\(\frac{12}{\left(\frac{6x}{x+6}+6\right)^2}\cdot x\)
Những câu hỏi liên quan
Tìm x
a, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
b,\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\frac{5}{12}.....\frac{30}{62}\cdot\frac{31}{64}=2^x\)
a)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\Leftrightarrow x\left(x-1\right)^{x+2}\left(x-2\right)=0\)
Do đó \(x\in\left\{0;1;2\right\}\)
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b)
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot...\cdot\frac{31}{64}=2^x\Leftrightarrow\frac{1\cdot2\cdot3\cdot...\cdot31}{4\cdot6\cdot8\cdot...\cdot64}=2^x\Leftrightarrow\frac{31!}{\left(2\cdot2\right)\cdot\left(2\cdot3\right)\cdot\left(2\cdot4\right)\cdot...\cdot\left(2\cdot31\right)\cdot64}=2^x\)
\(\frac{31!}{2^{30}\cdot31!\cdot2^6}=2^x\Leftrightarrow\frac{1}{2^{36}}=2^x\Leftrightarrow2^{-36}=2^x\Rightarrow x=-36\)
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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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TÌM x
\(\left(\left(\frac{3}{4}\cdot x+5\right)-\left(\frac{2}{3}\cdot x-4\right)-\left(\frac{1}{6}\cdot x+1\right)\right)=\left(\frac{1}{3}\cdot x+4\right)-\left(\frac{1}{3}-3\right)\)
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
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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!!!
Bài 1:a,left(x-frac{1}{2}right)cdotleft(frac{1}{2}+frac{1}{6}+frac{1}{12}+frac{1}{20}+......+frac{1}{90}right)frac{1}{3}b,frac{1}{1cdot6}+frac{1}{6cdot11}+.....+frac{1}{96cdot101}frac{1}{10cdot x}c,460+85cdot4frac{x+175}{5}+30d,left(x-5right)cdotleft(10-9frac{40}{41}right):left(1-frac{81}{82}right):left(1-frac{204}{205}right)2050
Đọc tiếp
Bài 1:
\(a,\left(x-\frac{1}{2}\right)\cdot\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{90}\right)=\frac{1}{3}\)
\(b,\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+.....+\frac{1}{96\cdot101}=\frac{1}{10\cdot x}\)
\(c,460+85\cdot4=\frac{x+175}{5}+30\)
\(d,\left(x-5\right)\cdot\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
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c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
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\(\frac{1}{\left(x+y\right)^2}\cdot\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^{\text{4}}}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)
Giúp vs cần gấp
Thiếu điều kiện xy = 1; x+y khác 0 nhá bn
Bài này tương tự câu 1 ở đây
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Help me:
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\cdot\left(6x+1\right)\)
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}.\left(6x+1\right)\)
\(\Rightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}.6x+\frac{2}{3}\)
\(\Rightarrow\frac{55-24}{60}-x=4x+\frac{2}{3}\)
\(\Rightarrow\frac{31}{60}-x=4x+\frac{2}{3}\)
\(\Rightarrow\frac{31}{60}-\frac{2}{3}=4x+x=5x\)
\(\Rightarrow5x=-\frac{11}{60}\)
\(\Rightarrow x=\frac{-11}{300}\)
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Bài 1:
a)frac{5}{6}-frac{2}{3}+frac{1}{4} b)1frac{11}{12}-frac{5}{12}cdotleft(frac{4}{5}-frac{1}{10}right):frac{-5}{12}
Bài 2:
a) frac{1}{2}cdot x-frac{2}{5}frac{1}{5} b)left(1-2cdot xright)cdotfrac{4}{3}left(-2right)^3
Đọc tiếp
Bài 1:
a)\(\frac{5}{6}\)-\(\frac{2}{3}+\frac{1}{4}\) b)\(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)
Bài 2:
a) \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\) b)\(\left(1-2\cdot x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)
Bài 1:
a) Ta có: \(\frac{5}{6}-\frac{2}{3}+\frac{1}{4}\)
\(=\frac{10}{12}-\frac{8}{12}+\frac{3}{12}\)
\(=\frac{2+3}{12}=\frac{5}{12}\)
b) Ta có: \(1\frac{11}{12}-\frac{5}{12}\cdot\left(\frac{4}{5}-\frac{1}{10}\right):\frac{-5}{12}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\left(\frac{8}{10}-\frac{1}{10}\right)\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{5}{12}\cdot\frac{7}{10}\cdot\frac{-12}{5}\)
\(=\frac{23}{12}-\frac{-7}{10}\)
\(=\frac{115}{60}+\frac{42}{60}=\frac{157}{60}\)
Bài 2:
a) Ta có: \(\frac{1}{2}\cdot x-\frac{2}{5}=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{2}\cdot x=\frac{1}{5}+\frac{2}{5}=\frac{3}{5}\)
\(\Leftrightarrow x=\frac{3}{5}:\frac{1}{2}=\frac{3}{5}\cdot2=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
b) Ta có: \(\left(1-2x\right)\cdot\frac{4}{3}=\left(-2\right)^3\)
\(\Leftrightarrow\left(1-2x\right)\cdot\frac{4}{3}=-8\)
\(\Leftrightarrow1-2x=-8:\frac{4}{3}=-8\cdot\frac{3}{4}=-6\)
\(\Leftrightarrow-2x=-6-1=-7\)
hay \(x=\frac{7}{2}\)
Vậy: \(x=\frac{7}{2}\)
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Bài 1:
a) Ta có: 56−23+1456−23+14
=1012−812+312=1012−812+312
=2+312=512=2+312=512
b) Ta có: 11112−512⋅(45−110):−51211112−512⋅(45−110):−512
=2312−512⋅(810−110)⋅−125=2312−512⋅(810−110)⋅−125
=2312−512⋅710⋅−125=2312−512⋅710⋅−125
=2312−−710=2312−−710
=11560+4260=15760=11560+4260=15760
Bài 2:
a) Ta có: 12⋅x−25=1512⋅x−25=15
⇔12⋅x=15+25=35⇔12⋅x=15+25=35
⇔x=35:12=35⋅2=65⇔x=35:12=35⋅2=65
Vậy: x=65x=65
b) Ta có: (1−2x)⋅43=(−2)3(1−2x)⋅43=(−2)3
⇔(1−2x)⋅43=−8⇔(1−2x)⋅43=−8
⇔1−2x=−8:43=−8⋅34=−6⇔1−2x=−8:43=−8⋅34=−6
⇔−2x=−6−1=−7⇔−2x=−6−1=−7
hay x=72x=72
Vậy: x=72
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Rút gọn:
\(\frac{1}{\left(x+y\right)^3}\cdot\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)
R/g\(\left[\left(x^3-1\right)-\frac{7-x^3}{3+x^3}.\frac{4}{x^5+3x^2}\right]:\left[\frac{3x^6-12}{x^9+6x^6+9x^3}.\frac{x}{3x^3+6}\right]\)