\(a,x=-2y\)

Thay vào M ta được \(M=\frac{-6y-4y}{3y+8y}=-\frac{10y}{11y}=-\frac{10}{11}\)

b,\(\frac{x}{3}=\frac{y}{12}=k\Rightarrow x=3k;y=12k\)

Thay vào M ta được \(M=\frac{9k-24k}{36k-12k}=\frac{-15k}{24k}=-\frac{15}{24}\)

c,\(3x+y=0\Rightarrow y=-3x\)

Thay vào M ta được \(M=\frac{3x+12x}{-9x-4x}=\frac{15x}{-13x}=-\frac{15}{13}\)

C = \(x^3\) + \(x^2\).y - 2\(x^2\) - \(xy\) - y\(^2\) + 3y + \(x\) - 1

C = (\(x^3\) + \(x^2\).y - 2\(x^2\)) - (\(xy\) + y\(^2\) - 2y) + (y + \(x\) - 2) + 1

C = \(x^2\).(\(x\) + y - 2) - y(\(x\) + y - 2) + (y + \(x\) - 2) + 1 (1)

Thay \(x+y-2\) vào biểu thức (1) ta có:

C = \(x^2\). 0 - y . 0 + 0 + 1

C = 0 - 0 + 0 + 1

C = 1

D = \(x\).(\(x^3\) - y)(\(x^3\) - 2y\(^2\))(\(x^3\) - 3y\(^2\))(\(x^3\) - 4y\(^4\)) (1)

Thay \(x\) = 2 và y = - 2 vào biểu thức (1) ta có:

D = 2.(2\(^3\)+2).(2\(^3\)- 2.(-2\()^2\)).(2\(^3\)-3(-2)\(^2\))(2\(^3\)- 4.(-2)\(^4\))

D = 2.(2\(^3\)+ 2).(8 - 8).(2\(^3\)- 3(-2)\(^2\))(2\(^3\)- 4.(-2)\(^4\))

D = 2.(2\(^3\)+ 2).0.(2\(^3\)- 3(-2)\(^2\))(2\(^3\)- 4.(-2)\(^4\))

D = 0

Thay x=1 và y= -1 vào là được.

\(C=\frac{x+5}{2x+\left(x-2y\right)}+\frac{y-5}{2y-\left(x-2y\right)}\)

\(=\frac{x+5}{2x+10}+\frac{y-5}{2y-10}=\frac{x+5}{2\left(x+5\right)}+\frac{y-5}{2\left(y-5\right)}=\frac{1}{2}+\frac{1}{2}=1\left(x\ne-5,y\ne5\right)\)

Trả lời :

Ta có :

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

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

C = \(\frac{2x+10}{6x-4y}+\frac{2y-10}{8y-2x}\)

Thay x - 2y = 10 . Ta được :

C = \(\frac{2x+x-2y}{6x-4y}+\frac{2y-x-2y}{8y-2x}\)

C = \(\frac{x\left(2+1\right)-2y}{6x-4y}+\frac{y\left(2+2\right)-x}{8y-2x}\)

C = \(\frac{3x-2y}{6x-4y}+\frac{4y-x}{8y-2x}\)

C = \(\frac{1}{2}+\frac{1}{2}\)

C = \(1\)

Vậy C = 1

Hok tốt

BÀI 1:

A) THAY X =2 ; Y= 1/3 VÀO BIỂU THỨC

2. ( 2+ 2) + 1/3. ( 1/3 + 3) + 2^2 + 2. (1/3) ^2 + 4. 1/3

= 2.4 + 1/3 . 10/3+ 4+ 2. 1/9 + 4/3

= 8+ 10/9 + 4+ 2/9 + 4/3

= 44/3

B) THAY X= 2; Y= 1/3 VÀO BIỂU THỨC

( 2 + 1/3 ) . 2 + ( 2- 1/3 ) . 2

= 7/3 . 2 + 5/3 . 2

= 7/6 + 10/3

= 9/2

C) TA CÓ: ( X+ Y) X + ( X+Y ) X= 2. X.( X+Y)

THAY X= 2; Y= 1/3 VÀO BIỂU THỨC

2. 2.( 2+ 1/3)

= 4. 7/3

= 28/3

BÀI 2:

A) \(\left(x^2+6x+5\right)+\left(-3x+9\right)=x^2+6x+5-3x+9\)

\(=\left(6x-3x\right)+\left(5+9\right)+x^2\)

\(=3x+14+x^2\)

B) \(\left(2x^2+3x+7\right)-\left(-2x+5\right)=2x^2+3x+7+2x-5\)

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

\(=2x^2+5x+2\)

C) \(\left(6x^2y-6xy^2\right)+\left(7xy+4xy^2-x^2y\right)=6x^2y-6xy^2+7xy+4xy^2-x^2y\)

\(=\left(6x^2y-x^2y\right)+\left(4xy^2-6xy^2\right)+7xy\)

\(=5x^2y+\left(-2xy^2\right)+7xy\)

CHÚC BN HỌC TỐT!!!!
 

Ta có : \(C=x^3+y^3-x^2y-xy^2+7x^2-7y^2+2016\)

\(=\left(x^3-x^2y+7x^2\right)-\left(xy^2-y^3+7y^3\right)+2016\)

\(=x^2\left(x-y+7\right)-y^2\left(x-y+7\right)+2016\)

\(=x^2\cdot0-y^2\cdot0+2016=2016\)

( Do \(x-y+7=0\))

Vậy : \(C=2016\)

\(C=\left(x^3-x^2y+7x^2\right)-\left(xy^2-y^3+7y^2\right)+2016=x^2\left(x-y+7\right)-y^2\left(x-y+7\right)+2016=0+0+2016=2016\)

Vậy C=2016