Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

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

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

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

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

\(a,=30x^3-6x^2-12x\\ b,=-6x^5-2x^3+x^2\)

a: \(=30x^3-6x^2-12x\)

b: \(=-6x^5-2x^3+x^2\)

giúp mk vs , mk đang cần gấp 

d

A

Chọn A

1) \(P=\dfrac{5-3\sqrt{x}}{\sqrt{x}-1}\left(đk:x\ge0,x\ne1\right)\)

\(=\dfrac{-3\left(\sqrt{x}-1\right)+2}{\sqrt{x}-1}=-3+\dfrac{2}{\sqrt{x}-1}\in Z\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Do \(x\ge0,x\ne1\) và x là số chính phương

\(\Rightarrow x\in\left\{0;4;9\right\}\)

2) \(3x^2-5x+1=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)

\(\Rightarrow C=\dfrac{2022}{3x^2-5x+1}\le2022:\left(-\dfrac{13}{12}\right)=-\dfrac{24264}{13}\)

\(minC=-\dfrac{24624}{13}\Leftrightarrow x=\dfrac{5}{6}\)

Ta có: \(A=\dfrac{1-0.5\cdot\left(3.84-2.4\right):0.8}{\dfrac{4}{5}-\left(1\dfrac{1}{3}-2\dfrac{1}{6}\right)-1.5}\)

\(=\dfrac{1-0.5\cdot1.44:0.8}{\dfrac{4}{5}-\left(\dfrac{4}{3}-\dfrac{13}{6}\right)-\dfrac{3}{2}}\)

\(=\dfrac{1-0.9}{\dfrac{4}{5}+\dfrac{5}{6}-\dfrac{3}{2}}=\dfrac{0.1}{\dfrac{2}{15}}=\dfrac{3}{4}\)

Giải:

A=1-0,5.(3,84-2,4):0,8 / 4/5-(1 1/3 - 2 1/6)-1,5

A=1-0,5.1,44:0.8 / 4/5-(4/3-13/6)-3/2

A=1-0.9 / 4/5-(-5/6)-3/2

A=0.1 / 2/15

A= 1/10 : 2/15

A=3/4

Chúc bạn học tốt!