A = 3x² - 5x + 1

= 3( x² - 5/3x + 25/36 ) - 13/12

= 3( x - 5/6 )² - 13/12 ≥ -13/12 ∀ x

Dấu "=" xảy ra khi x = 5/6

=> MinA = -13/12 <=> x = 5/6

B = 7x² + 21x + 32

= 7( x² + 3x + 9/4 ) + 65/4

= 7( x + 3/2 )² + 65/4 ≥ 65/4 ∀ x

Dấu "=" xảy ra khi x = -3/2

=> MinB = 65/4 <=> x = -3/2

C = \(\frac{5}{x-x^2}\)

Để C đạt Min => x - x² đạt Max

Ta có x - x² = -( x² - x + 1/4 ) + 1/4 = -( x - 1/2 )² + 1/4 ≤ 1/4 ∀ x

Dấu "=" xảy ra khi x = 1/2

=> Max x - x² = 1/4 khi x = 1/2

=> MinC = \(\frac{5}{\frac{1}{4}}=20\)khi x = 1/2

a) \(A=3x^2-5x+1=3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)-\frac{13}{12}\)

\(=3\left(x-\frac{5}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(A=-\frac{13}{12}\Leftrightarrow3\left(x-\frac{5}{6}\right)^2=0\)

\(\Rightarrow x=\frac{5}{6}\)

Vậy Min(A) = -13/12 khi x = 5/6

\(2x^2+5x^3+x^2y=x^2\left(2+5x+y\right)\)

\(2x^2+5x^3+x^2y\)

\(=2x^2+5x^2.5x+x^2y\)

\(=x^2\left(2x+5x+y\right)=x^2\left(7x+y\right)\)

Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\Rightarrow a^3+b^3⋮a+b\)

a) \(A=\left(1^3+10^3\right)+\left(2^3+9^3\right)+\left(3^3+8^3\right)+\left(4^3+7^3\right)+\left(5^3+6^3\right)\Rightarrow A⋮11\)

b) \(A=\left(1^3+9^3\right)+\left(2^3+8^3\right)+\left(3^3+7^3\right)+\left(4^3+6^3\right)+5^3+10^3\)\(\Rightarrow A⋮5\)

a) (x + 3)³ - x(3x + 1)²  + (2x + 1)(4x² - 2x + 1) = 28

=> x³ + 9x² + 27x + 27 - x(9x² + 6x + 1) +(2x + 1)[(2x)² - 2.x.1 + 1² ] = 28

=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + (2x)³ + 1³ = 28

=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

=> (x³ - 9x³  + 8x³) + (9x² - 6x²) + (27x - x) + (27 + 1) = 28

=> 3x² + 26x + 28 = 28

=> 3x² + 26x = 0

=> 3x² + 26x = 0

=> \(3x\left(x+\frac{26}{3}\right)=0\)

=> 3x = 0 hoặc x + 26/3 = 0

=> x = 0 hoặc x = -26/3

b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

=> \(x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)

=> \(x^6-3x^4+3x^2-1-x^6+1=0\)

=> \(\left(x^6-x^6\right)-3x^4+3x^2+\left(-1+1\right)=0\)

=> \(-3x^4+3x^2=0\)

=> \(-\left(3x^4-3x^2\right)=0\)

=> \(3x\left(x^3-x\right)=0\)

=> \(\orbr{\begin{cases}3x=0\\x^3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x\left(x^2-1\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Hãy tích cho tui đi!

Khi bạn tích cho tui

Tui ko tích lại bạn đâu.

Thanks

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

\(\Leftrightarrow x^3+x^2+3x+3-x=-1\Leftrightarrow x^3+x^2+2x+4=0\)

Vonghiem => Ko co x thoa man 

\(M=4.6\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)=\left(5^{16}-1\right)\left(5^{16}+1\right)=5^{32}-1\)

Vậy M <N

tyjfghukjl;

Sửa \(x^2\left(x-2y\right)-y^2\left(y-2x\right)=x^3-2x^2y-y^3+2xy^2\)

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

Ta có: \(2x^2+y^2+5=\left(x^2+y^2\right)+\left(x^2+1\right)+4\ge2xy+2x+4=2\left(xy+x+2\right)\Rightarrow\frac{x}{2x^2+y^2+5}\le\frac{x}{2\left(xy+x+2\right)}\)\(6y^2+z^2+6=\left(4y^2+z^2\right)+\left(2y^2+2\right)+4\ge4yz+4y+4=4\left(yz+y+1\right)\Rightarrow\frac{2y}{6y^2+z^2+6}\le\frac{y}{2\left(yz+y+1\right)}\)\(3z^2+4x^2+16=\left(z^2+4x^2\right)+\left(2z^2+8\right)+8\ge4zx+8z+8=4\left(zx+2z+2\right)\Rightarrow\frac{4z}{2z^2+4x^2+16}\le\frac{z}{zx+2z+2}\)Từ ba bất đẳng thức trên suy ra:\(\frac{x}{2x^2+y^2+5}+\frac{2y}{6y^2+z^2+6}+\frac{4z}{3z^2+4x^2+16}\le\frac{1}{2}\left(\frac{x}{xy+x+2}+\frac{y}{yz+y+1}+\frac{2z}{zx+2z+2}\right)=\frac{1}{2}\left(\frac{xz}{xyz+xz+2z}+\frac{xyz}{xyz^2+xyz+xz}+\frac{2z}{zx+2z+2}\right)=\frac{1}{2}\left(\frac{zx}{zx+2z+2}+\frac{2}{zx+2z+2}+\frac{2z}{zx+2z+2}\right)=\frac{1}{2}\)Đẳng thức xảy ra khi x = y = 1; z = 2