cả phần c bài 3 ạ

Bài 4 : 

a, \(D=\left(\frac{2}{x-3}+\frac{1}{x+3}\right):\frac{x+1}{x-3}\)ĐK : \(x\ne\pm3;-1\)

\(=\left(\frac{2x+6+x-3}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+1}{x-3}=\frac{3\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x+1\right)}=\frac{3}{x+3}\)

b, \(D=\frac{3}{x+3}=\frac{x}{6}\Rightarrow x^2+3x=18\Leftrightarrow\left(x-3\right)\left(x+6\right)=0\Leftrightarrow x=3\left(ktm\right);x=-6\left(tm\right)\)

c, \(D=\frac{3}{x+3}< \frac{x}{x+3}\Leftrightarrow\frac{3-x}{x+3}< 0\Leftrightarrow\frac{x-3}{x+3}>0\)

TH1 : \(\hept{\begin{cases}x-3>0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>3\\x>-3\end{cases}\Leftrightarrow x>3}}\)

TH2 : \(\hept{\begin{cases}x-3< 0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 3\\x< -3\end{cases}}\Leftrightarrow x< -3}\)

Vậy x > 3 ; x < -3 

d, Để \(\frac{3}{x+3}\in Z\Rightarrow x+3\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

3c làm tương tự 4d em nhé 

Trả lời:

Bài 1:

a, \(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)\)

\(=8x^3+36x^2+54x+27+8x^3-36x^2+54x-27-8x^2+18\)

\(=16x^3-8x^2+108x+18\)

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

\(=x^3+6x^2+12x+8+x^3-6x^2+12x-8+x^3-3x\left(x^2-4\right)\)

\(=3x^3+24x-3x^3+12x=36x\)

Bài 2:

a, \(A=\left(3x+2\right)^2+\left(2x-7\right)^2-2\left(3x+2\right)\left(2x-7\right)\)

\(=\left(3x+2-2x+7\right)^2=\left(x+9\right)^2\)

Thay x = - 19 vào A, ta có:

\(A=\left(-19+9\right)^2=\left(-10\right)^2=100\)

b, \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)

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

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

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

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

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

Thay x + y = 1 vào A, ta có:

\(A=2.1^3-3.1^2=-1\)

c, \(B=x^3+y^3+3xy\)

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

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

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

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

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

Thay x + y = 1 vào B, ta có:

\(B=1^3-3xy.\left(1-1\right)=1-3xy.0=1-0=1\)

d, \(C=8x^3-27y^3\)

\(=\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)\)

\(=\left(2x-3y\right)\left(4x^2-12xy+9y^2+6xy\right)\)

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

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

Thay xy = 4 và 2x + 3y = 5 vào C, ta có:

\(C\)\(=5^3+6.4.5=125+120=245\)

Trả lời:

Bài 3:

\(A=x^2+x-2=\left(x^2+x+\frac{1}{4}\right)-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\forall x\)

Dấu "=" xảy ra khi \(x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy GTNN của \(A=-\frac{9}{4}\Leftrightarrow x=-\frac{1}{2}\)

\(B=x^2+y^2+x-6y+2021\)

\(=x^2+y^2+x-6y+\frac{1}{4}+9+\frac{8047}{4}\)

\(=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-6y+9\right)+\frac{8047}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2+\frac{8047}{4}\)\(\ge\frac{8047}{4}\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+\frac{1}{2}=0\\y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}}\)

Vậy GTNN của B = \(\frac{8047}{4}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}\)

\(C=x^2+10y^2-6xy-10y+35\)

\(=x^2+9y^2+y^2-6xy-10y+25+10\)

\(=\left(x^2-6xy+9y^2\right)+\left(y^2-10y+25\right)+10\)

\(=\left(x-3y\right)^2+\left(y-5\right)^2+10\ge10\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-3y=0\\y-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=15\\y=5\end{cases}}}\)

Vậy GTNN của C = 10 <=> \(\hept{\begin{cases}x=15\\y=5\end{cases}}\)

\(D=4x-x^2+5\)

\(=-\left(x^2-4x-5\right)\)

\(=-\left(x^2-4x+4-9\right)\)

\(=-\left[\left(x-2\right)^2-9\right]\)

\(=-\left(x-2\right)^2+9\le9\forall x\)

Dấu "=" xảy ra khi x - 2 = 0 <=> x = 2

Vậy GTLN của D = 9 <=> x = 2

\(E=-x^2-4y^2+2x-4y+3\)

\(=-x^2-4y^2+2x-4y-1-1+5\)

\(=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+5\)

\(=-\left(x-1\right)^2-\left(2y+1\right)^2+5\le5\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-1=0\\2y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-\frac{1}{2}\end{cases}}}\)

Vậy GTLN của D = 5  <=> \(\hept{\begin{cases}x=1\\y=-\frac{1}{2}\end{cases}}\)

Ta có : \(x^5+x^3+x^2+1=x^2\left(x^3+1\right)+\left(x^3+1\right)=\left(x^2+1\right)\left(x^3+1\right)\)

\(=>\left(x^5+x^3+x^2+1\right):\left(x^3+1\right)=\left(x^2+1\right)\left(x^3+1\right):\left(x^3+1\right)=x^2+1\)