bài 2: so sánh
a)\(\left(\frac{1}{2}\right)^{225}\)và \(\left(\frac{1}{3}\right)^{100}\)
giải nhanh giúp với mai mk nộp rồi mk tick cho
bài 2: so sánh
a)\(\left(\frac{1}{2}\right)^{225}va\left(\frac{1}{3}\right)^{100}\)
giải nhanh giúp mk với nhé mk tick cho
Ta có
\(\left(\frac{1}{2}\right)^{225}\)=\(\left(\frac{1}{2}\right)^{9.25}\)=\(\left(\frac{1}{512}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}\)=\(\left(\frac{1}{3}\right)^{4.25}\)=\(\left(\frac{1}{81}\right)^{25}\)
Vì \(\frac{1}{512}\)<\(\frac{1}{81}\) => \(\left(\frac{1}{512}\right)^{25}\)<\(\left(\frac{1}{81}\right)^{25}\)
Hay \(\left(\frac{1}{2}\right)^{225}\)<\(\left(\frac{1}{3}\right)^{100}\)
Mong bạn tích cho mình nhé
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{81}\right)^{25}\)\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
vì \(\left(\frac{1}{81}\right)^{25}=\left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}=\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrowđpcm\)
Tính:
\(\left[\frac{1}{100}-1^2\right].\left[\frac{1}{100}-\left(\frac{1}{2}\right)^2\right].\left[\frac{1}{100}-\left(\frac{1}{3}\right)^2\right].....\left[\frac{1}{100}-\left(\frac{1}{20}\right)^2\right]\)
Giải nhanh lên giúp mk với! Rồi mk tick cho 3 cái
đây có chắc là toán lớp 7 không đấy
nếu có bài hình nào khó thì cho lên đấy nhé mình chuyên về toán lớp 7 hơn
AI GIẢI GIÚP MK BÀI NÀY VỚI. AI XONG NHANH NHẤT, GIẢI RÕ RÀNG NHẤT THÌ MK TICK CHO..!!!
CHO : \(B=\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{9}\right)\cdot\left(1-\frac{1}{19}\right)\cdot...\cdot\left(1-\frac{1}{81}\right)\cdot\left(1-\frac{1}{100}\right)\)
SO SÁNH B VỚI \(\frac{11}{19}\)
Bài 2: tính
a)\(\left(\frac{1}{5}\right)^5.5^5\)
giải nhanh giúp vai mk nộp rồi mk tick cho cảm ơn nhiều
a)\(\left(\frac{1}{5}\right)^5\).\(5^5\)=\(\frac{1}{3125}\).3125=1
a)\(\left(\frac{1}{5}\right)^5.5^5\)
=\(\left(\frac{1}{5}.5\right)^5\)= 15 =1
bài 4: so sánh
a) \(\left(\frac{-1}{5}\right)^{300}\) và \(\left(\frac{-1}{3}\right)^{500}\)
b) \(-\left(-2\right)^{300}\) và \(\left(-3\right)^{200}\)
giúp mk với mai mk nộp rồi thanks nhiều
Ta có : (-1/5)^300=(-1/5^3)100=(-1/125)^100
(-1/3)^500=(-1/3^5)^100=(-1/243)^100
vì (-1/243)^100<(-1/125)^100→(-1/5)^300>(-1/3)^500
b, ta có:-(-2)^300=(2^3)^100=8^100
(-3)^200=(-3^2)^100=9^100
vì 8^100<9^100→-(-2)^300<(-3)^200
Bài 4 tính\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
giúp mik với tối mik nộp bài rồi mik sẽ tick cho tất cả các bạn giải cả lời giải nữa nha
\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)
\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)
1.
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
CÓ 1 BÀI THUI NHA, MONG MN GIẢI NHANH GIÚP MK
AI NHANH MK CHO 1 TICK, MK ĐANG RẤT CẦN ĐẤY
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)
\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)
\(\Leftrightarrow\frac{38}{x+3}=16\)
\(\Leftrightarrow x+3=2,375\)
\(\Leftrightarrow x=-0,625\)
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)
\(\Leftrightarrow\frac{38}{x+3}=14\)
\(\Leftrightarrow\left(x+3\right)14=38\)
\(\Leftrightarrow14x+42=38\)
\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)
Vậy \(x=-\frac{2}{7}\)
tính
a)\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+3+...+20}\right)\)
b)\(\frac{\left(1+2+3+...+100\right).\left(12.3,4-6,86\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}\)
c)(18.123+9.436.2+3.5310.6):1+4+7+...+100-410)
giúp mk vs mai mk kiểm tra rồi . ai đúng mk tick nha
Bài 1: Tìm x :
a ) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
b) \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)
giải giúp mk với mai mk kiểm tra rồi
a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)
Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)
\(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)
\(x+\frac{1}{2}=x+x+3\\\)
\(x+\frac{1}{2}=x+\left(x+3\right)\)
\(\Rightarrow\frac{1}{2}=x+3\)
\(\Rightarrow x=\frac{1}{2}-3\)
\(\Rightarrow x=-\frac{5}{2}\)
Vậy \(x=-\frac{5}{2}\)
b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)
\(Ta\) \(có\)
\(x+\frac{1}{5}+x+\frac{2}{5}+x+1\frac{2}{5}\)\(=4x\)
\(3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)
\(3x+2=4x\)
\(3x+2=3x+x\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) Vì \(\left|x+\frac{1}{5}\right|\ge0;\left|x+\frac{2}{5}\right|\ge0;\left|x+1\frac{2}{5}\right|\ge0\forall x\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
Với \(x\ge0\) ta có:
\(\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+1\frac{2}{5}\right)=4x\)
\(\Rightarrow3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)
\(\Rightarrow2=4x-3x=x\)
Vậy x = 2