4.

a, (x+y)(x²+y²)=3.5

x³+x²y+y³+xy²=15

x³+y³+xy(x+y)=15

x³+y³+3xy=15  (1)

x+y=3 suy ra x²+2xy+y²=9 mà x²+y²=5 suy ra 5+2xy=9 suy ra 2xy=4 suy ra xy=2 (2)

thay (1) vào (2) ta có:

x³+y³+3.2=15

suy ra x³+y³=9

b, bạn làm tương tự như câu a

a, 

vì a+b=1 suy ra (a+b)³=1 suy ra a³+3a²b+3b²a+b³=1 suy ra  a³+3ba(a+b)+b³=1 suy ra  a³+3ab+b³=1

b,

a+b=1 suy ra a²+2ab+b²=1 suy ra 3a²+3b²-2a²+2ab-b²=1 suy ra 3(a²+b²)-2(a²+ab+b²)=1 suy ra  3(a²+b²)-2(a²+ab+b²)(a+b)=1

suy ra  3(a²+b²)-2(a³+b³)=1 suy ra 2(a³+b³)- 3(a²+b²)=-1

a) 20x³y² - 25x²y³ + 5x²y²

= 5x²y²(4x - 5y + 5) 

b) Ta có x³ - 25x = 0

<=> x(x² - 25) = 0

<=> x(x - 5)(x + 5) = 0

<=> x = 0 hoặc x - 5 = 0 hoặc x + 5 = 0

<=> x = 0 hoặc x = 5 hoặc x = -5

Vậy x \(\in\left\{0;5;-5\right\}\)là nghiệm phương trình

c) (x + 3)² = x + 3

<=> (x + 3)² - (x + 3) = 0

<=> (x + 3)(x + 3 - 1) = 0

<=> (x + 3)(x + 2) = 0

<=> \(\orbr{\begin{cases}x+3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-2\end{cases}}\)

Vậy x \(\in\left\{-3;-2\right\}\)

\(E=\frac{2a^4+6}{a^4+2a^2+3}=\frac{2a^4+4a^2+6-4a^2}{a^4+2a^2+3}=2-\frac{4a^2}{a^4+2a^2+3}\)

Vì \(4a^2\ge0\)

\(\Rightarrow\frac{4a^2}{a^4+2a^2+3}\ge0\)

\(\Rightarrow2-\frac{4a^2}{a^4+2a^2+3}\le2\)

Dấu = xảy ra

\(\Leftrightarrow4a^2=0\)

\(\Leftrightarrow a=0\)

Vậy ...

\(P=4a+\frac{9}{a-3}+6\)

\(=4\left(a-3\right)+\frac{9}{a-3}+18\)

Vì a>3

=> a-3>0

=> \(\frac{9}{a-3}>0\)

Áp dụng bđt Cô-sy cho 2 số ko âm ta có:

\(4\left(a-3\right)+\frac{9}{a-3}\ge2\sqrt{4\left(a-3\right)\cdot\frac{9}{a-3}}=2\sqrt{36}=12\)

\(\Rightarrow4\left(a-3\right)+\frac{9}{a-3}+18\ge30\)

Dấu = xảy ra

\(\Leftrightarrow4\left(a-3\right)=\frac{9}{a-3}\)

\(\Leftrightarrow4\left(a-3\right)^2=9\)

\(\Leftrightarrow\left(a-3\right)^2=\frac{9}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}a-3=\frac{3}{2}\\a-3=-\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=\frac{9}{2}\left(tm\right)\\a=\frac{3}{2}\left(kotm\right)\end{cases}}\)

Vậy ...

Cảm ơn a

\(B=\left(\frac{x}{2}+y\right)^3-6\left(\frac{x}{2}+y\right)^2.z+6\left(x+2y\right)z^2-8z^3\)

\(=\left(\frac{x}{2}+y\right)^3-3.\left(\frac{x}{2}+y\right)^2.2z+3.\left(\frac{x}{2}+y\right).\left(2z\right)^2-\left(2z\right)^3\)

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

\(C=\left(m-n\right)^3+15\left(m-n\right)^2.\left(m-p\right)-75\left(n-m\right)\left(p-m\right)^2-125\left(p-m\right)^3\)

\(=\left(m-n\right)^3+3.\left(m-n\right).\left[5\left(m-p\right)\right]+3.\left(m-n\right).\left[5\left(m-p\right)\right]^2+\left[5\left(m-p\right)\right]^3\)

\(=\left(m-n+5m-5p\right)^3=\left(6m-n-5p\right)^3\)

x+y = 47  ( đầu bài bạn)

!~ HỌC TỐT ~!

à khi sao chép nó bị thiếu í bạn là ( x+y ) mũ ba nha

\(a,y^4-14y^2+49\)

\(\left(y^2-7\right)^2\)

\(b,x^2-2\)

\(x^2-\left(\sqrt{2}\right)^2=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

\(c,y^2-13\)

\(y^2-\left(\sqrt{13}\right)^2=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)

\(d,-4x^2+9y^2\)

\(\left(3y\right)^2-\left(2x\right)^2\)

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

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

\(\left(x^2-2\right)^5=1\)

\(x^2-2=1\)

\(x^2=3\)

\(\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}}\)