a: a\(\perp\)HK

b\(\perp\)HK

Do đó: a//b

b: Ta có: \(\widehat{BAH}+45^0=180^0\)

=>\(\widehat{BAH}=180^0-45^0=135^0\)

Ta có: a//b

=>\(\widehat{BAH}+\widehat{ABK}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{ABK}+135^0=180^0\)

=>\(\widehat{ABK}=45^0\)

a: Hiệu vận tốc hai xe là 51-36=15(km/h)

Thời gian ô tô đuổi kịp xe máy là:

45:15=3(giờ)

b: Hai xe gặp nhau lúc:

8h30p+3h=11h30p

\(\dfrac{3}{1\text{x}2}+\dfrac{3}{2\text{x}3}+...+\dfrac{3}{99\text{x}100}\)

\(=3\text{x}\left(\dfrac{1}{1\text{x}2}+\dfrac{1}{2\text{x}3}+...+\dfrac{1}{99\text{x}100}\right)\)

\(=3\text{x}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=3\text{x}\left(1-\dfrac{1}{100}\right)=3\text{x}\dfrac{99}{100}=\dfrac{297}{100}\)

\(\dfrac{3}{1\times2}+\dfrac{3}{2\times3}+...+\dfrac{3}{99\times100}\)

\(=3\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\right)\)

\(=3\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=3\times\left(1-\dfrac{1}{100}\right)\)

\(=3\times\dfrac{99}{100}\)

\(=\dfrac{297}{100}\)

Số phần quả bóng còn lại so với tổng số bóng ban đầu là:

\(1-\dfrac{1}{7}-\dfrac{1}{5}=1-\dfrac{12}{35}=\dfrac{23}{35}\)

\(d.x^{11}+x^7+1\\ =x^{11}-x^2+x^7-x+x^2+x+1\\ =x^2\left(x^9-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3-1\right)\left(x^6+x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\\ =x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^6+x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\\=\left(x^2+x+1\right)\left[x^2\left(x-1\right)\left(x^6+x^3+1\right)+x\left(x-1\right)\left(x^3+1\right)+1\right]\\ =\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^6+x^3+1\right)+\left(x^2-x\right)\left(x^3+1\right)+1\right]\\ =\left(x^2+x+1\right)\left(x^9+x^6+x^3-x^8-x^5-x^2+x^5+x^2-x^4-x+1\right)\\ =\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^4+x^3-x+1\right)\) 

\(e.x^8+x+1\\ =x^8-x^2+x^2+x+1\\ =x^2\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\\ =x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left[x^2\left(x-1\right)\left(x^3+1\right)+1\right]\\ =\left(x^2+x+1\right)\left[\left(x^3-x^2\right)\left(x^3+1\right)+1\right]\\ =\left(x^2+x+1\right)\left(x^6+x^3-x^5-x^2+1\right)\)

\(\left(x-1\right)+\left(x-2\right)+...+\left(x-20\right)=150\\ x-1+x-2+...+x-20=150\\ \left(x+x+...+x\right)-\left(1+2+...+20\right)\\ 20\cdot x-\left[\left(20-1\right):1+1\right]\cdot\left(20+1\right):2=150\\ 20\cdot x-20\cdot21:2=150\\ 20\cdot x-210=150\\ 20\cdot x=150+210\\ 20\cdot x=360\\ x=360:20\\ x=18\)

Bài 1:

Quân còn số tiền là:

$50000-30000=20000$ (đồng)

Bài 2:

Số liền sau của số lớn nhất có 2 chữ số là: 100

Số hạng thứ hai là: $30-25=5$

$4^{x-2}+4^{x-1}=20$

$\Rightarrow 4^{x-2}+4^{x-2}.4=20$

$\Rightarrow 4^{x-2}.(1+4)=20$

$\Rightarrow 4^{x-2}.5=20$

$\Rightarrow 4^{x-2}=20:5$

$\Rightarrow 4^{x-2}=4$

$\Rightarrow x-2=1$

$\Rightarrow x=1+2=3$

Bài 1:

a) $x^2+6x+9$

$=x^2+2.x.3+3^2$

$=(x+3)^2$

b) $9x^2-6x+1$

$=(3x)^2-2.3x.1+1^2$

$=(3x-1)^2$

c) $x^2y^2+xy+\frac14$

$=(xy)^2+2.xy.\frac12+\left(\frac12\right)^2$

$=\left(xy+\frac12\right)^2$

d) $(x-y)^2+6(x-y)+9$

$=(x-y)^2+2.(x-y).3+3^2$

$=(x-y+3)^2$

Bài 2: 

