\(\left(\frac{1}{3}\right)^{^{30}}.x+\left(\frac{1}{3}\right)^{31}=\left(\frac{1}{3}\right)^{32}\)
1) Tính hợp lý
a)\(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{32}\right)\)
b)\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
c)\(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
d)\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right).\left(-\frac{5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
c)38/45-(8/45-17/51-3/11)
=38/45-8/45+17/51+3/11
=30/45+1/3+3/11
=2/3+1/3+3/11
=3/3+3/11
=1+3/11
=1 3/11.
d)(17/28+18/29-19/30-30/31).(-5/12+1/4+1/6)
=(17/28+18/29+19/30+30/31).(-5+3+2/12)
=(17/28+18/29+19/30+30/31).0
=0.
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
\(\left(4\frac{1}{6}x^2-\frac{2}{3}\right)\left(-0,75x-\frac{21}{32}\right)\left(\frac{5}{6}\left|x\right|-3\frac{1}{3}\right)\)\(\left(4\frac{1}{2}x^4+1\frac{1}{3}x\right)=0\)
Tìm x biết : \(\left(\frac{1}{3}\right)^x\left(\frac{1}{9}\right)^x\left(\frac{1}{27}\right)^x\left(\frac{1}{81}\right)^x\left(\frac{1}{243}\right)^x=\left(-\frac{1}{3}\right)^{30}\)
Bài 1: Tìm x biết: \(\left(-2\right)\cdot\left(x+1\right)-3\cdot\left(1-x\right)=4\)
Bài 2: Chứng minh rằng: \(\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}< \frac{4}{5}\)
Bài 1 :
\(\left(-2\right)\left(x+1\right)-3\left(1-x\right)=4\)
\(\Leftrightarrow-2x-2-3+3x=4\)
\(\Leftrightarrow x=4+2+3=9\)
Bài 2 :
Cho \(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)
\(\Leftrightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Rightarrow S< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)
\(+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(\Leftrightarrow S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)(1)
Lại có :
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\Leftrightarrow S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(\Leftrightarrow S>\frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)(2)
Từ (1) và (2) , ta có :
\(\frac{3}{5}< S< \frac{4}{5}hay\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}< \frac{4}{5}\)
Nguyen Ribi Nkok Ngok Khôi Bùi nguyễn ngọc dinh Phùng Tuệ Minh Akai Haruma buithianhtho ?Amanda? Nguyễn Thành Trương Nguyễn Ngô Minh Trí
3.Tính hợp lí:
a,\(\frac{17}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}\)
b,\(\left(\frac{11}{4}.\frac{5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)
c,\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
4.Tìm tích:
a,\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right).....\left(\frac{1}{99}+1\right)\)
b,\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{100}-1\right)\)
c,\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}....\frac{899}{30^2}\)
AI CÒN THỨC THÌ GIÚP MIK VS,MIK ĐANG CẦN GẤP
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)
\(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)+\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
minh k biet xin loi ban nha!
minh k biet xin loi ban nha!
minh k biet xin loi ban nha!
minh k biet xin loi ban nha!
Thực hiện phép tính sao cho hợp lí nhất:
a)\(2\frac{3}{4}.\left(-0.4\right)-1\frac{3}{5}.2,75+1,2:\frac{4}{11}\)
b)\(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).....\left(\frac{1}{31}+1\right)\)
c)\(\frac{930}{1+2+3+....+30}\)
a) \(2\frac{3}{4}\cdot\left(-0,4\right)-1\frac{3}{5}\cdot2,75+1,2:\frac{4}{11}\)
\(=2\frac{3}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}:\frac{4}{11}\)
\(=\frac{11}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}\cdot\frac{11}{4}\)
\(=\frac{11}{4}\left(-\frac{2}{5}-1\frac{3}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\left(-\frac{2}{5}-\frac{8}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\cdot\left(-\frac{4}{5}\right)=\frac{11}{1}\cdot\left(-\frac{1}{5}\right)=-\frac{11}{5}\)
b) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)....\left(\frac{1}{31}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{31}+\frac{31}{31}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{32}{31}\)
\(=\frac{3\cdot4\cdot5\cdot...\cdot32}{2\cdot3\cdot4\cdot...\cdot31}=\frac{32}{2}=16\)
c) Đặt \(C=1+2+3+...+30\)
Số số hạng là : \(\left(30-1\right):1+1=30\)(số)
Tổng của dãy số là : \(\frac{\left(1+30\right)\cdot30}{2}=465\)
Do đó : \(\frac{930}{C}=\frac{930}{465}=2\)
1. tính A= \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)
2. tính B= \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}\)
3. So sánh C= \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)với \(\frac{1}{21}\)
4. So sánh D= \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100}\right)\)với \(\frac{11}{19}\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)