mốt bạn lấy trên mạng thì ghi tham khảo vào 

mốt bạn lấy trên mạng thì ghi tham khảo vào 

mik thấy trên ấy ghi là \(\dfrac{1}{6}\) chứ có phải là\(\dfrac{1}{16}\) đâu bạn

we are fine, thank you

a, hệ số cao nhất là \(8x^4\)

    hệ số tự do của đa thức là 15

b, \(A\left(x\right)+B\left(x\right)=8x^4+8x^3-6x-15+8x^4+8x^3-4x^2-6x-10\)

\(A\left(x\right)+B\left(x\right)=\left(8x^4+8x^4\right)+\left(8x^3+8x^3\right)-4x^2-\left(6x-6x\right)-\left(15-10\right)\)

\(A\left(x\right)+B\left(x\right)=16x^4+16x^3-4x^2-5\)

c, \(B\left(x\right)-A\left(x\right)=8x^4+8x^3-4x^2-6x-10-\left(8x^4+8x^3-6x-15\right)\)

\(B\left(x\right)-A\left(x\right)=8x^4+8x^3-4x^2-6x-10-8x^4-8x^3+6x+15\)

\(B\left(x\right)-A\left(x\right)=\left(8x^4-8x^4\right)+\left(8x^3-8x^3\right)-4x^2-\left(6x+6x\right)-\left(10+15\right)\)

\(B\left(x\right)-A\left(x\right)=-4x^2-12x-25\)

\(A=\left(\dfrac{1}{4}x^2y^2z\right).\left(-4xyz^2\right)\)

a, \(A=-1x^3y^3z^3\)

b, thay x = -1; y = 1; z = -2 vào đơn thức A

ta có \(A=\left(-1\right).\left(-1\right)^3.1^3\left(-2\right)^3\)

        \(A=\left(-1\right).\left(-1\right).1.\left(-8\right)\)

        \(A=1.1.\left(-8\right)=-8\)

1 Xét △AMB và △ANC

ta có AB = AC

        AM = AN

        BM = CN

=> △AMB = △ANC

b, góc BAN = BACgóc - NACgóc

    góc CAM = BACgóc - BAMgóc

=> gócBAN = gócCAM

Xét △ANB = △AMC

ta có : AN = AM

          góc BAN = góc CAM

         AB = AC

=> △ANB = △AMC

\(a,\dfrac{11}{12}-\dfrac{3}{16}.\dfrac{8}{18}+\dfrac{1}{18}\)

\(=\dfrac{11}{12}-\dfrac{24}{288}+\dfrac{1}{18}\)

\(=\dfrac{11}{12}-\dfrac{1}{12}+\dfrac{1}{18}\)

\(=\dfrac{10}{12}+\dfrac{1}{18}\)

\(=\dfrac{5}{6}+\dfrac{1}{18}\)

\(=\dfrac{15}{18}+\dfrac{1}{18}=\dfrac{16}{18}=\dfrac{8}{9}\)

b,\(\left(7^2.13+7^2.36\right):7^3\)

\(=\left[7^2.\left(13+36\right)\right]:7^3\)

\(=\left(7^2.49\right):7^3\)

\(=\left(49.49\right):343\)

\(=2401:343\)

\(=7\)

\(\left(\dfrac{5}{3}x+2\right).\left(x+1\right)-\dfrac{5}{3}x\left(x+2\right)=7\)

\(\dfrac{5x^2}{3}+\dfrac{5x+6x+6}{3}-\dfrac{5}{3}x\left(x+2\right)=7\)

\(\dfrac{5x^2}{3}+\dfrac{11x+6}{3}=\dfrac{5}{3}x^2-\dfrac{5}{3}x.2=7\)

\(\dfrac{5x^2}{3}+\dfrac{11x+6}{3}-\dfrac{5x^2}{3}-\dfrac{5}{3}x.2=7\)

\(\dfrac{5x^2}{3}+\dfrac{11x+6}{3}-\dfrac{5x^2}{3}-\dfrac{10}{3}x=7\)

\(\dfrac{5x^2}{3}+\dfrac{11x+6}{3}-\dfrac{5x^2}{3}-\dfrac{10x}{3}=7\)

\(\dfrac{11x+6}{3}-\dfrac{10x}{3}=7\)

\(11x+6-\left(10x\right)=21\)

\(11x-\left(10x\right)+6=21\)

\(11x=\left(10x\right)=21-6=15\)

\(x=15\)