Đặt \(x^2=a;y^2=b\left(a,b\ge0\right)\)
Ta có

\(x^6+y^6=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=a^2-ab+b^2\)

\(\ge a^2-\frac{a^2+b^2}{2}+b^2=\frac{a^2+b^2}{2}\ge\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\)

Vậy Min = 1/4 khi \(x=y=\frac{1}{\sqrt{2}}\)
Ta có

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

+) Đặt x+y=S, xy = P, ta được: \(S^2-2P=1\)
+)\(x^6+y^6=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)=x^4-x^2y^2+y^4=\left(x^2+y^2\right)^2-3x^2y^2\)

\(=\left[\left(x+y\right)^2-2xy\right]^2-3x^2y^2=\left(S^2-2P\right)^2-3P^2=S^4-4S^2P+4P^2-3P^2\)

\(=S^4-4S^2P+P^2=\left(2P+1\right)^2-4\left(2P+1\right)P+P^2\)

\(=4P^2+4P+1-8P^2-4P+P^2=-3P^2+1\le1\)

Dấu = xảy ra khi \(\hept{\begin{cases}P=0\\S=1\end{cases}}\), khi đó x=1, y=0 hoặc x=0, y=1

Sửa đề:

Cách 1:

\(\left(5x-1\right)^2+2.\left(1-5x\right).\left(4+5x\right)+\left(5x+4\right)^2\)

\(=\left(1-5x\right)^2+2.\left(1-5x\right).\left(5x+4\right)+\left(5x+4\right)^2\)

\(=\left(1-5x+5x+4\right)^2\)

\(=5^2\)

\(=25\)

Cách 2:

\(\left(5x-1\right)^2+2.\left(1-5x\right).\left(4+5x\right)+\left(5x+4\right)^2\)

\(=\left(5x-1\right)^2-2.\left(5x-1\right).\left(5x+4\right)+\left(5x+4\right)^2\)

\(=\left(5x-1-5x-4\right)^2\)

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

\(=25\)

a) 3x(x + 7)² - 11x²(x + 7) + 9(x + 7) = (x + 7)[3x(x + 7) - 11x² + 9) = (x + 7)(3x² + 21x - 11x² + 9)

= (x + 7)(-8x² + 21x + 9)(-8x² + 24x - 3x + 9) = (x + 7)[-8x(x - 3) - 3(x - 3)] = -(x + 7)(8x + 3)(x - 3)

b) 3x(x - 9)² - (9 - x)³ = 3x(x - 9)² + (x - 9)³ = (x - 9)²(3x + x - 9) = (x - 9)²(4x - 9)

c) p^{m + 2}.q - p^{m + 1}.q³ - p².q^{n + 1} + p.q^{n + 3} = (p^{m + 2}.q - p².q^{n + 1}) - (p^{m + 1}.q³ - p.q^{n + 3})

= p².q(p^m - qⁿ) - p.q³(p^m - qⁿ) = pq(p^m - qⁿ)(p - q²)

d) x²y²z + xy²z² + x²yz = xyz(xy + yz + x)

a) \(3x\left(x+7\right)^2-11x^2\left(x+7\right)+9\left(x+7\right)\)

\(=\left(x+7\right)\left[3x\left(x+7\right)-11x^2+9\right]=\left(x+7\right)\left(3x^2+21x-11x^2+9\right)\)

\(=\left(x+7\right)\left(-8x^2+21x+9\right)=\left(x+7\right)\left[\left(-8x^2+24x\right)-\left(3x-9\right)\right]\)

\(=\left(x+7\right)\left[-8x\left(x-3\right)-3\left(x-3\right)\right]=-\left(x+7\right)\left(x-3\right)\left(8x+3\right)\)

b) \(3x\left(x-9\right)^2-\left(9-x\right)^3=3x\left(x-9\right)^2+\left(x-9\right)^3\)

\(=\left(x-9\right)^2\left(3x+x-9\right)=\left(x-9\right)^2\left(4x-9\right)\)

c) \(p^{m+2}.q-p^{m+1}.q^3-p^2.q^{n+1}+p.q^{n+3}\)

\(=p^{m+1}.q\left(p-q^2\right)-p.q^{n+1}\left(p-q^2\right)\)\(=p.q.\left(p-q^2\right).\left(p^m.q^n\right)\)

d) \(x^2y^2z+xy^2z^2+x^2yz=xyz\left(xy+yz+x\right)\)

\(a^2+a+1=0\Rightarrow\left(a+\frac{1}{2}\right)^2+\frac{3}{4}=0\Rightarrow a\in C\)

Vì vậy P không tồn tại

Lớp 8 nên làm như này nhé :))

Giá 10 quả trứng là: \(3000.10=30000\)

Giá 4 cân thịt là: \(50000.4=200000\)

Giá 5 gói bột canh là: \(5000.5=25000\)

Lan mua hết: \(30000+200000+25000=255000\)

Còn thừa: \(300000-255000=45000\)

                              Bài làm :

10 quả trứng có giá là :

3 000 x 10 = 30 000

Thịt 4 cân có giá là :

50 000 x 4 = 200 000

Bột canh 5 gói có giá :

5 000 x 5 = 25 000

Vậy Lan mua hết số tiền là :

30 000 + 200 000 + 25 000 = 255 000

Và còn thừa số tiền là :

300 000  - 255 000 = 45 000

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Bài làm:

Ta có: \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)(hằng đẳng thức đầu)

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

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

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

\(=\left[\left(x-y+z\right)+\left(y-z\right)\right]^2=\left(x-y+z+y-z\right)^2=x^2\)

Ta có: \(\hept{\begin{cases}a>c+d\\b>c+d\end{cases}\Leftrightarrow\hept{\begin{cases}a-c>d\\b-d>c\end{cases}\Rightarrow}\left(a-c\right)\left(b-d\right)>cd\Leftrightarrow ab-bc-ad+cd>cd}\Leftrightarrow ab>ad+bc\)