Các cặp là:

(3;5); (5;7);(11;13);(17;19);(29;31);(41;43);(59;61);(71;73)

3n+5 chia hết 2n-3

2(3n+5) chia hết 2n-3

6n+10 chia hết 2n-3

3.(2n-3)+19 chia hết 2n-3

Suy ra 19 chia hết 2n-3

Suy ra 2n-3=Ư(19)

2n-3={-19;-1;1;19}

n={-8;1;2;11}

\(4=a+b+2ab\ge2\sqrt{ab}+ab\Rightarrow ab+\sqrt{ab}-2\le0\)

\(\Rightarrow\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+2\right)\le0\)

\(\Rightarrow\sqrt{ab}-1\le0\)

\(\Rightarrow ab\le1\)

Lại có a;b ko âm nên \(ab\ge0\Rightarrow0\le ab\le1\)

\(P=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=\left(4-2ab\right)^3-3ab\left(4-2ab\right)\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(P=\left(4-2x\right)^3-3x\left(4-2x\right)=-8x^3+54x^2-108x+64\)

\(=64-\frac{x}{8}\left(64x^2-432x+864\right)=64-\frac{x}{8}.\left\lbrack\left(8x-27\right)^2+135\right\rbrack\)

Do \(\left(8x-27\right)^2+135>0;\forall x\Rightarrow\frac{x}{8}\left\lbrack\left(8x-27\right)^2+135\right\rbrack\ge0;\forall x\ge0\)

\(\Rightarrow P\le64\)

\(P_{max}=64\) khi x=0 \(\Rightarrow\left(a;b\right)=\left(0;4\right);\left(4;0\right)\)

Lại có:

\(P=2+\left(-8x^3+54x^2-108x+62\right)=2+2\left(1-x\right)\left(4x^2-23x+31\right)\)

Do \(1-x\ge0;\forall x\le1\)

Đồng thời \(4x^2-23x+31=4x^2+23\left(1-x\right)+8>0;\forall x\le1\)

\(\Rightarrow2\left(1-x\right)\left(4x^2-23x+31\right)\ge0;\forall x\le1\)

\(\Rightarrow P\ge2\)

\(P_{\min}=2\) khi x=1 =>a=b=1

a.

806 660 000

b.

3 005 080

1.

a. 3xy+6xz=3x(y+2z)

b. \(18x^2y-12x^3=6x^2\left(3y-2x\right)\)

\(8xy^2-2x^2y=2xy\left(4y-x\right)\)

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

2.

a. \(x\left(x+y\right)-7x-7y=x\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(x-7\right)\)

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

c. \(x\left(x+3y\right)-5x-15y=x\left(x+3y\right)-5\left(x+3y\right)=\left(x+3y\right)\left(x-5\right)\)

d. \(2x\left(x-5\right)+3x-15=2x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(2x+3\right)\)

e. \(x\left(x+y\right)+ax+ay=x\left(x+y\right)+a\left(x+y\right)=\left(x+y\right)\left(x+a\right)\)

f. \(a\left(x+y\right)-4x-4y=a\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(a-4\right)\)

3.

a.

\(\left(3x+1\right)^2-\left(x+1\right)^2=\left\lbrack\left(3x+1\right)-\left(x+1\right)\right\rbrack\left\lbrack\left(3x+1\right)+\left(x+1\right)\right\rbrack=2x\left(4x+2\right)\)

b.

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

c.

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

\(=\left\lbrack\left(3x-3y\right)-\left(2x+2y\right)\right\rbrack\left\lbrack\left(3x-3y\right)+\left(2x+2y\right)\right\rbrack=\left(x-5y\right)\left(5x-y\right)\)

d.

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

\(=\left(x+y\right)\left(5x-5y\right)=5\left(x-y\right)\left(x+y\right)\)

\(x^2-\left(y+1\right)x+y-2=0\)

\(\Leftrightarrow x^2-xy-x+y-2=0\)

\(\Leftrightarrow x^2-x-2=y\left(x-1\right)\)

Với x=1 ko phải nghiệm

Với x≠1

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

=>x-1 là Ư(2)

=>x-1={-2;-1;1;2}

=>x={-1;0;1;2}

CD là tiếp tuyến của gì em?

\(P=\sqrt{x^2+3^2}+\sqrt{y^2+5^2}\ge\sqrt{\left(x+y\right)^2+\left(3+5\right)^2}=10\)

\(P_{\min}=10\) khi x=9/4;y=15/4

Đề là \(\sqrt{\frac{2x-3}{x+5}}+\sqrt{\frac{y+5}{2x-3}}=2\) hay \(\sqrt{\frac{2x-3}{x+5}}+\sqrt{\frac{y+5}{2y-3}}=2\) em?

Thay \(3-xy=x+y\) vào pt dưới

\(\Rightarrow1+12\left(x+y\right)=7y^3+6xy\left(x+2y\right)\)

\(\Rightarrow x^3+y^3+1+12\left(x+y\right)=8y^3+6xy\left(y+2y\right)+x^3\)

\(\Rightarrow x^3+y^3+12\left(x+y\right)+1=\left(x+2y\right)^3\)

\(\Rightarrow x^3+y^3+3\left(x+y\right)+1+3.3\left(x+y\right)=\left(x+2y\right)^3\)

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

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

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

\(\Rightarrow x+y+1=x+2y\)

\(\Rightarrow y=1\)

Thay vào pt đầu \(\Rightarrow x=1\)