CM hộ mik với :v

a: Khi x=4 thì \(B=\dfrac{2}{2-5}=\dfrac{2}{-3}=\dfrac{-2}{3}\)

b: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\)

\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{x-9}=\dfrac{x-3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

a: Đặt 117=a; 119=b

\(A=3\frac{1}{117}\cdot4\frac{1}{119}-1\frac{116}{117}\cdot5\frac{118}{119}-\frac{5}{119}\)

\(=3\frac{1}{a}\cdot4\frac{1}{b}-\left(1+\frac{a-1}{a}\right)\cdot\left(5+\frac{b-1}{b}\right)-\frac{5}{b}\)

\(=\frac{3a+1}{a}\cdot\frac{4b+1}{b}-\frac{2a-1}{a}\cdot\frac{6b-1}{b}-\frac{5}{b}\)

\(=\frac{\left(3a+1\right)\left(4b+1\right)-\left(2a-1\right)\left(6b-1\right)-5a}{ab}\)

\(=\frac{12ab+3a+4b+1-\left(12ab-2a-6b+1\right)-5a}{ab}\)

\(=\frac{12ab+3a+4b+1-12ab+2a+6b-1-5a}{ab}=\frac{10b}{ab}=\frac{10}{a}\)

\(=\frac{10}{117}\)

b: Đặt 105=a; 651=b

\(B=2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)

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

\(=\frac{6a+1}{3a}\cdot\frac{1}{b}-\frac{1}{a}\cdot\frac{3b+b-1}{b}-\frac{4}{3ab}+\frac{4}{a}\)

\(=\frac{6a+1}{3ab}-\frac{4b-1}{ab}-\frac{4}{3ab}+\frac{4}{a}=\frac{6a+1-3\left(4b-1\right)-4+12b}{3ab}\)

\(=\frac{6a-3+12b-12b+3}{3ab}=\frac{6a}{3ab}=\frac{2}{b}=\frac{2}{651}\)

Ta có:$P=x^3+y^3+2xy=(x+y{)}^3-3xy(x+y)+2xy={201}^3-601xy$

Đặt $S=xy=x(201-x)$

Dễ có:$1\le x\le 200$

$S=200-(x-1)(x-200)\ge 0\Rightarrow S_{min}=200$

Không mất tính TQ giả sử $x\le y\Rightarrow x\le 100$

$S=100.101-$

Ta có:

A-B=2m^3+3m^3-4mn^2

TH1

Nếu m > n. Đặt m=n+x

óA-B=2(n+x)^3+3m^3-4(n+x)n^2

óA-B=2(n^3+3n^2x+2nx^2+x^3)=3m^3-4n^3-4n^2x

óA-B=n^3+2n^2x+6nx^2+2x^3>0

Vậy A>B

TH2                     

Nếu m < n. Đặt n=m+y

óA-B=2m^3+3(m+y)^3-4m(m+y)^2

óA-B=2m^3+3(m^3+3m^2y+3my^2+y^3)-4m^3-8m^2y-4my^2

óA-B=m^3+m^2y+5my^2+3y^3> 0

Vậy A > B

a) Thay m = -1 và n = 2 ta có:

3m - 2n = 3(-1) -2.2 = -3 - 4 = -7

b) Thay m = -1 và n = 2 ta được 

7m + 2n - 6 = 7.(-1) + 2.2 - 6 = -7 + 4 - 6 = -9.

Đặt :
\(\frac{1}{315}=a;\frac{1}{651}=b\)  thay vào A ta được :

\(A=\left(2+a\right)b-\left(3+1-b\right).3a-4ab+12a\)

\(\Leftrightarrow A=2b+ab-12a+3ab-4ab+12a\)

\(\Leftrightarrow A=2b\)

Thay \(b=\frac{1}{651}\) ta được :

\(A=\frac{2}{651}\)

Chúc bạn học tốt !!!

Bài 2:

a: a>=b

=>5a>=5b

=>5a+10>=5b+10

b: a>=b

=>-8a<=-8b

=>-8a-9<=-8b-9<-8b+3

x^5+y^5=xxxxx+yyyyy=aaaaa