a) Với x = 11 <=> 12 = x+1

\(A\left(x\right)=x^{17}-\left(x+1\right)x^{16}+\left(x+1\right)x^{15}-...+12x-1\)

\(A\left(x\right)=12x-11=12.11-1=120\)

b) \(B=6x-6y+10-3ax+3ay+15a\)

\(B=6\left(x-y\right)+10-3a\left(x-y\right)+15a\)

\(B=6.5+10-3.a.5+15a\)

\(B=40\)

c)\(C=\frac{x-y}{x+6}=\frac{x-y}{x+x-2y}=\frac{x-y}{2\left(x-y\right)}=\frac{1}{2}\left(x-2y=6\right)\)

\(C=\frac{2x+6}{3x-2y}+\frac{2y-6}{4y-x}\)

\(C=\frac{2x+1-2y}{3x-2y}+\frac{2y-x+2y}{4y-x}\)

\(C=1+1=2\)

d) ta có : x-y-x = 0

\(\Rightarrow\left\{{}\begin{matrix}x-z=y\\x-y=z\\x=y+z\end{matrix}\right.\).Thay vào B, ta có :

\(B=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)

\(B=\frac{y}{x}.\frac{\left(-z\right)}{y}.\frac{x}{z}\)

B= -1

A= 6x - 6y + 10 - 3ax +3ay +15 a

A= 6.( x-y ) - 3a . ( x-y) + 10+15a

thay x-y = 5 vào biểu thức A

A= 6.5 - 3a. 5 + 15 a+10

A=30 - (15a-15a)+10

A=30+10

A=40

CHÚC BN HỌC TỐT!!!!!

Ta có : A = 6.(x - y) -3a.(x-y) + 15a + 10

thay x - y = 5 vào A

A = 6.5 -3a.5 +15a +10

A = 40

a: A=(-x)^3+3*(-x)^2*2+3*(-x)*2^2+2^3=(-x+2)^3

=(28+2)^3=30^3=27000

b: \(C=\left(x+2y-2\right)^3=\left(20+2\cdot9-2\right)^3\)

=36^3

c: 11^3-1

=(11-1)(11^2+11+1)

=10*(121+12)

=1330

d: x^3-y^3=(x-y)^3+3xy(x-y)

=6^3+3*6*9

=216+162

=378

x-2y=6

nên x=2y+6

\(P=\dfrac{2y+6-y}{2y+6+6}=\dfrac{y+6}{2y+12}=\dfrac{1}{2}\)

\(Q=\dfrac{2x+6}{3x-2y}+\dfrac{2y-6}{4y-x}\)

\(=\dfrac{2\left(2y+6\right)}{3\left(2y+6\right)-2y}+\dfrac{2y-6}{4y-\left(2y+6\right)}\)

\(=\dfrac{4y+12}{6y+18-2y}+\dfrac{2y-6}{4y-2y-6}\)

\(=\dfrac{4y+12}{4y+18}+\dfrac{2y-6}{2y-6}\)

\(=\dfrac{8y^2-24y+24y-48+8y^2+36y-24y-108}{\left(4y+18\right)\left(2y-6\right)}\)

\(=\dfrac{16y^2+12y-156}{4\left(2y+9\right)\cdot\left(y-3\right)}\)

\(=\dfrac{4y^2+3y-34}{\left(2y+9\right)\left(y-3\right)}\)

\(1,ĐK:x\ne0;x\ne\pm6\)

\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right].\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}.\frac{1}{12\left(x^2+1\right)}\)

\(=\frac{12\left(x^2+1\right)}{x}.\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)

\(2,A=\frac{1}{x}=\frac{1}{\frac{1}{\sqrt{9+4\sqrt{5}}}}=\sqrt{9+4\sqrt{5}}\)

Cho tam giác ABC vuông tại B có góc B₁=B₂  ; Â=60^o, kẻ BH vuông góc với AC (H thuộc AC). Qua B kẻ đường thẳng d song song với AC.

a) Tính góc ABH.

b) Chứng minh đường thẳng d vuông góc với BH.