Bài 1 Phân tích đa thức thành nhân tử
\(x^4-4x^2+12x-9\)
Bài 2 Cho x,y là các số thỏa mãn \(\left(\sqrt{x^2+2013+x}\right)\left(\sqrt{y^2+2013+y}\right)=2013\)
Hãy tính giá trị của biểu thức x+y
Bài 3
\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
a) rút gọn
b) tính A khi a=3+2\(\sqrt{2}\)
Bài 3:
a) \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a+1}\right)}\right)\)
\(=\dfrac{a-1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{1}{\sqrt{a}-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
\(=\dfrac{a-1}{\sqrt{a}}\)
b) Thay \(a=3+2\sqrt{2}\) vào biểu thức A:
Ta có: \(\dfrac{3+2\sqrt{2}-1}{\sqrt{3+2\sqrt{2}}}=\dfrac{2+2\sqrt{2}}{\sqrt{\left(1+2\sqrt{2}\right)^2}}=\dfrac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=2\)
Vậy giá trị biểu thức A tại \(a=3+2\sqrt{2}\)
Bài 1:
Sửa đề: (theo mình là như vậy)
\(x^4-4x^2-12x-9\)
\(=x^4+x^3-x^3-x^2-3x^2-3x-9x-9\)
\(=\left(x^4+x^3\right)-\left(x^3+x^2\right)-\left(3x^2+3x\right)-\left(9x+9\right)\)
\(=x^3.\left(x+1\right)-x^2.\left(x+1\right)-3x.\left(x+1\right)-9.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3-x^2-3x-9\right)\)
\(=\left(x+1\right).\left(x^3-3x^2+2x-6x+3x-9\right)\)
\(=\left(x+1\right).\left[\left(x^3-3x^2\right)+\left(2x-6x\right)+\left(3x-9\right)\right]\)
\(=\left(x+1\right).\left[x^2.\left(x-3\right)+2x.\left(x-3\right)+3.\left(x-3\right)\right]\)
\(=\left(x+1\right).\left(x-3\right).\left(x^2+2x+3\right)\)
Chúc bạn học tốt!!!
À nhầm không cần sửa đề!
Bài 1:
\(x^4-4x^2+12x-9\)
\(=x^4-x^3+x^3-x^2-3x^2+3x+9x-9\)
\(=\left(x^4-x^3\right)+\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-9\right)\)
\(=x^3.\left(x-1\right)+x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-1\right)\)
\(=\left(x-1\right).\left(x^3+x^2-3x+9\right)\)
\(=\left(x-1\right).\left(x^3+3x^2-2x^2-6x+3x+9\right)\)
\(=\left(x-1\right).\left[\left(x^3+3x^2\right)-\left(2x^2+6x\right)+\left(3x+9\right)\right]\)
\(=\left(x-1\right).\left[x^2.\left(x+3\right)-2x.\left(x+3\right)+3.\left(x+3\right)\right]\)
\(=\left(x-1\right).\left(x+3\right).\left(x^2-2x+3\right)\)
Chúc bạn học tốt!!!
