lỗi

2. 

Gọi quãng đường AB là x(km) ( x>0 )

Thời gian đi là \(\dfrac{x}{20}\)

Thời gian về là \(\dfrac{x}{15}\)

Theo đề bài, ta có:

\(\dfrac{x}{15}-\dfrac{x}{20}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{4x-3x}{60}=\dfrac{10}{60}\)

\(\Leftrightarrow x=10\left(tm\right)\)

Vậy quãng đường AB dài 10km

Bài 1:

\(a,\)

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

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

\(=8x^2-8x+2-x^2-2x-1+3\)

\(=7x^2-10x+4\)

\(=7\left(x^2-\dfrac{10}{7}x+\dfrac{4}{7}\right)\)

\(=7\left[x^2-2.x.\dfrac{5}{7}+\left(\dfrac{5}{7}\right)^2-\left(\dfrac{5}{7}\right)^2+\dfrac{4}{7}\right]\)

\(=7\left[\left(x-\dfrac{5}{7}\right)^2+\dfrac{3}{49}\right]\)

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

Vì \(7\left(x-\dfrac{5}{7}\right)^2\ge0\forall x\)

\(\Rightarrow7\left(x-\dfrac{5}{7}\right)^2+\dfrac{3}{7}\ge\dfrac{3}{7}\forall x\)

\(\Rightarrow P_{min}=\dfrac{3}{7}\Leftrightarrow7\left(x-\dfrac{5}{7}\right)^2=0\Leftrightarrow x=\dfrac{5}{7}\)

\(b,\)

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

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

\(=3x^2+12x+12-x^2+4\)

\(=2x^2+12x+16\)

\(=2\left(x^2+6x+8\right)\)

\(=2\left(x^2+6x+9-1\right)\)

\(=2\left[\left(x+3\right)^2-1\right]\)

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

Vì \(2\left(x+3\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x+3\right)^2-2\ge-2\forall x\)

\(\Rightarrow Q_{min}=-2\Leftrightarrow2\left(x+3\right)^2=0\Leftrightarrow x=-3\)

\(c,\)

Tương tự: \(M=\left(x+1\right)^3-\left(x-2\right)^3-5\)

\(=9x^2-9x+4\)

\(=9\left(x^2-x+\dfrac{4}{9}\right)\)

\(=9\left[x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+\dfrac{4}{9}\right]\)

\(=9\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{36}\right]\)

\(=9\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)

Ta có: \(9\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

\(\Rightarrow M_{min}=\dfrac{7}{4}\Leftrightarrow9\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)

Bài 2:

\(a,\)

\(M=-3x^2-4x+1\)

\(=-3\left(x^2+\dfrac{4}{3}x-\dfrac{1}{3}\right)\)

\(=-3\left[x^2+2.x.\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2-\left(\dfrac{2}{3}\right)^2-\dfrac{1}{3}\right]\)

\(=-3\left[\left(x+\dfrac{2}{3}\right)^2-\dfrac{7}{9}\right]\)

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

Ta có: \(-3\left(x+\dfrac{2}{3}\right)^2\le\dfrac{7}{3}\forall x\)

\(\Rightarrow M_{max}=\dfrac{7}{3}\Leftrightarrow-3\left(x+\dfrac{2}{3}\right)^2=0\Leftrightarrow x=-\dfrac{2}{3}\)

\(b,\)

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

\(=-2x^2+10x+4\)

\(=-2\left(x^2-5x-2\right)\)

\(=-2\left[\left(x-\dfrac{5}{2}\right)^2-\dfrac{33}{4}\right]\)

\(=-2\left(x-\dfrac{5}{2}\right)^2+\dfrac{33}{2}\)

Tương tự, \(Q_{max}=\dfrac{33}{2}\Leftrightarrow x=\dfrac{5}{2}\)

\(c,\)

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

\(=-x^2-11x-20\)

\(=-\left(x^2+11x+20\right)\)

\(=-\left(x+\dfrac{11}{2}\right)^2+\dfrac{41}{4}\)

Tương tự, \(Q_{max}=\dfrac{41}{4}\Leftrightarrow x=-\dfrac{11}{2}\)

2:

a: =-3(x^2+4/3x-1/3)

=-3*(x^2+2*x*2/3+4/9-7/9)

=-3(x+2/3)^2+7/3<=7/3

Dấu = xảy ra khi x=-2/3

b: =x^2+4x+4-3x^2+6x-3+3

=-2x^2+10x+4

=-2(x^2-5x-2)

=-2(x^2-5x+25/4-33/4)

=-2(x-5/2)^2+33/2<=33/2

Dấu = xảy ra khi x=5/2

c: =x^2+x-2-2x^2-12x-18

=-x^2-11x-20

=-(x^2+11x+20)

=-(x^2+11x+121/4-41/4)

=-(x+11/2)^2+41/4<=41/4

Dấu = xảy ra khi x=-11/2

2:

a: \(A⋮B\)

=>\(3x^4-12x^3+5x^3-20x^2+31x^2-124x+125x-500+500-a⋮x-4\)

=>500-a=0

=>a=500

b: A chia hết cho B

=>x^4-x^3+5x^2+x^2-x+5+a-5 chia hết cho x^2-x+5

=>a-5=0

=>a=5

1:

a: A chia hết cho B

=>2n-n-3>=0 và 6-n-n>=0

=>n>=3 và n<=3

=>n=3

b: C chia hết cho D

=>9-n>=0 và n+2-9>=0 và n+1-n>=0 và n-9>=0

=>n=9

a: =x^2-2xy+y^2+y^2+4y+4-2

=(x-y)^2+(y+2)^2-2>=-2

Dấu = xảy ra khi x=y=-2

c: =(x+y)^2-4(x+y)+4+1

=(x+y-2)^2+1>=1

Dấu = xảy ra khi x+y=2

b: =x^2-3x+9/4+4y^2-4y+1+23/4

=(x-3/2)^2+(2y-1)^2+23/4>=23/4

Dấu = xảy ra khi x=3/2 và y=1/2

Help cái j

Ảnh lỗi

Bài 10:

a: Xét tứ giác AEMF có 

AE//MF

AF//ME

Do đó: AEMF là hình bình hành

Bài 1: 

\(VT=1\cdot\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\cdot\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)

\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)

\(=\left(a^4-b^4\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)

\(=a^{32}-b^{32}\)