4x2/5x3/4x1/6x5x6
A=5x3-7x2-(-3x3+4x2)+(x2-x3+5x-1)
B=(3x2+5x3-7x4)-(5x3-4x2+x4-3)
\(A=5x^3-7x^2+3x^3-4x^2+x^2-x^3+5x-1=7x^3-10x^2+5x-1\)
\(B=5x^3+3x^2-7x^4-5x^3+4x^2-x^4+3=-8x^4+7x^2+3\)
\(A=7x^3-10x^2+5x-1\)
\(B=-8x^4+7x^2+3\)
Tìm x biết:
(5x3 + 4x2 + 3x + 3) - (4 - x - 4x2 + 5x3)= 5
=>5x^3+4x^2+3x+3-4+x+4x^2-5x^3=5
=>8x^2+4x-1-5=0
=>8x^2+4x-6=0
=>4x^2+2x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{4}\)
Thu gọn đa thức ( 5 x 3 + 4 x 2 - 1 ) - ( 4 x 3 - 4 x 2 + 1 ) ta được
A. 0
B. x 3 + 8 x 2 - 2
C. - x 3 + 8 x 2 - 2
D. - x 3 - 8 x 2 - 2
(5x3-4x2):2x2
`(5x^3 -4x^2) :2x^2`
`= 5x^3 : 2x^2 - 4x^2 :2x^2`
`= 5/2 x - 2`
2/5x3/4+1/4x2/5
\(\dfrac{2}{5}\times\dfrac{3}{4}+\dfrac{1}{4}\times\dfrac{2}{5}\\ =\dfrac{2}{5}\times\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =\dfrac{2}{5}\times1\\ =\dfrac{2}{5}\)
Cho pt 2x2 +3x --1 =0 có 2 nghiệm x1, x2
Tính B= (4x1-1)/x2 + (4x2-1)/x1
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{3}{2}\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)
\(B=\dfrac{4x_1-1}{x_2}+\dfrac{4x_2-1}{x_1}=\dfrac{4x_1^2-x_1+4x_2^2-x_2}{x_1x_2}\)
\(=\dfrac{4\left(x_1+x_2\right)^2-8x_1x_2-\left(x_1+x_2\right)}{x_1x_2}=\dfrac{4.\left(-\dfrac{3}{2}\right)^2-8.\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{2}\right)}{-\dfrac{1}{2}}=-29\)
Tìm các đạo hàm sau: y = 2 x - 4 x 2 + 5 x 3 - 6 7 x 4
tìm x biết
5x3+2x4-x2+3x2-x3-x4+1-4x2
Tìm đạo hàm của hàm số sau: y = 2 x - 4 x 2 + 5 x 3 - 6 7 x 4