\left(4a-4b\right)^2\)
\(=\left(4a-2b-4a+4b\right)\left(4a-2b+4a-4b\right)\)
\(=2b\left(8a-6b\right)\)
=4b(4a-3b)
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13 tháng 11 2023 lúc 21:17
so sánh: \(A=26^2-24^2\) và \(B=27^2-25^2\)
tìm x, biết:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
Bài 1:
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)
\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)
=>A<B
Bài 2:
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
=>4x+13=11
=>4x=-2
=>\(x=-\dfrac{1}{2}\)
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Tính:
\(\left(30^2+28^2+29^2+....+4^2+2^2\right)-\left(29^2+27^2+25^2+....+3^2+1^2\right)\)
5dfrac{5}{27}+dfrac{7}{23}-0,5.dfrac{5}{27}+dfrac{16}{23}
45dfrac{1}{6}:left(dfrac{-4}{5}right)-35dfrac{1}{6}:left(dfrac{-4}{5}right)
25.left(dfrac{-1}{5}right)^3+dfrac{1}{5}-2.left(dfrac{-1}{2}right)^2-dfrac{1}{2}
left(3,1-2,5right)-left(-2,5-3,1right)
dfrac{3}{8}.dfrac{7}{5}-dfrac{7}{5}.dfrac{1}{8}+dfrac{13}{20}
CẦN GẤP NGAY VÀ LUÔN :)))))
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\(5\dfrac{5}{27}+\dfrac{7}{23}-0,5.\dfrac{5}{27}+\dfrac{16}{23}=\)
\(45\dfrac{1}{6}:\left(\dfrac{-4}{5}\right)-35\dfrac{1}{6}:\left(\dfrac{-4}{5}\right)=\)
\(25.\left(\dfrac{-1}{5}\right)^3+\dfrac{1}{5}-2.\left(\dfrac{-1}{2}\right)^2-\dfrac{1}{2}=\)
\(\left(3,1-2,5\right)-\left(-2,5-3,1\right)=\)
\(\dfrac{3}{8}.\dfrac{7}{5}-\dfrac{7}{5}.\dfrac{1}{8}+\dfrac{13}{20}=\)
CẦN GẤP NGAY VÀ LUÔN :)))))
Giải PT
a,\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
b, \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
c, \(\frac{x+1}{2012}+\frac{x+2}{2011}=\frac{x+3}{2010}+\frac{x+4}{2009}\)
\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)
\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))
\(⇔x=23\)
\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)
\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)
\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))
\(⇔x=-100\)
\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)
\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)
\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)
\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))
\(⇔x=-2013\)
1.tính
a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
b)\(\left(\dfrac{-5}{24}+\dfrac{3}{4}-\dfrac{7}{12}\right):\left(\dfrac{-5}{16}\right)\)
c)\(\dfrac{6}{7}+\dfrac{5}{4}:\left(-5\right)-\dfrac{-1}{28}.\left(-2\right)^2\)
ai giải đc mik sẽ tick
a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)
\(=1-\dfrac{-5}{2}\)
\(=\dfrac{2}{2}-\dfrac{-5}{2}\)
\(=\dfrac{7}{2}\)
dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!
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Giải:
a)(4-12/5).25/8-2/5:-4/25
=8/5.25/8-(-5/2)
=5+5/2
=15/2
b)(-5/24+3/4-7/12):(-5/16)
=-1/24:(-5/16)
=2/15
c)6/7+5/4:(-5)-(-1/28).(-2)2
=6/7+(-1/4)-(-1/28).4
=6/7-1/4-(-1/7)
=6/7-1/4+1/7
=(6/7+1/7)-1/4
=1-1/4
=3/4
Chúc bạn học tốt!
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Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
\(a,\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(b,\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
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tìm x biết :
a) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
c) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
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\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
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Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)
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