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1: =>7/3x=3+1/3-8-2/3=-5-1/3=-16/3
=>x=-16/3:7/3=-7/16
2: =>1/3|x-2|=4/5+3/7=28/35+15/35=43/35
=>|x-2|=129/35
=>x-2=129/35 hoặc x-2=-129/35
=>x=199/35 hoặc x=-59/35
Đáp án nè:
Đặt A=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{99}}\)
3A=\(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
3A+A=\(\left(\dfrac{1}{1}-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)\)
4A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)
4A bé hơn(sorry tớ không thấy dấu bé hơn)\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
Đặt B=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
3B=\(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
4B=\(3-\dfrac{1}{3^{99}}\) bé hơn 3 \(\Rightarrow\)B bé hơn \(\dfrac{3}{4}\)
\(\Rightarrow\) 4A bé hơn\(\dfrac{3}{4}\Rightarrow\)A bé hơn \(\dfrac{3}{16}\)
Tick cho mình nha , ngồi đánh máy tính mỏi cả mắt lun
Chúc học tốt
D=\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
=>3D=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>3D-D=(1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\))-(\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\))
=>2D=1-\(\dfrac{1}{3^{100}}< 1\)
=>D<\(\dfrac{1}{2}\)
Vậy...
1/2^2=4
1/3^2<1/2.3
.................
1/100^2<1/99.100
A<1/4+1/2.3+...+1/99.100
A<1/4+1/2-1/100
A<1/4<3/4
Vậy A<3/4(dpcm).CHÚC BẠN HỌC TỐT!
a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
Kiyoko Vũ
a, xét từng đoạn 1 , 1/2 ,1/2^3 ,1/2^4 ,1/2^5 ,1/2^6
ta có
1 = 1
1/2 + 1/3 < 1/2 + 1/2 = 1
1/4 + 1/5 + .. + 1/7 < 1/4 +..+ 1/4 = 4/4 = 1
1/8 + 1/9 + .. + 1/15 < 1/8 + .. + 1/8 = 8/8 = 1
tương tự
1/16 +1/17 + .. + 1/31 < 1
1/32 + 1/33 + .. + 1/63 < 1
=> cộng lại => A < 6
b, Câu hỏi của trịnh quỳnh trang - Toán lớp 6 - Học toán với OnlineMath
Ta có :
\(D=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+......................+\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}\)
\(3D=1+\dfrac{2}{3}+\dfrac{3}{3^2}+.....................+\dfrac{100}{3^{99}}\)
\(3D-D=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...................+\dfrac{101}{3^{101}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+..............+\dfrac{100}{3^{99}}\right)\)\(2D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6D=3+1+\dfrac{1}{3}+................+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6D-2D=\left(3+1+\dfrac{1}{3}+.............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)-\left(1+\dfrac{1}{3}+..........+\dfrac{1}{3^{99}}-\dfrac{100}{3^{99}}\right)\)\(4D=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow D< \dfrac{3}{4}\rightarrowđpcm\)
~ Chúc bn học tốt ~