\(4.\left(3x+y\right)^2+\left(x+y\right)^2\) 

\(=3x^2+6xy+y^2+x^2-2xy+y^2\) 

\(=9x^2+6xy+y^2+x^2-2xy+y^2\)

\(=10x^2-4xy+2y^2\) 

\(7.\left(x-4\right)^2+\left(x+4y\right)\) 

\(=x^2-8x+16+x+4y\) 

\(=x^2-7x+16+4y\) 

\(10.\left(2x+7\right)^2+\left(-2x-3\right)^2\) 

\(=4x^2+28x+49+4x^2+12x+9\) 

\(=8x^2+40x+58\)

\(12.-\left(x+1\right)^2-\left(x-1\right)^2\) 

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

\(=-x^2-2x-1+x^2+2x-1\)  

\(=4x\) 

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

\(=-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\) 

\(=-x^2-10-25+x^2+6x-9\) 

\(=-16x-16\) 

\(8.-\left(-2x+3\right)^2-\left(5x-3\right)^2\) 

\(=4x^2+12x+9-25x^2+30x-9\) 

\(=-21x^2+42x\)

\(11.-\left(2x-y\right)^2-\left(x+3y\right)^2\) 

\(=-4x^2+4xy-y^2-\left(x^2+6xy+9y^2\right)\) 

\(=-4x^2+4xy-y^2-x^2-6xy-9y^2\) 

\(=-5x^2-2xy-10y^2\)

4: =9x^2+6xy+y^2+x^2-2xy+y^2

=10x^2+4xy+2y^2

5: =-x^2-10x-25-x^2+6x-9

=-4x-34

7; \(=x^2-8xy+16y^2+x+4y\)

10: \(=4x^2+28x+49+4x^2+12x+9\)

=8x^2+40x+58

11: =-4x^2+4xy-y^2-x^2-6xy-9y^2

=-5x^2-2xy-10y^2

Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)

(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)

b.

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)

c.

\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)

d.

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)

a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)

\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)

\(A=9x\)

Thay x = 15 vào, ta có: 

\(A=9.15=135\)

b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(B=5x^2-20xy-4y^2+20xy\)

\(B=5x^2-4y\)

Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có: 

\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)

c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)

\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(C=9x^2y^2-xy^3-8x^3\)

Thay \(x=\frac{1}{2};y=2\) vào, ta có:

\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)

d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)

\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)

\(D=18x^2+12x-7\)

Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

+) Với x = -2

\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)

+) Với x = 2

\(D=18.2^2+12.2-7=89\)

mấy bài này em chưa hc,xin lỗi,em ms hc lớp 6 thui

a) (x+2)²+x(x-4)

   =x²+4x+4+x²-4x

   =2x²+4

b)(x-3)²-(x+3)(x-4)

  =x²-6x+9-x²+4x-3x+12

  =-5x+12

c) (3x+1)²+3x(2-4x)

   =9x²+6x+1+6x-12x²

   =-3x²+12x+1

d) (2x-4y)²-(2x-3)(2x-3y)

  =4x²-16xy+16y²-4x²+6xy+6x-9y

  =16y²-10xy+6x-9y

2:

a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)

\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)

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

\(=\left(x-2y\right)\left(x+2y\right)-\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+2y-1\right)\)

c: \(=\left(y^2+10y+25\right)-9z^2\)

\(=\left(y+5\right)^2-\left(3z\right)^2\)

\(=\left(y+5+3z\right)\left(y+5-3z\right)\)

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

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

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

1:

a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)

b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)

\(=2y\left(5y-6\right)+4\left(5y-6\right)\)

\(=2\left(5y-6\right)\left(y+2\right)\)

c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)

\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)

\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)

d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)

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

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

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

Bài 1

a) x(3 - 4x) + 5(3 - 4x)

= (3 - 4x)(x + 5)

b) 2y(5y - 6) - 4(6- 5y)

= 2y(5y - 6) + 4(5y - 6)

= (5y - 6)(2y + 4)

= 2(5y - 6)(y + 2)

c) 27(x - 2)³ - 3x(2 - x)²

= 27(x - 2)³ - 3x(x - 2)²

= 3(x - 2)²[9(x - 2) - x]

= 3(x - 2)²(9x - 18 - x)

= 3(x - 2)²(8x - 18)

= 6(x - 2)²(4x - 9)

d) 6y(x² - y²) - 8y(x + y)²

= 6y(x - y)(x + y) - 8y(x + y)²

= 2y(x + y)[3(x - y) - 4(x + y)]

= 2y(x + y)(3x - 3y - 4x - 4y)

= 2y(x + y)(-x - 7y)

= -2y(x + y)(x + 7y)

a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)

Thay x = 15 vào bt A ta có

A = 9 . 15 = 135

b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)

Thay x = -1/5 ; y = - 1/2 vào bt B ta có

\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)

\(=9x^2y^2-xy^3-8x^3\)

Thay x = 1/2 ; y = 2 vào bt C ta có

\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)

d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)

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

\(\left|x\right|=2\Rightarrow x=\pm2\)

Thay x = 2 vào bt D có

\(D=12.4+12.2-3=69\)

Thay x = - 2 vào bt D ta có

\(D=12.4-12.2-3=21\)

Bài 1:

\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)

Bài 2:

\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)