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a) Để A lớn nhất thì \(\frac{15}{4.\left|3x+7\right|+3}\) lớn nhất hay 4.|3x + 7| + 3 nhỏ nhất
Có: \(4.\left|3x+7\right|+3\ge3\forall x\)
Dấu "=" xảy ra khi |3x + 7| = 0
=> 3x + 7 = 0
=> 3x = -7
\(\Rightarrow x=\frac{-7}{3}\)
Với x = \(\frac{-7}{3}\) thay vào đề bài ta được A = 10
Vậy \(A_{Max}=10\) khi x = \(\frac{-7}{3}\)
b) Để B lớn nhất thì \(\frac{21}{8.\left|15x-21\right|+7}\) lớn nhất hay 8.|15x - 21| + 7 nhỏ nhất
Có: \(8.\left|15x-21\right|+7\ge7\forall x\)
Dấu "=" xảy ra khi |15x - 21| = 0
=> 15x - 21 = 0
=> 15x = 21
\(\Rightarrow x=\frac{21}{15}=\frac{7}{5}\)
Với \(x=\frac{7}{5}\) thay vảo đề bài ta tìm được B = \(\frac{8}{3}\)
Vậy \(B_{Max}=\frac{8}{3}\) khi x = \(\frac{7}{5}\)
c) Có: \(\begin{cases}\left|x+1\right|\ge x+1\\\left|3x-4\right|\ge4-3x\\\left|2x-1\right|\ge2x-1\end{cases}\)\(\forall x\)
\(\Rightarrow C\ge\left(x+1\right)+\left(4-3x\right)+\left(2x-1\right)+5\)
hay \(C\ge9\)
Dấu "=" xảy ra khi \(\begin{cases}x+1\ge0\\3x-4\le0\\2x-1\ge0\end{cases}\)\(\Rightarrow\begin{cases}x\ge-1\\3x\le4\\2x\ge1\end{cases}\)\(\Rightarrow\begin{cases}x\ge-1\\x\le\frac{3}{4}\\x\ge\frac{1}{2}\end{cases}\)\(\Rightarrow\frac{1}{2}\le x\le\frac{3}{4}\)
Vậy \(C_{Max}=9\) khi \(\frac{1}{2}\le x\le\frac{3}{4}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
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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) =
a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b)\(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.\left(-14\right)\)
\(=-6\)
c)\(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)
\(=9.\left(-\frac{1}{3}\right)^3+9.\frac{1}{27}\)
\(=9.\left[\left(-\frac{1}{3}\right)^3+\frac{1}{27}\right]\)
\(=9.0=0\)
d)\(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)
\(=\left(15\frac{1}{4}-25\frac{1}{4}\right):\left(-\frac{5}{7}\right)\)
\(=\left(-10\right):\left(-\frac{5}{7}\right)\)
\(=14\)
a) \(1\frac{3}{19}+\frac{8}{21}-\frac{3}{19}+0.5+\frac{13}{21}\)
\(=\left(1\frac{3}{19}-\frac{3}{19}\right)+\left(\frac{8}{21}+\frac{13}{21}\right)+0.5\)
\(=1+1+0.5=2.5\)
b) \(\left(-\frac{3}{4}+\frac{2}{7}\right):\frac{3}{7}+\left(\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=0:\frac{3}{7}=0\)
a)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5=2,5\)
b)\(\frac{3}{7}\cdot19\frac{1}{3}-\frac{3}{7}\cdot33\frac{1}{3}\)
\(=\frac{3}{7}\left(19\frac{1}{3}-33\frac{1}{3}\right)=\frac{3}{7}\cdot\left(-14\right)=-6\)
c) \(9\left(-\frac{1}{3}\right)^3+\frac{1}{3}=9\cdot\left(\frac{-1}{27}\right)+\frac{1}{3}=-\frac{1}{3}+\frac{1}{3}=0\)
d) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)=-\frac{7}{5}:\left(15\frac{1}{4}-25\frac{1}{4}\right)=-\frac{7}{5}:\left(-10\right)=14\)
a) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(\frac{27}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b) \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\frac{58}{3}-\frac{3}{7}.\frac{100}{3}\)
\(=\frac{3}{7}.\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{7}.\left(-14\right)\)
\(=-6\)
c) \(9.\left(-\frac{1}{3}\right)^3+\frac{1}{3}\)
\(=9.\left(-\frac{1}{27}\right)+\frac{1}{3}\)
\(=-\frac{1}{3}+\frac{1}{3}\)
\(=0\)
d) \(15\frac{1}{4}:\left(-\frac{5}{7}\right)-25\frac{1}{4}:\left(-\frac{5}{7}\right)\)
\(=\frac{61}{4}:\left(-\frac{5}{7}\right)-\frac{101}{4}:\left(-\frac{5}{7}\right)\)
\(=\left(\frac{61}{4}-\frac{101}{4}\right):\left(-\frac{5}{7}\right)\)
\(=\left(-10\right):\left(-\frac{5}{7}\right)\)
\(=14\)
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