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\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
c) C = ( 1 - 2 ) + ( 3 - 4 ) + ... + ( 79 - 80 )
C = ( -1 ) + ( -1 ) + ... + ( -1 )
C = ( -1 ) x ( 80 - 1 + 1 ) : 2
C = ( -1 ) x 80 : 2
C = ( -40 )
a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)
= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)
=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)
=\(\frac{1}{2}-1-1\)
=\(\frac{-3}{2}\)
b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)
= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)
= \(-12:\left\{\frac{-1}{12}\right\}^2\)
= \(-12:\frac{1}{144}\)
= \(-12.144\)
= -1728
c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)
= \(\frac{-7}{6}\)
d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)
= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)
= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)
= \(10.\frac{7}{5}\)
= 14
e) (1+23−14).(0,8−34)2
= (1+23−14).(\(\frac{4}{5}\)−34)2
= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
= \(\frac{17}{20}.\frac{1}{400}\)
= \(\frac{17}{8000}\)
1.a) \(\left(31\frac{6}{13}+5\frac{9}{41}\right)-36\frac{6}{13}=\left(31+\frac{6}{13}+5+\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)\)
\(=\left(36+\frac{6}{13}-\frac{9}{41}\right)-\left(36+\frac{6}{13}\right)=\left(36+\frac{6}{13}\right)-\left(36+\frac{6}{13}\right)-\frac{9}{41}=-\frac{9}{41}\)
b) \(\frac{5}{3}+\left(-\frac{2}{7}\right)-\left(-1,2\right)-\left|1.4-0,2\right|\)
\(=\frac{5}{3}-\frac{2}{7}+1,2-1,2=\frac{29}{21}\)
c) \(0,25+\frac{3}{5}-\left(\frac{1}{8}-\frac{2}{5}+1\frac{1}{4}\right)+\left|\frac{3}{5}\right|\)
\(=\frac{1}{4}+\frac{3}{5}-\frac{1}{8}+\frac{2}{5}-1-\frac{1}{4}+\frac{3}{5}\)
\(=\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{3}{5}+\frac{2}{5}-1\right)+\frac{3}{5}-\frac{1}{8}=\frac{19}{40}\)
2) \(-\frac{3}{5}-x=0,75\)
=> \(-\frac{3}{5}-x=\frac{3}{4}\)
=> \(x=-\frac{3}{5}-\frac{3}{4}=\frac{-27}{20}\)
b) \(x+\frac{1}{3}=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
=> \(x+\frac{1}{3}=\frac{2}{5}+\frac{1}{3}\)
=> \(x=\frac{2}{5}\)
c) |2x - 4| + 1 = 5
=> |2x - 4| = 4
<=> \(\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)
Giúp mình với nha cả nhả :<
Cả nhà làm vài ý thui cx được ạ :<