a,\(=x^3+x^2-\left(31x^2+31x\right)\)

\(=x^2\left(x+1\right)-31x\left(x+1\right)\)

\(=\left(x^2-31x\right)\left(x+1\right)=\left(31^2-31^2\right)\left(31+1\right)=0\)

b, Phân tích 3 số hạng đầu ta có:\(=x^5-x^4-\left(14x^4-14x^3\right)=\left(x^4-14x^3\right)\left(x-1\right)=\left(14^4-14^4\right)\left(x-1\right)=0\)

Thay x= 14 vào ta có: \(-29.14^2+13.14=-5502\)

c, do x=9 => x+1=10; Thay vào ta có:

\(C=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-...+\left(x+1\right)x^2-\left(x+1\right)x+10\)

\(C=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-....+x^3+x^2-x^2-x+10\)

\(C=-x+10=-9+10=1\)

CHÚC BẠN HỌC TỐT.....

hình như ban ghi dau - thành + ở chỗ x^3-x^2-x^2-x+10

\(B=x^5-15x^4+16x^3-29x^2+13x\)

\(=x^5-14x^4-x^4+14x^3+2x^3-28x^2-x^2+14x-x+14-14\)

\(=x^4\left(x-14\right)-x^3\left(x-14\right)+2x^2\left(x-14\right)-x\left(x-14\right)-\left(x-14\right)-14\)

\(=\left(x^4-x^3+2x^2-x-1\right)\left(x-14\right)-14\)

Thay x = 14 => B = -14

Vậy...

phần còn lại tách ra làm tương tự nhé

cu tao to

Ta có: x = 9 => x - 9 = 0

\(Q\left(x\right)=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

\(=x^{14}-9x^{13}-x^{13}+9x^{12}+x^{12}-9x^{11}+...-x^3+9x^2+x^2-9x-x+9+1\)

\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+...-x^2\left(x-9\right)+x\left(x-9\right)-\left(x-9\right)+1\)

\(=0+1=1\)

\(A=6xy\left(xy-y^2\right)-8x^2.\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)

\(A=6x^2y^2-6xy^3-8x^3+8x^2y^2+5y^2x^2-5xy^3\)

\(A=19x^2y^2-11xy^3-8x^3\)

Tại x=1/2, y=2

\(A=19.\frac{1}{4}.2^2-11.\frac{1}{2}.2^3-8\left(\frac{1}{2}\right)^3=19-44-1=-26\)

\(A=x^3-30x-31x+1\)

=\(x^3-31x^2+x^2-31x+1\)

=\(x^2\left(x-31\right)+x\left(x-31\right)+1\)

=1(do x=31)

\(B= x^4 -17x^3 +17x^2 -17x + 20 tại x= 16\)

\(=x^4-16x^3-x^3+16x^2+x^2-16x-x+20\)

=\(x^3\left(x-16\right)+x^2\left(x-16\right)+x\left(x-16\right)-x+20\)

=-16+20=4

Thay 30 = x - 1, 2 câu kia tương tự

Help me !!!!!

Bài 1:

a) \(\dfrac{15xy}{10x^2y}\)

= \(\dfrac{3.5xy}{2.5xyx}\)

= \(\dfrac{3}{2x}\)

d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)

= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)

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

à....cái đó thì mình chưa tính ra được

\(A=x^2+3x+7\)

\(=x^2+2.1,5x+2,25+4,75\)

\(=\left(x+1,5\right)^2+4,75\ge4,75\)

Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)

\(B=2x^2-8x\)

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

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

\(=2\left[\left(x-2\right)^2-4\right]\)

\(=2\left(x-2\right)^2-8\ge-8\)

Vậy \(B_{min}=-8\Leftrightarrow x=2\)