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câu 1: A rectangle has a length of 60cm and a width of 30cm. It is cut into 2 indentical squares, 2 identical rectangles and a shaded small square. Find the area of the shaded square. Find the area of the shaded square. câu 2.The number of ordered pairs (x; y) where x, y ∈ N* such that x2y2 - 2(x + y) is perfect square is .......... câu 3.Let ABCD be the square with the side length 56cm. If E and F lie on CD, C respectively such that CF = 14cm and EAF = 45o then CE = ........cm. câu...
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câu 1: A rectangle has a length of 60cm and a width of 30cm. It is cut into 2 indentical squares, 2 identical rectangles and a shaded small square. Find the area of the shaded square.
Find the area of the shaded square.
Đề thi violympic toán tiếng anh lớp 8 vòng 10
câu 2.The number of ordered pairs (x; y) where x, y ∈ N* such that x2y2 - 2(x + y) is perfect square is ..........

câu 3.Let ABCD be the square with the side length 56cm. If E and F lie on CD, C respectively such that CF = 14cm and EAF = 45o then CE = ........cm.

câu 4.

Given P(x) = (x2 - 1/2 x - 1/2)1008
If P(x) = a2016x2016 + a2015x2015 + ..... + a1x + a0
then the value of the sum a0 + a2 + a4 + .... + a2014 is ...........

Write your answer by decimal in simplest form
câu 5.Let ABC be an isoceles triangle (AB = AC) and its area is 501cm2. BD is the internal bisector of the angle ABC (D ∈ AC), E is a point on the opposite ray of CA such that CE = CB. I is a point on BC such that CI = 1/2 BI. The line EI meets AB at K, BD meets KC at H. Find the area of the triangle AHC.
câu 6. all roots of the polynomial P(x) = x2 + 5x - 1 are also roots of the polynomial Q(x) = x3 + ax2 + bx + c then the value of a + b + 6c is ............
câu 7.Suppose that the polynomial f(x) = x5 - x4 - 4x3 + 2x2 + 4x + 1 has 5 solutions x1; x2; x3; x4; x5. The other polynomial k(x) = x2 - 4.
Find the value of P = k(x1) x k(x2) x k(x3) x k(x4) x k(x5)
câu 8.The smallest value of Đề thi violympic toán tiếng anh lớp 8 vòng 10 is .............
câu 9.Let ABCD be a trapezoid with bases AB, CD and O be the intersection of AC and BD. If the areas of triangle OAB, triangle OCD are 16cm2, 40cm2respectively and M is the midpoint of BD, then the area of the triangle AMD is .........cm2.
câu 10.Bottle A contains 15% syrup. Bottle B contains 40% syrup. When these 2 bottles of syrup are mixed, the syrup content is 30% and the total volume is 600ml. How much syrup is in the bottle A at first?
câu 11.Let ABCD.A'B'C'D' be a cube with AC' = √3cm. Find the total surface area of this cube.
câu 12.Let ABC be a triangle with AB = 3cm, AC = 7cm. The internal bisector of the angle BAC intersects BC at D. The line passing through D and parallel to AC cuts AB at E. Find the measure of DE.
câu 12.Given two numbers x, y such that (4y2 - 12y + 25)(4x2 + 6x + 4) = 28
The ratio of y to x is ........
câu 13.Mr.Joseph drives car from A to B at a constant speed. If the speed of the car is increased by 20%, it takes him one hour less than the usual time. If he drives at the constant speed for the first 100km before increasing the speed by 30%, it also takes him one hour less than the usual. The distance of AB is ..........km.
câu 14.A triangle ABC has  = 120o and the bisector AD (D ∈ BC). If AB = 40cm, AD = 30cm, then AC = ..... cm.
câu 15.As shown in the figure, the length of BE is .............
Đề thi violympic toán tiếng anh lớp 8 vòng 10
câu 16.

Let f(x) the polynomial given by f(x) = (1 + 2x + 3x2 + 4x3 + 5x4 + 84x5)
Suppose that f(x) = ao + a1x + a2x2 + ..... + ..... + a50x50.
The value of T = a1 + a2 + .... + a50 is .........

  • a. 9910 - 1
  • b. 9910
  • c. 10010
  • d. 10010 - 1
  • câu 17.Find the area of the trapezoid ABCD, BC // AD, AB = CD = 5cm, BC = 10cm, AD = 16cm.
    The area of the trapezoid ABCD is .........cm2.
  • câu 18.Find the least possible value of A = 4x2 - 3x + 1/4x + 2015, where x varies in the set of positive real numbers. The least possible value of A is
  • câu 19.
0
24 tháng 5 2016

Phương trình đúng là 

x- 2(m + 1)x + m2 = 0 

19 tháng 4 2020

a) PT hoành độ giao điểm (d) (P)

mx-n+1=x2

<=> x2-mx+m-1=0

\(\Delta=m^2-4\left(m-1\right)=m^2-4m+4=\left(m-2\right)^2\ge0\forall m\)

Vậy (d); (P) luôn cắt nhau tại 2 điểm phân biệt

b) \(x_1^2x_2+x_2^2x_1=2\)

\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)=2\)

\(\Leftrightarrow\left(m-1\right)m=2\)

<=> m2-m-2=0

\(\Leftrightarrow\orbr{\begin{cases}m=2\\m=-1\end{cases}}\)

19 tháng 4 2020

a) phương trình hoành độ giao điểm của (d)và (P) là:

\(x^2=mx-m+1\)\(\Leftrightarrow x^2-mx+m-1=0\)

TA CÓ: a=1, b'=\(\frac{-m}{2},\)c= m-1

\(\Rightarrow\)\(\Delta'\)=\(\left(b'\right)^2-ac=\left(\frac{-m}{2}\right)^2-\left(m-1\right).1\)\(=\frac{m^2}{4}-m+1\)

\(=\)\(\frac{m^2}{4}-2.\frac{m}{2}.1+1=\left(\frac{m}{2}-1\right)^2\)

\(\text{ để đường thẳng d và parabol ( P) cắt nhau tại 2 điểm phân biệt}:\)

\(\Delta'>0\Leftrightarrow\)\(\left(\frac{m}{2}-1\right)^2>0\Leftrightarrow m\ne2\)

vậy với m \(\ne2\) thì ......