\(\left(\frac{3}{5^{ }}\right)^{2x^{ }}=\frac{9}{25}\)
Những câu hỏi liên quan
\(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\left(2x+\frac{3}{5}\right)^{^2}-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)+\frac{1}{9}=0\)
\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
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\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)
\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)
\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)
\(\Rightarrow2x-1=\frac{-4}{63}\)
\(\Rightarrow2x=-\frac{4}{63}+1\)
\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)
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\(\left[2x+\frac{3}{5}\right]^2-\frac{9}{25}=0\)
\(\Rightarrow\left[2x+\frac{3}{5}\right]^2=\frac{9}{25}\)
\(\Rightarrow\left[2x+\frac{3}{5}\right]^2=\left[\frac{9}{25}\right]^2\)
\(\Rightarrow2x+\frac{3}{5}=\pm\frac{9}{25}\)
\(\Rightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{9}{25}\\2x+\frac{3}{5}=-\frac{9}{25}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{25}\\x=-\frac{12}{25}\end{cases}}\)
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\(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
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bài 2 giúp đi mọi người
a,\(\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)
b,\(\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+:\frac{9}{25}\)
c,\(\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)
d,\(\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)
\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)
\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)
\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)
\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)
\(\Leftrightarrow\text{ }x=\frac{53}{45}\)
\(\text{Vậy }x=\frac{53}{45}\)
\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)
\(\text{TH 1: Dấu chia}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)
\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)
\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)
\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)
\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)
\(\text{Vậy }\text{}x=\frac{-1}{5}\)
\(\text{TH 2:Dấu cộng}\)
\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)
\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)
\(\Leftrightarrow2x=\frac{-3}{25}\)
\(\Leftrightarrow x=\frac{-3}{25}:2\)
\(\Leftrightarrow x=\frac{-3}{50}\)
\(\text{Vậy }x=\frac{-3}{50}\)
\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)
\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)
\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)
\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)
\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)
\(\Leftrightarrow x=\frac{5}{4}\)
\(\text{Vậy }x=\frac{5}{4}\)
Tính:
a)\(A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\left( { - 4,6} \right);\) b) \(B = \left( {\frac{{ - 7}}{9}} \right).\frac{{13}}{{25}} - \frac{{13}}{{25}}.\frac{2}{9}\)
a)
\(\begin{array}{l}A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\left( { - 4,6} \right)\\A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\frac{{ - 23}}{5}\\A = \frac{{5.\left( { - 3} \right).11.\left( { - 23} \right)}}{{11.23.5.5}}\\A = \frac{3}{5}\end{array}\)
b)
\(\begin{array}{l}B = \left( {\frac{{ - 7}}{9}} \right).\frac{{13}}{{25}} - \frac{{13}}{{25}}.\frac{2}{9}\\B = \frac{{13}}{{25}}.\left( {\frac{{ - 7}}{9} - \frac{2}{9}} \right)\\B = \frac{{13}}{{25}}.(-1)\\B = \frac{{-13}}{{25}}.\end{array}\)
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\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
từ phần sau số 0 thứ nhất là câu mới nha
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tìm x biết :
a) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
c) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
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\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
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Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)
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Tìm x
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
giải nhanh hộ mình với, mai mình nộp rồi
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
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\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
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a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=0+\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\frac{3^2}{5^2}\)
\(\Rightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
b) \(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)
\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}:3\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1^3}{3^3}\right)\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{6}:3\)
\(\Rightarrow x=\frac{1}{18}\)
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Giải phương trình:
1. frac{2x+3}{4}-frac{5x+3}{6}frac{3-4x}{12}
2. frac{3.left(2x+1right)}{4}-1frac{15x-1}{10}
3. frac{2x-1}{5}-frac{x-2}{3}frac{x+7}{15}
4. frac{x+3}{2}-frac{x-1}{3}frac{x+5}{6}+1
5. frac{x-4}{5}-frac{3x-2}{10}-xfrac{2x-5}{3}-frac{7x+2}{6}
6. frac{left(x+2right)left(x+10right)}{3}-frac{left(x+4right)left(x+10right)}{12}frac{left(x-2right)left(x+4right)}{4}
7. frac{left(x+2right)^2}{8}-2left(2x-1right)25+frac{left(x-2right)^2}{8}
8.frac{7x^2-14x-5}{5}frac{left(2x+1right)^...
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Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(100-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{27}}+\left(1,2.0,5\right):\frac{3}{5}\)