1D

2C

1.

Phương trình đường thẳng có dạng:

\(2\left(x-2\right)-1\left(y-1\right)=0\Leftrightarrow2x-y-3=0\)

2.

Do d song song \(\Delta\) nên nhận \(\left(2;-3\right)\) là 1 vtpt

Phương trình: \(2\left(x-1\right)-3\left(y-1\right)=0\Leftrightarrow2x-3y+1=0\)

3.

Do đường thẳng vuông góc d nên nhận \(\left(3;4\right)\) là 1vtpt

\(3\left(x-2\right)+4\left(y-3\right)=0\Leftrightarrow3x+4y-18=0\)

a: \(=\left(-\dfrac{6}{2}\right)\cdot\dfrac{x^3}{x}\cdot\dfrac{y^2}{y^2}=-3x^2\)

b: \(=\left(-\dfrac{1}{4}:\dfrac{1}{2}\right)\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^3}{y^2}=-\dfrac{1}{2}xy\)

c: \(=\dfrac{8}{4}\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^5}{y^4}=2xy\)

\(a,-6x^3y^2:2xy^2=-3x^2\)

\(b,-\dfrac{1}{4}x^4y^3:\dfrac{1}{2}x^3y^2=-\dfrac{1}{2}xy\)

\(c,8x^4y^5:4x^3y^4=2xy\)

#Urushi

Bài 1:

a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Rightarrow2\left[4\left(x-2\right)-\left(3x-4\right)\right]=2\)

\(\Rightarrow4\left(x-2\right)-3x+4=0\)

\(\Rightarrow4x-8-3x+4=0\)

\(\Rightarrow x-4=0\)

\(\Rightarrow x=4\)

b) \(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Rightarrow5\left[2\left(3x-2\right)+11-4x\right]-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(6x-4+11-4x\right)-3\left(5x+2\right)=25\)

\(\Rightarrow5\left(2x+7\right)-3\left(5x+2\right)=25\)

\(\Rightarrow10x+35-15x-6=25\)

\(\Rightarrow-5x+29=25\)

\(\Rightarrow-5x=25-29\)

\(\Rightarrow-5x=-4\)

\(\Rightarrow x=\dfrac{4}{5}\)

c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Rightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Rightarrow x+4=0\)

\(\Rightarrow x=-4\)

d) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Rightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Rightarrow-x-21=0\)

\(\Rightarrow x=-21\)

Bài 2:

a) \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(P=8x^2y-6y^2-9x^2y+12y^2\)

\(P=-x^2y+6y^2\)

Thay x = -1 ; y = 2 vào P ta được

\(P=-\left(-1\right)^2.2+6.2^2\)

\(P=-2+24=22\)

b) \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(Q=20x^3-12x^2y-4x^3-x^2y\)

\(Q=16x^3-13x^2y\)

Thay x = -1 ; y = 2 vào Q ta được

\(Q=16\left(-1\right)^3-13\left(-1\right)^2.2\)

\(Q=-16-26\)

\(Q=-42\)

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

\(H=x^4-xy+x^2y-x^4-x^2y+3xy\)

\(H=2xy\)

Thay x = 1/4 ; y = 2012 vào H ta được

\(H=2.\dfrac{1}{4}.2012\)

\(H=1006\)

1.a)\(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Leftrightarrow8x-16-6x+8=2\)

\(\Leftrightarrow2x-8=2\Leftrightarrow2x=10\Leftrightarrow x=5\)

b)\(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)

\(\Leftrightarrow30x-20-15x-6+55-20x=25\)

\(\Leftrightarrow-5x+29=25\Leftrightarrow-5x=-4\Leftrightarrow x=\dfrac{4}{5}=0,8\)

\(c)2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

\(d)4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Leftrightarrow-x-21=0\Leftrightarrow-x=21\Leftrightarrow x=-21\)

2.

a)\(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(\Leftrightarrow8x^2y-6y^2-9x^2y-12y^2\)

\(\Leftrightarrow x^2y-18y^2\)

tại x=-1 , y=2

ta có:\(x^2y-18y^2=\left(-1\right)^2.2-18.2^2=2-72=-70\)

vậy \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y=-70\) tại x=-1,y=2

b)\(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)

\(\Leftrightarrow20x^3-12x^2y-4x^3-x^2y\)

\(\Leftrightarrow17x^3-13x^2y\)

tại x=-1,y=2

ta có:\(17x^3-13x^2y=17\left(-1\right)^3-13\left(-1\right)^2.2=-17-26=-43\)

vậy \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)=-43\)

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

\(\Leftrightarrow x^4-xy+x^2y-x^3-x^2y+3xy\)

\(\Leftrightarrow x^4+2xy-x^3\)

tại x=1/4 và y=2012

ta có:\(x^4+2xy-x^3=\left(\dfrac{1}{4}\right)^4+2.\dfrac{1}{4}.2012-\left(\dfrac{1}{4}\right)^3\approx1006\)

a. 8(x-2)-2(3x-4)=2

8x-16-6x-8 =2

(8x-6x)-(16-8)=2

2x-2 =2

2x =2+2

2x =4

x =\(\dfrac{4}{2}\)=2

\(\frac{4x-3}{3}=\frac{3y+1}{7}=\frac{4x+3y-2}{5y}\)

\(=\frac{4x-3+3y+1-\left(4x+3y-2\right)}{3+7-5y}\)

\(=\frac{4x-3+3y+1-4x-3y+2}{10-5y}\)

\(=\frac{\left(4x-4x\right)+3y-3y-3+1+2}{10-5y}=0\)

\(\Rightarrow\hept{\begin{cases}4x-3=0\Leftrightarrow x=\frac{3}{4}\\3y+1=0\Leftrightarrow y=-\frac{1}{3}\end{cases}}\)

Vậy \(x=\frac{3}{4};y=-\frac{1}{3}\).

Câu trả lời đúng là :

x = 3/4

y = -1/3

Đáp số : ...

Bài 1:

           \(A=x^2-6x+13=\left(x-3\right)^2+4\ge4\)

Vậy  \(Min\)\(A=4\)\(\Leftrightarrow\)\(x=3\)

        \(B=2x^2+8x=2\left(x^2+4x+4\right)-8=2\left(x+2\right)^2-8\ge-8\)

Vậy  \(Min\)\(B=-8\)\(\Leftrightarrow\)\(x=-2\)

        \(C=4x^2+20x=\left(2x+5\right)^2-25\ge-25\)

Vậy  \(Min\)\(C=-25\)\(\Leftrightarrow\)\(x=-\frac{5}{2}\)

Bài 3:

a)   \(x^2+12x+39=\left(x+6\right)^2+3>0\) 

b)   \(4x^2+4x+3=\left(2x+1\right)^2+2>0\)