a)\(\left(\frac{1}{13}\right)^{13}:\left(\frac{1}{3}\right)^{x-2}=\frac{1}{81}\)
b)\(\left(0,4\right)^{x-1}:\left(\frac{2}{5}\right)^2=\frac{8}{125}\)
Những câu hỏi liên quan
1. tinh` giá trị biểu thức ( tính nhanh nếu có thế )
a)frac{-6}{11}.frac{5}{13}+frac{-6}{11}.frac{8}{13}-left(frac{-2}{5}right)^0 b)left(2frac{2}{3}+3frac{1}{2}right);left(4frac{3}{4}-2frac{1}{6}right)+frac{19}{31} c)2,4:left(-2right)^3+left(3-frac{9}{11}right).1frac{3}{8}
d)left(-frac{3}{4}right)^2:frac{-3}{8}+frac{1}{2}-frac{3}{4}-left(frac{-78}{57}right)^0
2. tìm x
a)x+frac{-1}{5}left(-frac{3}{4^{ }}right)^2 b)left|frac{5}{2}x+frac{2}{3}right|-frac{1}{4}0...
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1. tinh` giá trị biểu thức ( tính nhanh nếu có thế )
\(a)\frac{-6}{11}.\frac{5}{13}+\frac{-6}{11}.\frac{8}{13}-\left(\frac{-2}{5}\right)^0\) \(b)\left(2\frac{2}{3}+3\frac{1}{2}\right);\left(4\frac{3}{4}-2\frac{1}{6}\right)+\frac{19}{31}\) \(c)2,4:\left(-2\right)^3+\left(3-\frac{9}{11}\right).1\frac{3}{8}\)
\(d)\left(-\frac{3}{4}\right)^2:\frac{-3}{8}+\frac{1}{2}-\frac{3}{4}-\left(\frac{-78}{57}\right)^0\)
2. tìm x
\(a)x+\frac{-1}{5}=\left(-\frac{3}{4^{ }}\right)^2\) \(b)\left|\frac{5}{2}x+\frac{2}{3}\right|-\frac{1}{4}=0\) \(c)\frac{2}{3}x-\frac{1}{2}=\frac{5}{12}+\frac{1}{2}x\) \(d)\left(x-\frac{1}{4}\right)^4=\frac{1}{81}\)
\(e)4x+3\frac{1}{4}=x-\frac{1}{4}\) \(g)\left(x-\frac{1}{3}\right)^3=\frac{1}{27}\)
a) \(\frac{\left(-1\right)^3}{15}+\left(-\frac{2}{3}\right):2\frac{2}{3}-\left|-\frac{5}{6}\right|\)
b) \(1\frac{5}{13}-0,\left(3\right)-\left(1\frac{4}{9}+\frac{18}{13}-\frac{1}{3}\right)\)
c) \(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{5}-125\frac{1}{3}\)
d) \(\frac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
a, left(12-frac{7}{18}-10frac{13}{18}right):x-1frac{7}{13}1frac{2}{3}
b,left[left(6frac{3}{7}-frac{0,75x-2}{0,35}right).2,8+1,75right]:0,05235
c.frac{3-x}{5-x}left(frac{3}{5}right)^2
d,left(1-frac{3}{10}-xright):left(frac{19}{10}-1-frac{2}{5}right)+frac{4}{5}1
e,left(12frac{7}{18}-10frac{13}{18}right):x-1frac{7}{33}:frac{8}{11}1frac{2}{3}
các bạn ơi giúp tớ với
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a, \(\left(12-\frac{7}{18}-10\frac{13}{18}\right):x-1\frac{7}{13}=1\frac{2}{3}\)
b,\(\left[\left(6\frac{3}{7}-\frac{0,75x-2}{0,35}\right).2,8+1,75\right]:0,05=235\)
c.\(\frac{3-x}{5-x}=\left(\frac{3}{5}\right)^2\)
d,\(\left(1-\frac{3}{10}-x\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1\)
e,\(\left(12\frac{7}{18}-10\frac{13}{18}\right):x-1\frac{7}{33}:\frac{8}{11}=1\frac{2}{3}\)
các bạn ơi giúp tớ với
a)\(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
b)\(\left(\frac{4}{5}+x\right).\left(x-\frac{8}{13}\right)=0\)
c)\(\left(2x-\frac{1}{2}\right).\left(x-3\right)=0\)
d)\(x+3\frac{1}{2}x+x=\frac{1}{2}\)
a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
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tìm x,biết:a)frac{2}{left(x+2right)left(x+4right)}+frac{4}{left(x+4right)left(x+8right)}+frac{6}{left(x+8right)left(x+14right)}frac{x}{left(x+2right)left(x+14right)}b)frac{x+1}{10}+frac{x+1}{11}+frac{x+1}{12}frac{x+1}{13}+frac{x+1}{14}c)left(x+2right)^2frac{38}{25}+frac{9}{10}-frac{11}{15}+frac{13}{21}-frac{15}{28}+frac{17}{36}-...+frac{197}{4851}-frac{199}{4950}giúp tớ với,huhu
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tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
