Câu 4: 

\(=\dfrac{a\left(a-b\right)-c\left(a-b\right)}{a\left(a+b\right)-c\left(a+b\right)}=\dfrac{a-b}{a+b}\)

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

cmm

Bài 2:

a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)

\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)

\(=2x^3+6x\)

b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

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

\(=27x-55\)

Ta có:

$(x+4)^2-x^2(x+12)=16$

$x^2+8x+16-x^3-12x^2=16$

$-x^3-11x^2+8x=0$

$-x(x^2+11x-8)=0$

Suy ra:

$x=0$ hoặc $x^2+11x-8=0$

Giải phương trình bậc hai:

$x=\dfrac{-11\pm\sqrt{121+32}}{2}$

$=\dfrac{-11\pm\sqrt{153}}{2}$

$=\dfrac{-11\pm3\sqrt{17}}{2}$

Vậy: $x=0,\quad x=\dfrac{-11+3\sqrt{17}}2,\quad x=\dfrac{-11-3\sqrt{17}}2$

a. x(2x2 – 3) – x2(5x + 1) + x2

      = 2x3 – 3x – 5x3 – x2 + x2 = -3x – 3x3

b. 3x(x – 2) – 5x(1 – x) – 8(x2 – 3)

      = 3x2 – 6x – 5x + 5x2 – 8x2 + 24

      = - 11x + 24

c. 1/2 x2(6x – 3) – x( x2 + 1/2 (x + 4)

      = 3x3 - 3/2 x2 – x3 - 1/2 x + 1/2 x + 2

      = 2x3 - 3/2 x2 + 2

a, x(2x2-3)-x2(5x+1)x2

=2x3-3x-5x3- x2+x2=-3x-3x3

học tốt nhé!!

a: \(A=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6\left(x+1\right)\cdot x}{3\left(x+1\right)\left(x-2\right)}=\dfrac{2x}{x-2}\)

\(a,x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

$(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$

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

$=[(6x+1)-(6x-1)]^2$

$=2^2$

$=4$

$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

$=3\cdot\dfrac{2^4-1}{2^2-1}\cdot\dfrac{2^8-1}{2^4-1}\cdot\dfrac{2^{16}-1}{2^8-1}\cdot\dfrac{2^{32}-1}{2^{16}-1}$

$=3\cdot\dfrac{2^{32}-1}{2^2-1}$

$=3\cdot\dfrac{2^{32}-1}{3}$

$=2^{32}-1$

$x(2x^2-3)-x^2(5x+1)+x^2$

$=2x^3-3x-5x^3-x^2+x^2$

$=2x^3-5x^3-3x$

$=-3x^3-3x$

$=-3x(x^2+1)$

$3x(x-2)-5x(1-x)-8(x^2-3)$

$=3x^2-6x-5x+5x^2-8x^2+24$

$=3x^2+5x^2-8x^2-11x+24$

$=-11x+24$

$=24-11x$