\(\left(15+5x^2-3x^3-9x\right):\left(3+x^2\right)=\left(-3x^3+5x^2-9x+15\right):\left(x^2+3\right)\)

x^2 + 3 -3x^3 + 5x^2 - 9x +15 -3x -3x^3 - 9x -3x^3 -9x -3x^3 - 9x - 5x^2 +15 + 5 - 5x^2 +15 0

viết thẳng hàng vào nhá, mk viết hơi lệch

\(a+b+c=\frac{3}{2}\)

\(\Leftrightarrow\left(a+b+c\right)^2=\frac{9}{4}\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=\frac{9}{4}\)

\(\Leftrightarrow a^2+b^2+c^2=\frac{9}{4}-2\left(ab+bc+ca\right)\) 

Mặt khác: 

Ta sẽ c/m \(\frac{9}{4}-2\left(ab+bc+ca\right)=\frac{3}{2}\) (1)

\(\Leftrightarrow2\left(ab+bc+ca\right)=\frac{9}{4}-\frac{3}{2}=\frac{3}{4}\)

\(\Leftrightarrow4\left(ab+bc+ca\right)=\frac{3}{2}=a+b+c\)

Suy ra \(ab+bc+ca=\frac{a+b+c}{4}\)

Do đó: 

\(=\frac{9}{4}-2\left(\frac{a+b+c}{4}\right)=\frac{9}{4}-2\left(\frac{\left(\frac{3}{2}\right)}{4}\right)=\frac{9}{4}-2.\frac{3}{8}=\frac{3}{2}\) (2)

Từ (2) suy ra (1) đúng.

Do (1) đúng: suy ra: \(a^2+b^2+c^2=\frac{9}{4}-2\left(ab+bc+ca\right)=\frac{3}{2}^{\left(đpcm\right)}\)

Mình ghi thiếu một chỗ nên có nhiều bạn không hiểu: Chỗ hàng thứ 4 từ dưới đếm lên cho đến hết,bạn sửa thành:

"Do đó:

\(\frac{9}{4}-2\left(ab+bc+ca\right)=\frac{9}{4}-2\left(\frac{a+b+c}{4}\right)\)

\(=\frac{9}{4}-2\left(\frac{\left(\frac{3}{2}\right)}{4}\right)=\frac{9}{4}-2.\frac{3}{8}=\frac{3}{2}\) (2)

Từ (2) suy ra (1) đúng suy ra \(a^2+b^2+c^2=\frac{9}{4}-2\left(ab+bc+ca\right)=\frac{3}{2}^{\left(đpcm\right)}\)"

\(M=\left(\frac{1}{3}x-y\right)\left(x^2+3xy+9y^2\right)+9y^3-\frac{1}{3}x^3\)

\(=\frac{1}{3}x^3+x^2y+3xy^2-x^2y-3xy^2-9y^3+9y^3-\frac{1}{3}x^3\)

\(=\left(\frac{1}{3}x^3-\frac{1}{3}x^3\right)+\left(x^2y-x^2y\right)+\left(3xy^2-3xy^2\right)-\left(9y^3-9y^3\right)\)

\(=0\)

Vậy : Giá trị của M ko phụ thuộc vào biến x,y

=.= hk tốt!!

\(2x^2+98+28x-8y^2\)

\(=2\left(x^2+14x+49-4y^2\right)\)

\(=2\left[\left(x^2+2\cdot x\cdot7+7^2\right)-\left(2y\right)^2\right]\)

\(=2\left[\left(x+7\right)^2-\left(2y\right)^2\right]\)

\(=2\left(x-2y+7\right)\left(x+2y+7\right)\)

\(ay^2-4ay+4a-by^2+4by-4b\)

\(=\left(ay^2-4ay+4a\right)-\left(by^2-4by+4b\right)\)

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

\(=a\left(y-2\right)^2-b\left(y-2\right)^2\)

\(=\left(y-2\right)^2\left(a-b\right)\)

=(ay2-by2)+(-4ay+4by)+(4a-4b)

=y2(a-b)+ 4(a-b)

=(a-b)×(y2+4)

\(\frac{4x^4-20x^3+13x^2+30x+9}{\left(4x^2-1\right)^2}\)

\(=\frac{4x^3\left(x-3\right)-8x^2\left(x-3\right)-11x\left(x-3\right)-3\left(x-3\right)}{\left(4x^2-1\right)^2}\)

\(=\frac{\left(x-3\right)\left(4x^3-8x^2-11x-3\right)}{\left(4x^2-1\right)^2}\)

\(=\frac{\left(x-3\right)\left[4x^2\left(x-3\right)+4x\left(x-3\right)+\left(x-3\right)\right]}{\left[\left(2x-1\right)\left(2x+1\right)\right]^2}\)

\(=\frac{\left(x-3\right)^2\left(4x^2+4x+1\right)}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2\left(2x+1\right)^2}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2}{\left(2x-1\right)^2}\)

\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)

  \(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)

\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)

\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)

\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)

\(=\frac{1}{n}.\frac{n+1}{2}\)

\(=\frac{n+1}{2n}\)