13.

$(x+4)^2+(x+5)(x-5)-2x(x+1)$

$=(x^2+8x+16)+(x^2-25)-(2x^2+2x)$

$=x^2+8x+16+x^2-25-2x^2-2x$

$=(x^2+x^2-2x^2)+(8x-2x)+(16-25)=6x-9$

14.

$(x-1)^2-2(x+3)(x-3)+4x(x-4)$

$=(x^2-2x+1)-2(x^2-9)+(4x^2-16x)$

$=x^2-2x+1-2x^2+18+4x^2-16x$

$=(x^2-2x^2+4x^2)+(-2x-16x)+(1+18)=3x^2-18x+19$

15.

$(y-3)(y+3)(y^2+9)-(y^2+2)(y^2-2)$

$=(y^2-9)(y^2+9)-(y^4-4)$

$=(y^4-81)-(y^4-4)=-81+4=-77$

Đặt a+b=x ; b+c=y ; c+a=z, A=(a+b)³.....

Khi đó A= x³+y³+z³-3xyz= (x+y)³- 3xy(x+y) - 3xyz +z³

= (x+y+z)³- 3z(x+y)(x+y+z)- 3xy(x+y+z)

=(x+y+z)(x²+y²+z²+2xy+2yz+2xz-3xz-3zy-3xy)

= (x+y+z)(x²+y²+z²-xy-yz-xz)

tu day em thay vao nhe

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề hơn.

\(A=\left(\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\left(x>0;x\ne1\right)\)

\(A=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right]:\dfrac{1-\sqrt{x}}{\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot2+2^2}\)

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

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

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

\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)

Vậy: ... 

mấy giờ nộp để đây làm cho ! Chờ đây ăn cơm xong đã!

          \(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)

\(=\)\(2\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{3}-\sqrt{60}\)

\(=\)\(6+\sqrt{15}-\sqrt{60}\)

\(=\)\(6-\sqrt{15}\)

\(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)

=\(6+\sqrt{15}-\sqrt{2^2.15}\)

=\(6+\sqrt{15}-2\sqrt{15}\)

=\(6-\sqrt{15}\)

k mk nha

Nhưng cho mk hỏi nhân kiểu nào ra 6 dc v ạ

Sửa đề: \(-3x+\sqrt{25x^2}\)

\(=-3x+\sqrt{\left(5x\right)^2}\)

\(=-3x-5x\left(x< 0\right)\)

=-8x

=>Chọn C

P=1+5²+5⁴+5⁶+...+5⁹⁸

=>5².P=P=5²+5⁴+5⁶+...+5⁹⁸+5¹⁰⁰

=>25P-P=24P=P=5²+5⁴+5⁶+...+5⁹⁸+5¹⁰⁰-(1+5²+5⁴+5⁶+...+5⁹⁸)

=>24P=5¹⁰⁰-1

=>P=\(\frac{5^{100}-1}{24}.\)