Tìm x, biết:frac{3}{left(x+2right)left(x+5right)}+frac{5}{left(x+5right)left(x+10right)}+frac{7}{left(x+10right)left(x+17right)}frac{x}{left(x+2right)left(x+17right)}left(xnotin-2;-5;-10;-17right)frac{2}{left(x-1right)left(x-3right)}+frac{5}{left(x-3right)left(x-8right)}+frac{12}{left(x-8right)left(x-20right)}-frac{1}{x-20}-frac{3}{4}Với xnotin1;3;8;20frac{x+1}{10}+frac{2+1}{11}frac{x+1}{12}frac{x+1}{13}+frac{x+1}{14}frac{x-10}{30}+frac{x-14}{43}+frac{x-5}{95}+frac{x-148}{8}0
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Tìm x, biết:
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\left(x\notin-2;-5;-10;-17\right)\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
Với \(x\notin1;3;8;20\)
\(\frac{x+1}{10}+\frac{2+1}{11}\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
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Tìm x
\(\frac{5}{2}-\left(\frac{3}{2}-2\frac{1}{3}+x\right)=\frac{8}{15}-\left(\frac{1}{4}-\frac{7}{10}\right)\)
\(1\frac{2}{3}-1\frac{3}{5}+x=\frac{2}{5}-\left|\frac{3}{4}-\frac{7}{8}\right|\)
\(2-\left(\frac{2}{3}-3\frac{1}{4}+x\right)=1\left|\frac{1}{6}-\frac{13}{12}\right|\)
TÍNH:
a. \(\frac{\left(\frac{1}{9}\right)^6.\left(\frac{3}{8}\right)^7}{\left(\frac{1}{3}\right)^{13}.\left(\frac{1}{2}\right)^{22}.3^6}\)
b. \(\frac{5^{-6}.\left(\frac{1}{2}\right)^{13}}{\left(\frac{1}{5}\right)^7.\left(\frac{1}{8}\right)^4}\)
Giúp mik vs các bạn nha!!!
a, \(\frac{\left(\frac{1}{9}\right)^6\cdot\left(\frac{3}{8}\right)^7}{\left(\frac{1}{3}\right)^{13}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{\left(\frac{1}{\left(3^2\right)^6}\right)\cdot\left(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot3\right)^7}{\left(\frac{1}{3}\right)^{13}.\left(\frac{1}{2}\right)^{22}.3^6}=\frac{\frac{1}{3^{12}}\cdot\left(\frac{1}{2}\right)^{21}\cdot3^7}{\frac{1}{3^{13}}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{3}{\frac{1}{3}\cdot\frac{1}{2}}=3\div\frac{1}{6}=3.6=18\)
b, Làm tương tự nha bn
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Giải các phương trình:a) frac{1}{x-1}-frac{3x^2}{x^3-1}frac{2x}{x^2+x+1}b) frac{3}{left(x-1right)left(x-2right)}+frac{2}{left(x-3right)left(x-1right)}frac{1}{left(x-2right)left(x-3right)}c) 1+frac{1}{x+2}frac{12}{8+x^3}d) frac{13}{left(x-3right)left(2x+7right)}+frac{1}{2x+7}frac{6}{left(x-3right)left(x+3right)}
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Giải các phương trình:
a) \(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
b) \(\frac{3}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(x-3\right)\left(x-1\right)}=\frac{1}{\left(x-2\right)\left(x-3\right)}\)
c) \(1+\frac{1}{x+2}=\frac{12}{8+x^3}\)
d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
a)\(\frac{1}{x-1}\)-\(\frac{3x2}{x3-1}\)=\(\frac{2x}{x2+x+1}\)
<=> \(\frac{1}{x-1}\)-\(\frac{3x2}{\left(x-1\right)\left(x2+x+1\right)}\)=\(\frac{2x}{x2+x+1}\) ĐKXĐ: x khác 1
<=> x2+x+1 - 3x2 = 2x(x-1)
<=>x2+x+1 - 3x2 = 2x2-2x
<=>x2-3x-1=0( đoạn này làm nhanh nhé)
<=>x2-2*\(\frac{3}{2}\)x +\(\frac{9}{4}\)-\(\frac{9}{4}\)-1=0
<=>(x-\(\frac{3}{2}\))2-\(\frac{13}{4}\)=0
<=>(x-\(\frac{3-\sqrt{13}}{2}\))(x-\(\frac{3+\sqrt{13}}{2}\))=0
\(\begin{cases}x=\frac{3+\sqrt{13}}{2}\\x=\frac{3-\sqrt{13}}{2}\end{cases}\)
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b) pt... đkxđ x khác 1;2;3
<=> 3(x-3) +2(x-2)=x-1
<=> 3x-9 +2x-4 = x-1
<=> 4x= 12
<=> x=3 ( ko thỏa đk)
vậy pt vô nghiệm
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c) 1+\(\frac{1}{x+2}\)=\(\frac{12}{\left(x+2\right)\left(x2+2x+4\right)}\)đkxđ : x khác -2
<=> x3+8 + x2+2x+4 = 12
<=> x3+x2+2x=0
<=> x2+x+2=0( chia cả 2 vế cho x)
pt này chắc chắn vô nghiệm nhé bạn
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