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

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

= \(x^3+2^3\)-\(x^3+2\)

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

=10

b) B= (a+2)(a-2)(\(a^2+2a+4\))(\(a^2-2a+4\))

= \(a^2-2^2\)+\(a^2+\left(2a\right)^2+4^2\)

=\(a^2-4+a^2+4a^2+16\)

=(\(a^2+a^2+4a^2\))+(-4+16)

=\(6a^2\)+12

a, ĐKXĐ : x + 3 khác 0 => x khác - 3

b, x^2-9/x+3 = 5

=> x^2 - 9 = 5(x + 3)

=> x^2 - 9 = 5x + 15

=> x^2 - 5x - 9 - 15 = 0

=> x^2 - 5x - 24 = 0

=> x^2 + 3x - 8x - 24 = 0

=> x(x + 3) - 8(x + 3) = 0

=> (x - 8)(x + 3) = 0

=> x = 8 hoặc x = -3

c, x^2-9/x+3 = -6

=> x^2 - 9 = -6(x+3)

=> x^2 - 9 = -6x - 18

=> x^2 + 6x - 9 + 18 = 0

=> x^2 + 6x + 9 = 0

=> (x + 3)^2 = 0

=> x + 3 = 0

=> x = -3  (ktm)

vậy không có....

Đặt \(A=\frac{x^2-9}{x+3}\)

a) A xác định khi \(x+3\ne0\Leftrightarrow x\ne-3\)

b) A=\(\frac{x^2-9}{x+3}=\frac{\left(x-3\right)\left(x+3\right)}{x+3}=x-3\)

Để A=5 => x-3=5 => x=8 (TMĐK)
c) Có A=x-3 \(\left(x\ne-3\right)\)

\(\Rightarrow x+3=-6\)

\(\Rightarrow x=-9\)(TMĐK)
Vậy có gt của x để A nhận giá trị bằng -6

Answer:

Câu 1:

\(\left(5x-x-\frac{1}{2}\right)2x\)

\(=\left(4x-\frac{1}{2}\right)2x\)

\(=4x.2x-\frac{1}{2}.2x\)

\(=8x^2-x\)

\(\left(x^3+4x^2+3x+12\right)\left(x+4\right)\)

\(=x\left(x^3+4x^2+3x+12\right)+4\left(x^3+4x^2+3x+12\right)\)

\(=x^4+4x^3+3x^2+12x+4x^3+16x^2+12x+48\)

\(=x^4+\left(4x^3+4x^3\right)+\left(3x^2+16x^2\right)+\left(12x+12x\right)+48\)

\(=x^4+8x^3+19x^2+24x+48\)

Ta thay \(x=99\) vào phân thức \(\frac{x^2+1}{x-1}\): \(\frac{\left(99\right)^2+1}{99-1}=\frac{9802}{98}=\frac{4901}{49}\)

Ta thay \(x=4\) vào phân thức \(\frac{x^2-x}{2\left(x-1\right)}\) : \(\frac{4^2-4}{2.\left(4-1\right)}=\frac{12}{6}=2\)

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

\(= (x²+2xy+y²)-(x²-2xy+y²)\)

\(= x²+2xy+y²-x²+2xy-y²\)

\(= 4xy\)

\(4x^2+4x+1=\left(2x+1\right)^2=\left(2.2+1\right)^2=25\)

Câu 2:

\(x^2+x=0\)

\(\Rightarrow x\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

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

\(\Rightarrow x^2.\left(x-1\right)+4\left(1-x\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^2-4\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\)

Trường hợp 1: \(x-1=0\Rightarrow x=1\)

Trường hợp 2: \(x-2=0\Rightarrow x=2\)

Trường hợp 3: \(x+2=0\Rightarrow x=-2\)

Câu 3: Bạn xem lại đề bài nhé.