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

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

\(=\left[2^2+2.3\right]^2-2.\left(-3\right)^2\)

\(=\left[2^2+6\right]^2-2.9\)

\(=10^2-18\)

\(=100-18=82\)

Mọi người giúp mình trả lời nhé, nay mình kiểm tra 1 tiết toán nên cần gấp đáp án ạ!

Mọi người giúp mình đi ạ! Mình đang cần gấp ạ!

\(M=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}-\frac{2a}{a^3-a^2+a-1}\right)\)

\(=\left(\frac{a^2+1}{a^2+1}+\frac{a}{a^2+1}\right):\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{a^2\left(a-1\right)+\left(a-1\right)}\right)\)

\(=\frac{a^2+a+1}{a^2+1}:\left(\frac{a^2+1}{\left(a-1\right)\left(a^2+1\right)}-\frac{2a}{\left(a^2+1\right)\left(a-1\right)}\right)\)

\(=\frac{a^2+a+1}{a^2+1}:\frac{a-1}{a^2+1}=\frac{a^2+a+1}{a-1}\)

2x(2x - 5) - 4x² = -5x + 25

=> 4x² - 10x - 4x² + 5x = 25

=> -10 + 5x = 25

=> 5x = 25 + 10

=> 5x = 35

=> x = 35/5 = 7

2x.(2x-5)-4x^2=-5x+25

<=>4x^2-10x-4x^2=-5x+25

<=>-10x+5x=25

<=>-5x=25

<=>x=-5

a) (x + 3y) (2x²y - 6xy²)

= (x + 3y) + 2xy (x - 3y)

= 2xy [(x + 3y) (x - 3y)]

= 2xy (x² - 3y²)

b) (6x⁵y² - 9x⁴y³ + 15x³y⁴) : 3x³y²

= (6x⁵y² : 3x³y²) + (-9x⁴y³ : 3x³y²) + (15x³y⁴ : 3x³y²)

= [(6 : 3) (x⁵ : x³) (y² : y²)] + [(-9 : 3) (x⁴ : x³) (y³ : y²)] + [(15 : 3) (x³ : x³) (y⁴ : y²)]

= 2x² + (-3xy) + 5y²

= 2x² - 3xy + 5y²

#Học tốt!!!

\(A=5-4x^2+4x\)

\(=-\left(4x^2-4x+1\right)+6\)

\(=-\left(2x-1\right)^2+6\le6\)

Dấu "=" xảy ra tại \(x=\frac{1}{2}\)

\(B=12-x^2-2x=-\left(x^2+2x+1\right)+13\)

\(=-\left(x+1\right)^2+13\le13\)

Dấu"=" xảy ra khi \(x=-1\)

\(a^2+4a+5⋮a+3\)

\(\Rightarrow\left(a+1\right)\left(a+3\right)+1⋮a+3\)

\(\Rightarrow1⋮a+3\)

Làm nốt

\(\left(a^2+4a+5\right)⋮\left(a+3\right)\)

\(\Leftrightarrow\left[a\left(a+3\right)+\left(a+3\right)+2\right]⋮\left(a+3\right)\)

Vì \(\hept{\begin{cases}\left[a\left(a+3\right)\right]⋮\left(a+3\right)\\\left(a+3\right)⋮\left(a+3\right)\end{cases}}\Rightarrow2⋮\left(a+3\right)\)

\(\Rightarrow a+2\in\left\{\pm1;\pm3\right\}\)

Lập bảng:

Vậy \(a\in\left\{-1;-3;1;-5\right\}\)