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(2/3×x-1/3)=2/3+1/3
(2/3×x-1/3)=3/3
2/3×x=3/3+1/3
2/3×x=4/3
x=4/3:3/2
x=4/3×2/3
x=8/9
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x=0+\frac{2}{5}\)
\(\Leftrightarrow x\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(\Leftrightarrow x\left(\frac{5}{15}+\frac{6}{15}\right)=\frac{2}{5}\)
\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}\)
\(\Leftrightarrow x=\frac{6}{11}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{49}{50}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{49}{50}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{50}\times\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{50}{100}-\frac{49}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Leftrightarrow x+1=100\)
\(\Leftrightarrow x=100-1\)
\(\Leftrightarrow x=99\)
D = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{100^2}-1.\right)\)
=>\(-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{100^2}.\right)\)
=>\(-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{100^2-1}{100^2}\)
=>\(-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}\right)\)
=>\(-\left(\frac{1.2.3...99}{2.3.4....100}\right)\left(\frac{3.4.5....101}{2.3.4....100}\right)\)
=>\(-\left(\frac{1}{100}.\frac{101}{2}\right)\)
=>\(D=-\frac{101}{200}\)
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(C=(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}):(\frac{5}{12}+1-\frac{7}{11})\)
\(=\left(\frac{88}{132}-\frac{33}{132}+\frac{60}{132}\right):\left(\frac{55}{132}+\frac{132}{132}-\frac{84}{132}\right)=\left(\frac{115}{132}\right):\frac{103}{132}=\frac{115}{132}.\frac{132}{103}=\frac{115}{103}\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}=\frac{1}{3}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}=\frac{1}{3}+\frac{1}{2}-\frac{1}{8}=\frac{8+12-3}{24}=\frac{17}{24}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0=\frac{1}{4}-4-1=\frac{1-16-4}{4}=\frac{-19}{4}\)
\(25\cdot\left(-\frac{1}{5}\right)^3+\frac{1}{5}-2\cdot\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\left(-\frac{1}{125}\right)+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=0-\frac{1}{2}-\frac{1}{2}=-1\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=-1\)