Xét 2 trường hợp

TH1: a = -1,5

Ta có \(B=\dfrac{-1,5}{2}-\dfrac{2}{-0,75}\)\(=\dfrac{23}{12}\)

TH2: a = 1,5

Ta có \(B=\dfrac{1,5}{2}-\dfrac{2}{-0,75}=\dfrac{41}{12}\)

mk mới lp 6

\(\left|a\right|=1,5\Rightarrow\orbr{\begin{cases}a=1,5\\a=-1,5\end{cases}}\)

\(Th1:a=1,5;b=-0,75\)hay \(a=\frac{3}{2};b=\frac{-3}{4}\)

\(\Rightarrow M=\frac{3}{2}+2.\frac{3}{2}.\frac{-3}{4}-\frac{-3}{4}\)

\(=\frac{6}{4}+\frac{-9}{4}+\frac{3}{4}=0\)

\(Th2:a=-1,5;b=0,75\)hay \(a=\frac{-3}{2};b=\frac{-3}{4}\)

​\(\Rightarrow M=\frac{-3}{2}+2.\frac{-3}{2}.\frac{-3}{4}-\frac{-3}{4}\)​

\(=\frac{-6}{4}+\frac{9}{4}+\frac{3}{4}=\frac{3}{2}\)

Vì |a| = 1,5

\(\Rightarrow\orbr{\begin{cases}a=1,5\\a=-1,5\end{cases}}\)

Thay a;b vào M ta có :

\(\orbr{\begin{cases}M=1,5+2.\left(1,5\right).\left(-0,75\right)+0,75\\M=-1,5+2.\left(-1,5\right).\left(-0,75\right)+0,75\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}M=1,5+3.\frac{-3}{4}+0,75\\M=-1,5+2.\left(1,5\right).\left(0,75\right)+0,75\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}M=1,5-2,25+0,75\\M=-1,5+3.\frac{3}{4}+0,75\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}M=1,5-\left(2,25+0,75\right)\\M=-1,5+2,25+0,75\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}M=1,5-3\\M=-1,5+\left(2,25+0,75\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}M=-1,5\\M=-1,5+3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}M=-1,5\\M=1,5\end{cases}}\)

Vậy M = -1,5 hoặc M = 1,5

N=1,5 : 2 - 2 : -0,75

  = 3/2 X 1/2 - 2 x -4/3

  = 3/4 x -8/3

=-2

Thay A= 1,5 ; B=-0,75 vào biểu thức N ta được

N=1,5/2-2/-0,75=0,75-2,75= (-2)

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

\(=\dfrac{4}{3}:\left(1+\dfrac{1}{4}\right)=\dfrac{4}{3}:\dfrac{5}{4}=\dfrac{16}{15}\)

Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)

\(=\left(x+y\right)^3=1^3=1\)

Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

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

Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)

\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)

\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)

\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)

\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)

a. Tại x=\(\frac{-1}{2}\), ta có:

 \(\left(\frac{-1}{2}\right)^2+4.\left(\frac{-1}{2}\right)+3=\frac{1}{4}+\left(-2\right)+3=\frac{5}{4}\)

b. Ta có:

 \(x^2+4x+3=0\)

\(\Rightarrow x^2+x+3x+3=0\)

\(\Rightarrow\left(x^2+x\right)+\left(3x+3\right)=0\)

\(\Rightarrow x\left(x+1\right)+3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x+3\right)=0\)

\(\Rightarrow\hept{\begin{cases}x+1=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-3\end{cases}}}\)

Vậy \(x=-1;x=-3\)

1=13500

2=103500

bạn xài cái này gõ công thức ra đi

a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)

\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)

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

b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :

\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)