2a) \(\frac{3^6+45^4-15^3.4^5}{27^4.25^3+45^6}\)

= \(\frac{3^6+\left(3^2.5\right)^4-\left(3.5\right)^3.\left(2^2\right)^5}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}\)

= \(\frac{3^6+3^8.5^4-3^3.5^3.4^{10}}{3^{12}.5^6-3^{12}.5^6}=\frac{3^3.\left(3^3+3^5.5^4-5^3.4^{10}\right)}{0}\)(xem lại đề)

b)  \(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{16}{3}\right)^3:\left(\frac{4}{9}\right)^3}{2^7.5^2+512}\)

= \(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{16}{3}:\frac{4}{9}\right)^3}{2^7.5^2+2^9}\)

= \(\frac{2^7+12^3}{2^7\left(5^2+2^2\right)}\)

= \(\frac{2^7+\left(2^2.3\right)^3}{2^7.29}\)

= \(\frac{2^7+2^6.3^3}{2^7.29}\)

= \(\frac{2^6\left(1+27\right)}{2^7.29}=\frac{28}{2.29}=\frac{14}{29}\)

mk xin lỗi bn nhé...mk vt nhầm đề...mk vt lại nha:

2a,\(\frac{3^6+45^4-15^3.9^5}{27^4.25^3+45^6}\)

a)Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x-1\right|+\left|3+x\right|=\left|1-x\right|+\left|3+x\right|\ge\left|1-x+3+x\right|=4\)

\(\Rightarrow VT\ge VP."="\Leftrightarrow-3\le x\le1\)

b) \(\hept{\begin{cases}\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\\\frac{8}{2\left(y-5\right)^2+2}\le4\end{cases}}\Leftrightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le\frac{1}{2}\\y=5\end{cases}}\)

c Tương tự b

2) \(\frac{1}{x}+\frac{1}{y}=5\Leftrightarrow x+y-5xy=0\Leftrightarrow5x+5y-25xy=0\Leftrightarrow5x\left(1-5y\right)-\left(1-5y\right)=-1\)

\(\Leftrightarrow\left(5x-1\right)\left(1-5y\right)=-1\)

Xét ước

Bài 1 và Bài 2 dễ, bn có thể tự làm được!

Bài 3:

a) ta có: 10²⁰ = (10²)¹⁰ = 100¹⁰

=> 100¹⁰>9¹⁰

=> 10²⁰>9¹⁰

b) ta có: (-5)³⁰ = 5³⁰ =( 5³)¹⁰ = 125¹⁰ ( vì là lũy thừa bậc chẵn)

(-3)⁵⁰ = 3⁵⁰ = (3⁵)¹⁰= 243¹⁰

=> 125¹⁰ < 243¹⁰

=> (-5)³⁰ < (-3)⁵⁰

c) ta có: 64⁸ = (2⁶)⁸= 2⁴⁸

16¹² = ( 2⁴)¹² = 2⁴⁸

=> 64⁸ = 16¹²

d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)

\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)

\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

3.a) Ta có: 9¹⁰=(3²)¹⁰=3²⁰

Mà 10²⁰<3²⁰

Nên 10²⁰<9¹⁰

c)Ta có:64⁸ =(8²)⁸=8¹⁶

16¹²=(2³)¹²=8³⁶

vì 8¹⁶<8³⁶

Nên 64⁸<16²

Bài 2:
a) \(x+\frac{1}{3}=\frac{3}{4}\)

\(\Rightarrow x=\frac{5}{12}\)

Vậy \(x=\frac{5}{12}\)

b) \(\left|x+\frac{1}{8}\right|-\frac{1}{6}=0\)

\(\Rightarrow\left|x+\frac{1}{8}\right|=\frac{1}{6}\)

\(\Rightarrow x+\frac{1}{8}=\frac{1}{6}\) hoặc \(x+\frac{1}{8}=\frac{-1}{6}\)

+) \(x+\frac{1}{8}=\frac{1}{6}\Rightarrow x=\frac{1}{24}\)

+) \(x+\frac{1}{8}=\frac{-1}{6}\Rightarrow x=\frac{-7}{24}\)

Vậy \(x\in\left\{\frac{1}{24};\frac{-7}{24}\right\}\)

c) \(\frac{x}{27}=\frac{-2}{36}\)

\(\Rightarrow\frac{x}{27}=\frac{-1}{18}\)

\(\Rightarrow18x=-27\)

\(\Rightarrow x=\frac{-3}{2}\)

Vậy \(x=\frac{-3}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\) hoặc \(x+\frac{1}{2}=\frac{-1}{4}\)

+) \(x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{-1}{4}\)

+) \(x+\frac{1}{2}=\frac{-1}{4}\Rightarrow x=\frac{-3}{4}\)

Vậy \(x\in\left\{\frac{-1}{4};\frac{-3}{4}\right\}\)

a)\(x+\frac{1}{3}=\frac{3}{4}\)

\(\Rightarrow x=\frac{3}{4}-\frac{1}{3}\)

\(\Rightarrow x=\frac{5}{12}\)

b)\(\left|x+\frac{1}{8}\right|-\frac{1}{6}=0\)

\(\Rightarrow\left|x+\frac{1}{8}\right|=\frac{1}{6}\)

\(\Rightarrow x+\frac{1}{8}=\frac{1}{6}\) hoặc \(x+\frac{1}{8}=-\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{6}-\frac{1}{8}\) hoặc \(x=-\frac{1}{6}-\frac{1}{8}\)

\(\Rightarrow x=\frac{1}{24}\) hoặc \(x=-\frac{7}{24}\)

c)\(\frac{x}{27}=-\frac{2}{36}\)

\(\Rightarrow x=\frac{\left(-2\right)\cdot27}{36}=-\frac{3}{2}\)

d)\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\) hoặc \(x+\frac{1}{2}=-\frac{1}{4}\)

\(\Rightarrow x=\frac{1}{4}-\frac{1}{2}\) hoặc \(x=-\frac{1}{4}-\frac{1}{2}\)

\(\Rightarrow x=-\frac{1}{4}\) hoặc \(x=-\frac{3}{4}\)

\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)

\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)

\(=\frac{7}{2}-2\)

\(=\frac{7}{2}-\frac{4}{2}\)

\(=\frac{3}{2}\)

\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)

\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)

\(=\frac{3}{7}.\left(2-9\right)\)

\(=\frac{3}{7}.\left(-7\right)\)

\(=-3\)

\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )

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) = 

\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)

\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)

\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)

\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)

\(=\frac{2005\times2010-6}{2005\times2011}\)

\(=\frac{2004}{2005}\)

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

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