a: \(=8x^3+y^3-8x^3+y^3=2y^3\)

b: \(=25-x^4\)

c: \(=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)

d: \(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^2\)

\(=6a^2b+2b^3-2b^2\)

e: \(=\left(x-1\right)^3\)

Trả lời:

a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)

d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)

e, \(E=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)

a,<=> 3x+1/4-2x-3/5=1

<=> x-7/20=1

<=> x= 27/20

a, \(\left(3x+\frac{1}{4}\right)-\frac{1}{3}\left(6x+\frac{9}{5}\right)=1\)

\(3x+\frac{1}{4}-\frac{6}{3}x-\frac{3}{5}=1\)

\(x-\frac{7}{20}=1\Leftrightarrow x=\frac{27}{20}\)

b,ĐKXĐ : x \(\ne\)-1/2 ; 1/2 

 \(\left(\frac{5}{2x+1}\right)-\left(\frac{2x}{1-2x}\right)=1-\left(\frac{6-4x}{4x^2-1}\right)\)

\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{6-4x}{4x^2-1}\)

\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{2\left(3-2x\right)}{\left(2x+1\right)\left(2x-1\right)}\)

\(\frac{5\left(1-2x\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2x\left(2x+1\right)^2\left(2x-1\right)}{\left(1-2x\right)\left(2x+1\right)^2\left(2x-1\right)}=\frac{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2\left(3-2x\right)\left(2x+1\right)\left(1-2x\right)}{\left(2x+1\right)\left(2x-1\right)^2\left(2x-1\right)\left(1-2x\right)}\)

\(22x-5-20x^2-8x^3=18x-7-8x^3-4x^2\)

lm nốt nha,bị troll rồi ko vt đc nữa.

a)(2x²+1)(3x³-2x²+3

= 6x⁵-4x⁴+6x²+3x³-2x²+3

= 6x⁵-4x⁴+3x³+4x²+3

b)(-3x+1)(4x⁴-x^³+x)

= -12x⁵+3x⁴-3x²+4x⁴-x^³+x

= -12x⁵+7x⁴-x³-3x²+x

\(=3x^2\left(x^2+3x+1\right)\)

=3x^4+9x^3+3x^2

\(a,\left(2x^2+1\right)\left(3x^3-2x^2+3\right)\)

\(=6x^5-4x^4+6x^2+3x^3-2x^2+3\)

\(=6x^5-4x^4+4x^2+3x^3+3\)

\(b,\left(-3x+1\right)\left(4x^4-x^3+x\right)\)

\(=-12x^5+3x^4-3x^2+4x^4-x^3+x\)

\(=-12x^5+7x^4-3x^2-x^3+x\)

3.x=2.y , 4.x=2.z  và x² + 

Làm giống như bạn Should A Person

\(a)\) Ta có : 

\(3x=2y\)\(\Leftrightarrow\)\(\frac{x}{2}=\frac{y}{3}\)

\(4x=2z\)\(\Leftrightarrow\)\(\frac{x}{2}=\frac{z}{4}\)

\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)\(\Leftrightarrow\)\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2+y^2+2z^2}{4+9+32}=\frac{108}{45}=\frac{12}{5}\)

Do đó : 

\(x^2=\frac{12}{5}.4=\frac{48}{5}\)\(\Rightarrow\)\(x=\sqrt{\frac{48}{5}}\) hoặc \(-\sqrt{\frac{48}{5}}\)

\(y^2=\frac{12}{5}.9=\frac{108}{5}\)\(\Rightarrow\)\(y=\sqrt{\frac{108}{5}}\) hoặc \(y=-\sqrt{\frac{108}{5}}\)

\(2z^2=\frac{12}{5}.32=\frac{384}{5}\)\(\Rightarrow\)\(z=\sqrt{\frac{192}{5}}\) hoặc \(z=-\sqrt{\frac{192}{5}}\)

Vì \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) nên x, y, z cùng dấu 

Vậy \(x=\sqrt{\frac{48}{5}}\)\(;\)\(y=\sqrt{\frac{108}{5}}\)\(;\)\(z=\sqrt{\frac{192}{5}}\) hoặc \(x=-\sqrt{\frac{48}{5}}\)\(;\)\(y=-\sqrt{\frac{108}{5}}\)\(;\)\(z=-\sqrt{\frac{192}{5}}\)

Chúc bạn học tốt ~ 

a/ => (6x - 3)(3x - 1) - (2x - 3)(9x - 1) = 0

=> 18x² - 15x + 3 -  18x² + 29x - 3 = 0

=> 14x = 0 => x = 0

b/ ghi dấu rõ vào tớ mới giải đc