a) $-x^3+3x^2-3x+1$

$=1^3-3.1^2.x+3.1.x^2-x^3$

$=(1-x)^3$

b) $x^3+x^2+\frac13 x+\frac{1}{27}$

$=x^3+3.x^2.\frac13+3.x.\left(\frac13\right)^2+\left(\frac13\right)^3$

$=\left(x+\frac13\right)^3$

c) $x^6-3x^4y+3x^2y^2-y^3$

$=(x^2)^3-3.(x^2)^2.y+3.x^2.y^2-y^3$

$=(x^2-y)^3$

d) $(x-y)^3+(x-y)^2+\frac13 (x-y)+\frac{1}{27}$

$=(x-y)^3+3.(x-y)^2.\frac13+3.(x-y).\left(\frac13\right)^2+\left(\frac13\right)^3$

$=\left(x-y+\frac13\right)^3$

Bài 3:

a) $x^3+27$

$=x^3+3^3$

$=(x+3)(x^2-x.3+3^2)$

$=(x+3)(x^2-3x+9)$

b) $x^3-\frac18$

$=x^3-\left(\frac12\right)^3$

$=\left(x-\frac12\right)\left[x^2-x.\frac12+\left(\frac12\right)^2\right]$

$=\left(x-\frac12\right)\left(x^2-\frac12 x+\frac14\right)$

c) $8x^3+y^3$

$=(2x)^3+y^3$

$=(2x+y)[(2x)^2-2x.y+y^2]$

$=(2x+y)(4x^2-2xy+y^2)$

d) $8x^3-27y^3$

$=(2x)^3-(3y)^3$

$=(2x-3y)[(2x)^2+2x.3y+(3y)^2]$

$=(2x-3y)(4x^2+6xy+9y^2)$

Bài 4:

a) \(101^2=\left(100+1\right)^2\)

\(=100^2+2.100.1+1^2\)

\(=10000+200+1=10201\)

b) \(75^2-50.75+25^2\)

\(=75^2-2.75.25+25^2\)

\(=\left(75-25\right)^2\)

\(=50^2=2500\)

c) \(103.97\)

\(=\left(100+3\right).\left(100-3\right)\)

\(=100^2-3^2\\ =10000-9=9991\)

Bài 5:

a) \(\left(x+3y\right)^2-\left(x-3y\right)^2\)

\(=\left(x+3y-x+3y\right)\left(x+3y+x-3y\right)\\ =6y.2x=12xy\)

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

\(=\left(x-y\right)^2-2.\left(x-y\right).2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\\ =\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\\ =\left(x-y-2x-4y\right)^2\\ =\left(-x-5y\right)^2\)

c) \(A=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(=\left(x+2+x-2\right)\left[\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right]-2x\left(x^2+12\right)\\ =2x\left(x^2+4x+4-x^2+4+x^2-4x+4\right)-2x\left(x^2+12\right)\\ =2x\left(x^2+12\right)-2x\left(x^2+12\right)=0\)

d) \(B=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)

\(=\left(xy+2\right)^3-3.\left(xy+2\right)^2.2+3.\left(xy+2\right).2^2-2^3\\ =\left(xy+2-2\right)^3\\ =\left(xy\right)^3=x^3y^3\)

e) \(A=\left(x-3\right)\left(x^2+3x+9\right)-\left(x^3+3\right)\)

\(=\left(x-3\right)\left(x^2+x.3+3^2\right)-x^3-3\\ =x^3-3^3-x^3-3\\ =-27-3=-30\)

Bài 6:

\(a,VT=\left(a-b\right)^2=a^2-2ab+b^2\\ =\left(a^2+2ab+b^2\right)-4ab\\ =\left(a+b\right)^2-4ab=VP\\ b,VT=\left(x+y\right)^2+\left(x-y\right)^2\\ =x^2+2xy+y^2+x^2-2xy+y^2\\ =2x^2+2y^2\\ =2\left(x^2+y^2\right)=VP\\ c,VT=\left(x+y\right)^2-\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]\\ =\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y.2x=4xy=VP\\ d,VT=\left(x-y\right)^2+\left(x+y\right)^2+2\left(x^2-y^2\right)\\ =\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\\ =\left[\left(x-y\right)+\left(x+y\right)\right]^2\\ =\left(x-y+x+y\right)^2\\ =\left(2x\right)^2=4x^2=VP\)

Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}ab\cdot sinC=\dfrac{1}{2}\cdot7\cdot23\cdot sin130^o=61,7\) (đvdt)