a,1/4

b,8

a, 1/4
b,8

A

C

a: \(x-\frac25+\frac{7}{12}=\frac12\cdot\frac34\cdot\frac{-9}{16}\)

=>\(x-\frac{24}{60}+\frac{35}{60}=\frac38\cdot\frac{-9}{16}=\frac{-27}{128}\)

=>\(x=-\frac{27}{128}-\frac{11}{60}=\frac{-27\cdot15-11\cdot32}{1920}=-\frac{757}{1920}\)

b: \(\frac49x-\frac12=-\frac59\)

=>\(\frac49x=-\frac59+\frac12=\frac{-10+9}{18}=\frac{-1}{18}\)

=>\(x=-\frac{1}{18}:\frac49=-\frac{1}{18}\cdot\frac94=\frac{-1}{2\cdot4}=-\frac18\)

c: \(-\frac23x+\frac58=\frac{7}{12}\)

=>\(-\frac23x=\frac{7}{12}-\frac58=\frac{14}{24}-\frac{15}{24}=-\frac{1}{24}\)

=>\(x=\frac{1}{24}:\frac23=\frac{1}{24}\cdot\frac32=\frac{1}{8\cdot2}=\frac{1}{16}\)

d: \(\frac25x:\frac73\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac25\cdot\frac37\cdot\frac18=\frac{29}{15}\)

=>\(x\cdot\frac{3}{4\cdot5\cdot7}=\frac{29}{15}\)

=>\(x=\frac{29}{15}:\frac{3}{140}=\frac{29}{15}\cdot\frac{140}{3}=\frac{29\cdot28}{3\cdot3}=\frac{812}{9}\)

\(A=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\\ A=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)

\(B=\dfrac{7a-7b+8a+8b-16b}{\left(a+b\right)\left(a-b\right)}=\dfrac{15a-15b}{\left(a-b\right)\left(a+b\right)}\\ B=\dfrac{15\left(a-b\right)}{\left(a-b\right)\left(a+b\right)}=\dfrac{15}{a+b}\)

Câu 1: C

Câu 2: A

a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(\Leftrightarrow9x+7=16\)

\(\Leftrightarrow9x=9\)

hay x=1

a) (x - 1)(x + 1)(x² + 1)(x⁴ + 1)(x⁸ + 1)

= (x² - 1)(x² + 1)(x⁴ + 1)(x⁸ + 1)

= (x⁴ - 1)(x⁴ + 1)(x⁸ + 1)

= (x⁸ - 1)(x⁸ + 1)

= x¹⁶ - 1

b) (a² - 2b)(a² + 2b)(a⁴ + 4b²)(a⁸ + 16b⁴)

= (a⁴ - 4b²)(a⁴ + 4b²)(a⁸ + 16b⁴)

= (a⁸ - 16b⁴)(a⁸ + 16b⁴)

= a¹⁶ - 256b⁸

a) \(\dfrac{x^4-2x^3}{2x^2-x^3}=\dfrac{x^3\left(x-2\right)}{x^2\left(2-x\right)}=\dfrac{-x^3}{x^2}=-x\)

Thay x vào ta có biểu thức đã cho bằng\(-\left(\dfrac{-1}{2}=\dfrac{1}{2}\right)\)

a, x² + 4y² + 4xy - 16

= (x + 2y)² - 4²

= (x + 2y - 4).(x + 2y + 4)

b, 5x² - 10xy + 5y²

= 5.(x² - 2xy + y²)

= 5.(x - y)²

a)-2.x=-16

x= -16:(-2)=8

Vậy....

b)x.-12=-42

x=-42:(-12)=3,5

Vậy....

c)-17-x=13

x= -17-13=-30

Vậy....

d)x+5=-14

x=-14-5=-19

Vậy....

e)x.(-2)=10

a, -2x=-16

       x= (-16):(-2)

       x= 8

vậy...

b,

x.(-12)=-42

x= (-42):(-12)

x= 7/2

Vậy...

c,

-17-x=13

 x= (-17)-13

x=(-30)

Vậy....

d,  x+5=-14

     x= -14-5

     x=-19

Vậy....

e, x.(-2)=10

     x= 10 : (-2)

     x= -5

Vậy...