\(1+3+5+...+\left(2x+1\right)=2601\)

số số hạng: \(\left[\left(2x+1\right)-1\right]:2+1=x+1\)

tổng: \(\left(2x+1\right)+1:2x\left(x+1\right)=\left(x+1\right)^2\)

\(\left(x+1\right)^2=2601\)

\(\Rightarrow\orbr{\begin{cases}x+1=51\\x+1=-51\end{cases}\Rightarrow\orbr{\begin{cases}x=50\\x=-52\end{cases}}}\)

2x + 1 = 101

=> x= 50

k co minh nhe

số các số hạng có đc là ( 2x-1+1):2+1 = x+1 số hạng

theo ta có : (2x+2)(x+1)/2 =2601

                  2(x+1)(x+1)/2=2601  

                  (x+1)²           =2601

             \(\Rightarrow\)    \(\hept{\begin{cases}x+1=\sqrt{2601}\\\\x+1=-\sqrt{2601}\end{cases}}\)

              vậy x=50 hoặc x=-52

2:

a: A=(x-1)(x+3)

=x^2+3x-x-3

=x^2+2x-3

=x^2+2x+1-4

=(x+1)^2-4>=-4

Dấu = xảy ra khi x=-1

b: B=(x-2)(x+5)

=x^2+5x-2x-10

=x^2+3x-10

=x^2+3x+9/4-49/4

=(x+3/2)^2-49/4>=-49/4

Dấu = xảy ra khi x=-3/2

c: C=(x+2)(x+3)

=x^2+5x+6

=x^2+5x+25/4-1/4

=(x+5/2)^2-1/4>=-1/4

Dấu = xảy ra khi x=-5/2

d: D=(x-2)(x-4)

=x^2-6x+8

=x^2-6x+9-1

=(x-3)^2-1>=-1

Dấu = xảy ra khi x=3

e: E=(x+3)(x-2)

=x^2-2x+3x-6

=x^2+x-6

=x^2+x+1/4-25/4

=(x+1/2)^2-25/4>=-25/4

Dấu = xảy ra khi x=-1/2

g: G=(x+1)(x+5)

=(x+3-2)(x+3+2)

=(x+3)^2-4>=-4

Dấu = xảy ra khi x=-3

Diện tích là:

\(70\cdot70=4900\left(m^2\right)\)

Câu 10: A

Câu 11: B

A

C

C

C

A

C

B

a,c,c,c,a,c,b

\(1,=\left(3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5\right):\left(3x^2-2x+1\right)\\ =\left(3x^2-2x+1\right)\left(x^2-2x-5\right):\left(3x^2-2x+1\right)\\ =x^2-2x-5\\ 2,=\left(x^4+x^2+2x^3+2x-x^2-1+8x-24\right):\left(x^2+1\right)\\ =\left[\left(x^2+1\right)\left(x^2+2x-1\right)+8x-24\right]:\left(x^2+1\right)\\ =x^2+2x-1\left(dư.8x-24\right)\\ 3,=\left(3x^4-8x^3-10x^2+8x-5\right):\left(3x^2-2x+1\right)\\ =x^2-2x-5\left(câu.1\right)\\ 4,=\left(x^3-5x^2-x^2+5x-3x+15\right):\left(x-5\right)\\ =\left(x-5\right)\left(x^2-x-3\right):\left(x-5\right)\\ =x^2-x-3\)

\(5,=\left(-12x^2+13x-5\right):\left(3x^2-2x+1\right)\\ =\left(-12x^2+8x+4+5x-9\right):\left(3x^2-2x+1\right)\\ =\left(3x^2-2x+1\right)\left(-4\right)-9\\ =-4\left(dư.-9\right)\\ 6,=\left(x^3+2x^2-5x^2-10x+12x+24\right):\left(x+2\right)\\ =\left(x+2\right)\left(x^2-5x+12\right):\left(x+2\right)\\ =x^2-5x+12\\ 7,=\left(2x^3-x^2+x+6x^2-3x+3\right):\left(2x^2-x+1\right)\\ =\left(2x^2-x+1\right)\left(x+3\right):\left(2x^2-x+1\right)\\ =x+3\\ 8,=\left(2x^3+4x^2-x^2-2x+x+2\right):\left(x+2\right)\\ =\left(x+2\right)\left(2x^2-x+1\right):\left(x+2\right)\\ =2x^2-x+1\)

\(9,=\left(9x^4-16x^2+4\right):\left(3x^2+2x-2\right)\\ =\left(9x^4+6x^3-6x^2-6x^3-4x^2+4x-6x^2-4x+4\right):\left(3x^2+2x-2\right)\\ =\left(3x^2+2x-2\right)\left(3x^2-2x-2\right):\left(3x^2+2x-2\right)\\ =3x^2-2x-2\\ 10,=\left(2x^3+4x^2-7x^2-14x+15x+30\right):\left(x+2\right)\\ =\left(x+2\right)\left(2x^2-7x+15\right):\left(x+2\right)\\ =2x^2-7x+15\)

ĐKXĐ: \(y\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y=3\\xy\left(x+y\right)=2\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=3\\uv=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=2\end{matrix}\right.\) (loại) hoặc \(\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Rightarrow x=y=1\)