Bài 1:

a) \(\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}\)

\(=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}\)

\(=\dfrac{5}{3}\)

b) \(\left(0,2\right)^2\cdot5-\dfrac{2^3\cdot27}{4^6\cdot9^5}\)

\(=0,2\cdot5\cdot0,2-\dfrac{2^3\cdot3^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=\dfrac{1}{5}-\dfrac{2^3\cdot3^3}{2^{12}\cdot3^{10}}\)

\(=\dfrac{1}{5}-\dfrac{1}{2^9\cdot3^7}\)

\(=\dfrac{2^9\cdot3^7}{2^9\cdot3^7\cdot5}-\dfrac{5}{2^9\cdot3^7\cdot5}\)

\(=\dfrac{2^9\cdot3^7-5}{2^9\cdot3^7\cdot5}\) 

c) \(\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6\cdot\left(1+2^2+2^3\right)}{26\cdot5^6}\)

\(=\dfrac{1+2^2+2^3}{26}\)

\(=\dfrac{1+4+8}{26}\)

\(=\dfrac{13}{26}\)

\(=\dfrac{1}{2}\)

Bài 2: 

Theo đề ta có:

\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):-\dfrac{1}{4}=-\dfrac{15}{4}\)

\(\Rightarrow\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right)=-\dfrac{15}{4}\cdot-\dfrac{1}{4}\)

\(\Rightarrow a\cdot\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{15}{16}\)

\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{15}{16}-\dfrac{3}{4}\)

\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{3}{16}\)

\(\Rightarrow a=\dfrac{3}{16}:\dfrac{1}{2}\)

\(\Rightarrow a=\dfrac{3}{8}\) 

1:

a: \(=\dfrac{5^{16}\cdot3^{21}}{3^{22}\cdot5^{15}}=\dfrac{1}{3}\cdot5=\dfrac{5}{3}\)

b: \(=0.04\cdot5-\dfrac{2^3\cdot3^3}{3^6\cdot2^{12}}\)

\(=0.2-\dfrac{1}{3^3\cdot2^9}=\dfrac{1}{5}-\dfrac{1}{3^3\cdot2^9}=\dfrac{3^3\cdot2^9-5}{5\cdot3^3\cdot2^9}\)

c: \(=\dfrac{5^6+4\cdot5^6+2^3\cdot5^6}{26\cdot5^6}=\dfrac{1+4+8}{26}=\dfrac{13}{26}=\dfrac{1}{2}\)

2:

Theo đề, ta có:

\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):\dfrac{-1}{4}=\dfrac{-15}{4}\)

=>\(\dfrac{1}{2}a+\dfrac{3}{4}=\dfrac{15}{16}\)

=>1/2a=15/16-12/16=3/16

=>a=3/8

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

\(a,\left(x-1\right)\left(2x-1\right)\)

\(=2x^2-x-2x+1\)

\(=2x^2-3x+1\)

\(b,\left(9x^4+12x^3-15x^2-3x\right):3x\)

\(=3x^3+4x^2-5x-1\)

a) 54 – 19i;

b) -15 + i.

bai EZ​ qua

mà bài dài quá sao làm hết dc

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

\(=x^3+2^3-2\left(x^2-1\right)\)

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)

\(=2x^3-x+x^2-y\)

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

\(a,\dfrac{x^2-4}{2x}:\dfrac{3x-6}{6}=\dfrac{\left(x-2\right)\left(x+2\right)}{2x}.\dfrac{6}{3\left(x-2\right)}=\dfrac{6\left(x-2\right)\left(x+2\right)}{6x\left(x-2\right)}=\dfrac{x+2}{x}\)

a, \(\dfrac{7}{15}\)+\(\dfrac{18}{54}\)
=  \(\dfrac{7}{15}\)+ \(\dfrac{1}{3}\)
=  \(\dfrac{7}{15}\)+\(\dfrac{5}{15}\)
=    \(\dfrac{12}{15}\) = \(\dfrac{4}{5}\) =0,8
b, \(\dfrac{12}{25}\)+\(\dfrac{23}{75}\)-\(\dfrac{1}{4}\)
=     \(\dfrac{59}{75}\)   -\(\dfrac{1}{4}\)
=         \(\dfrac{161}{300}\)