Giải phương trình:
\(\frac{7}{10}-\frac{5}{8}\cdot x-\frac{7}{6}=\frac{5}{6}\cdot x\)
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Giải phương trình:
\(\frac{7}{10}-\frac{5}{8}\cdot x-\frac{7}{6}=\frac{5}{6}\cdot x\)
22 tháng 6 2016 lúc 20:38
Giải phương trình sau:
\(\frac{7}{8}-\frac{1}{3}\cdot x=\frac{7}{10}+\frac{2}{3}\cdot x\)
Ta có: \(\frac{7}{8}-\frac{1}{3}x=\frac{7}{10}+\frac{2}{3}x\)
<=> \(\frac{7}{8}-\frac{7}{10}=\frac{2}{3}x+\frac{1}{3}x\)
<=>\(\frac{7}{40}=x\)
Vậy x=7/40
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\(\frac{7}{8}-\frac{1}{3}x=\frac{7}{10}+\frac{2}{3}\)
\(\Leftrightarrow-\frac{1}{3}x-\frac{2}{3}x=\frac{7}{10}-\frac{7}{8}\)
\(\Leftrightarrow-x=-\frac{7}{40}\)
\(\Leftrightarrow x=\frac{7}{40}=0,175\)
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Ta có: \(\frac{7}{8}-\frac{1}{3}.x=\frac{7}{10}+\frac{2}{3}.x\)
\(\Rightarrow\frac{7}{8}-\frac{7}{10}=\frac{2}{3}.x+\frac{1}{3}.x\)
\(\Rightarrow\frac{35}{40}-\frac{28}{40}=\left(\frac{2}{3}+\frac{1}{3}\right).x\)
\(\Rightarrow\frac{7}{40}=x\)
Vậy \(x=\frac{7}{40}\)
Chúc bạn hok tốt! 
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TÌM x biết:
a) \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\frac{5}{12}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{62}=4^x\)
b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
c)\(\left|4x+3\right|-\left|x-1\right|=7\)
a.4^7
b.8^5
c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu
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7 tháng 1 2020 lúc 20:48
tìm x biết
a, \(\frac{1}{2}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=4^x\)
b, \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
c,\(\left|4x+3\right|-\left|x-1\right|=7\)
mong các bạn giúp !!!
b) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=\frac{4^5.\left(1+1+1+1\right)}{3^5.\left(1+1+1\right)}.\frac{6^5.\left(1+1+1+1+1+1\right)}{2^5.\left(1+1\right)}\)
\(=\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=\frac{4^6}{3^6}.\frac{6^6}{2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)
Ta có: \(2^{12}=\left(2^3\right)^4=8^4\)
Vậy x= 4
bài 1: tìm x, biết
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)
bài 2:
cho: p = \(\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+5\right)}}}\)
tính p(x)=7
giúp mk vs!!!!!
mk cần gấp!!
Bài làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
mk cần cả giải thích
giúp mk vs!!!
1. Tính tổng
\(\frac{1}{2}\cdot\frac{1}{3}\cdot+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
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\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
* LÀM NỐT *
#Louis
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1/2.1/3+1/3.1/4+1/4.1/5+...+1/8.1/9
=1/2.3=1/3.4+1/4.5+...+1/8.9\
=1/2-1/3+1/3-1/4=1/4-1/5+...+1/8.1/9
=1/2-1/9
=9/18-2/18
=7/18
HỌC TỐT NHA BẠN
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Bai1:tìm x biết a, (frac{1}{7}cdot x -frac{2}{7})(-frac{1}{5}x+frac{3}{5})(frac{1}{3}cdot x+frac{4}{3})b, frac{1}{6}cdot x+frac{1}{10}cdot x-frac{4}{5}cdot x+10c,(x-frac{2}{9})^3(frac{2}{3})^6d, (x-7)^x+1-(x-7)^x+110
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Bai1:tìm x biết
a, (\(\frac{1}{7}\)\(\cdot\) x -\(\frac{2}{7}\))(-\(\frac{1}{5}x+\frac{3}{5}\))(\(\frac{1}{3}\cdot x+\frac{4}{3}\))
b, \(\frac{1}{6}\cdot x+\frac{1}{10}\cdot x-\frac{4}{5}\cdot x+1=0\)
c,(\(x-\frac{2}{9}\))^3=(\(\frac{2}{3}\))^6
d, (x-7)^x+1-(x-7)^x+11=0
Tìm X
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\frac{5}{12}\cdot.......\cdot\frac{30}{62}\cdot\frac{31}{64}=4^x\)
1/4.2/6.3/8.4/10.........30/62.31/64=4x
=1/2.1/2.1/2.1/2.............1/2.1/64=4^x
=1/2^30.1/2^6=4^x
=1/2^36=4^x
=1/4^18=4^x
=>x=-18
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\(\frac{5}{6}+\frac{1}{3}\cdot x=\frac{8}{9}b,120+x\cdot2=7\cdot x+45\)