a: Ta có: \(x^2-4-\left(x+2\right)^2\)

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

=-4x-8

b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)

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

=2x-1

c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)

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

\(=-3x+6\)

d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

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

=4

e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)

\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)

\(=29a^2-45a-3-36a^2+24a-4\)

\(=-7a^2-21a-7\)

g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)

\(=25y^2-9-25y^2+40y-16\)

=40y-25

h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)

\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)

\(=35x^3+15x^2+15x\)

i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=16x^2\)

a: \(\left\{{}\begin{matrix}x+2y=3\\4x+5y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+8y=12\\4x+5y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3y=6\\x+2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3-2y=3-2\cdot2=-1\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}x+y=5\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y+2x-y=5+4\\x+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=9\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=5-3=2\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}x+2y=5\\x-5y=-9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+2y-x+5y=5+9=14\\x+2y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7y=14\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=5-2y=1\end{matrix}\right.\)

\(a,\left(6x+5y\right)\left(6x-5y\right)\)

\(=\left(6x\right)^2-\left(5y\right)^2\)

\(=36x^2-25x^2\)

\(b,\left(-4xy-5\right)\left(5-4xy\right)\)

\(=-\left(5+4xy\right)\left(5-4xy\right)\)

\(=-[5^2-\left(4xy\right)^2]\)

\(=-\left(25-16xy^2\right)\)

\(c,\left(3x-4\right)^2+2.\left(3x-4\right).\left(4-x\right)+\left(4-x\right)^2\)

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

\(=\left(3x-4\right)\left(3x-2\right)\left(4-x\right)\left(5-x\right)\)

không thấy gì ngoài vạch đen

??

Đề đâu bn

Ủa gì z?:))

khó quá Lê Ngọc Diệp

bn có máy tính thì vào eqn rùi vào un.. j đó nhấn 2 rùi giải thui ak, mk mới lớp 6 nhưng mk bít cách giải hệ phương trình 2 ẩn rùi

Hệ gì lạ thế >? bạn viết rõ ra mình mới giúp được 

`a)5y(2y-1)-(3y+2)(3-3y)`

`=10y^2-5y+(3y+2)(3y-3)`

`=10y^2-5y+9y^2-9y+6y-6`

`=19y^2-8y-6`

`b)(6x+1)^2-2(6x+1)(6x-1)+(6x-1)^2`

`=(6x+1-6x+1)^2`

`=2^2=4`

`c)(2x+3)^2-2(2x+3)(x-20+(x-2)^2`

`=(2x+3-x+2)^2`

`=(x+5)^2`

`=x^2+10x+25`

Có H = x² + 5y² + 4xy - 6x + 5y - 9

         = [(x² + 4xy + 4y²) - 6x - 12y + 9] + (y² + 17y + \(\frac{289}{4}\)) - \(\frac{361}{4}\)

         = [(x + 2y)² - 2(x + 2y).3 + 3²] + (y² + 2.y.\(\frac{17}{2}\)+ \(\left(\frac{17}{2}\right)^2\)) - \(\frac{361}{4}\)

         = (x + 2y - 3)² + \(\left(y+\frac{17}{2}\right)^2\) - \(\frac{361}{4}\)

Thấy (x + 2y - 3)² ≥ 0 với mọi x; y

         \(\left(y+\frac{17}{2}\right)^2\ge0\) với mọi y

=> (x + 2y - 3)² + \(\left(y+\frac{17}{2}\right)^2\) ≥ 0 với mọi x; y

=> (x + 2y - 3)² + \(\left(y+\frac{17}{2}\right)^2\) - \(\frac{361}{4}\) ≥ \(\frac{-361}{4}\) với mọi x; y

=> H ≥ \(\frac{-361}{4}\) với mọi x; y

Dấu "=" xảy ra khi ...

Bn tự giải tiếp.

P/s: ko chắc đúng