Ta có: \(A=\overline{5a7,34}+\overline{bc,1}+\overline{14,2d}\)

=500+10a+7+0,34-10b-c-0,1+14,2+0,01d

=10a-10b-c+0,01d+521,44

\(B=527,9+\overline{ab,cd}\overline{}\)

=527,9+10a+b+0,1c+0,01d

B-A

=527,9+10a+b+0,1c+0,01d-10a+10b+c-0,01d-521,44

=11b+1,1c+6,46>0

=>B>A

2:

a: =>x^2(5x^2+2)+2=0

x^2>=0

5x^2+2>=2

=>x^2(5x^2+2)>=0 với mọi x

=>x^2(5x^2+2)+2>=2>0 với mọi x

=>PTVN

b: x^4-12x^2+24=0

=>x^4-12x^2+36-12=0

=>(x^2-6)^2-12=0

=>(x^2-6-2căn 3)(x^2-6+2căn 3)=0

=>x^2=6+2căn 3 hoặc x^2=6-2căn 3

=>\(x=\pm\sqrt{6+2\sqrt{3}};x=\pm\sqrt{6-2\sqrt{3}}\)

\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)

cảm ơn bạn minh nhiều nha

Bài 22:

a: \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)

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

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

\(=\left(-x-4\right)\left(2x^2+3x+4\right)\)

c: \(-16+\left(x-3\right)^2\)

\(=\left(x-3\right)^2-16\)

=(x-3-4)(x-3+4)

=(x-7)(x+1)

d: \(64+16y+y^2=y^2+2\cdot y\cdot8+8^2=\left(y+8\right)^2\)

Bài 21:

a: \(\left(\frac12+x\right)^2=x^2+2\cdot x\cdot\frac12+\left(\frac12\right)^2=x^2+x+\frac14\)

\(\left(2x+1\right)^2=\left(2x\right)^2+2\cdot2x\cdot1+1^2=4x^2+4x+1\)

b: \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)

\(\left(xy+0,01\right)^2=\left(xy\right)^2+2\cdot xy\cdot0,01+\left(0,01\right)^2\)

\(=x^2y^2+0,02xy+0,0001\)

c: \(\left(\frac12-x\right)^2=\left(\frac12\right)^2-2\cdot\frac12\cdot x+x^2=x^2-x+\frac14\)

\(\left(2x-1\right)^2=\left(2x\right)^2-2\cdot2x\cdot1+1^2=4x^2-4x+1\)

d: \(\left(2x-3y\right)^2=\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2=4x^2-12xy+9y^2\)

\(\left(xy-0,01\right)^2=\left(xy\right)^2-2\cdot xy\cdot0,01+\left(0,01\right)^2\)

\(=x^2y^2-0,02xy+0,0001\)

e: (x+1)(x-1)=x^2-1

g: (x+y+z)(x-y-z)

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

\(=x^2-y^2-z^2-2yz\)

f: (x-2y)(x-2y)

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

=2001

\(\left|2x-3\right|=3-2x\)

\(ĐK:x\le\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)

Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)

Ta có: \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}\)

            \(2y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{6}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{6}=\dfrac{x+z}{20+6}=\dfrac{52}{26}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=20.2=40\\y=15.2=30\\z=6.2=12\end{matrix}\right.\)

Bài 2:

a) \(\dfrac{2}{15}-\dfrac{7}{10}=\dfrac{4}{30}-\dfrac{21}{30}=-\dfrac{17}{30}\)

b) \(\dfrac{-3}{14}+\dfrac{2}{21}=\dfrac{-9}{42}+\dfrac{4}{42}=\dfrac{-5}{42}\)

c) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-96}{144}+\dfrac{-108}{144}=\dfrac{-204}{144}=-\dfrac{17}{12}\)

Bài 3: 

a) \(\dfrac{3}{8}+\dfrac{-5}{6}=\dfrac{3}{8}-\dfrac{5}{6}=\dfrac{18}{48}-\dfrac{40}{48}=-\dfrac{22}{48}=-\dfrac{11}{24}\)

b) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-24}{54}-\dfrac{30}{54}=\dfrac{-54}{54}=-1\)

c) \(\dfrac{2}{21}-\dfrac{-1}{28}=\dfrac{8}{84}-\dfrac{-3}{84}=\dfrac{11}{84}\)