Cho hình tang ABCD với AB // CD. Giả sử AD = BD. Gọi M là trung điểm của đường chéo AC. CMR góc MDC = góc BCD
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4x(x2-2x+3)-3x(x+1)(x-2)+(2x-5)(x-7)-4x(x2-x+1)
=4x3-8x2+12x-3x(x2-2x+x-2)+2x2-14x-5x+35-4x3+4x2-4x
=4x3-8x2+12x-3x3+6x2-3x2+6x+2x2-14x-5x+35-4x3+4x2-4x
=-3x3+x2-5x+35
#H
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Ta có: x3 - x2 - 4 = x2 - 2x2 + x2 - 2x + 2x - 4 = x2(x - 2) + x(x - 2) + 2(x -2) = (x - 2)(x2 + x + 2)
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(4x-1)(3x+1)-5x2(x-3)-(x-4)(x-3)-7(x3-2x2+x-1)
=12x2+4x-3x-1-5x3+15x2-x2+3x+4x-12-7x3+14x2-7x+7
=-12x340x2+x-6
#H
(4x-1)(3x+1)-5x2(x-3)-(x-4)(x-3)-7(x3-2x2+x-1)
= 12x2+4x-3x-1-5x3+15x2-x2-3x-4x+12-7x3+14x2-7x-1
=-12x3+40x2-13x+10
![](https://rs.olm.vn/images/avt/0.png?1311)
5x(x-3)(2x+4)-(x+7)(x-3)+(5x-2)(3x+4)
=5x(2x2+4x-6x-12)-(x2-3x+7x-21)+15x2+20x-6x-8
=10x3+20x2-30x2-60x-x2+3x-7x+21+15x2+20x-6x-8
=10x3+4x2-50x+13
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5) a. A = -x2 + 2x + 5 = -x2 + 2x - 1 + 6 = -(x - 1)2 + 6 \(\le\)6
=> Max A = 6
Dấu "=" xảy ra <=> x - 1 = 0
<=> x = 1
Vậy Max A = 6 <=> x = 1
b) Sửa đề B = -(2x)2 + 3x + 1 = \(-\left(2x\right)^2+3x-\frac{9}{16}+\frac{25}{16}=-\left(2x-\frac{3}{4}\right)^2+\frac{25}{16}\le\frac{25}{16}\)
=> Max B = 25/16
Dấu "=" xảy ra <=> 2x - 3/4 = 0 <=> x = 3/8
Vậy Max B = 25/16 <=> x = 3/8
6) a. Ta có C = x2 + y2 - 2x + 4y + 1
= (x2 - 2x + 1) + (y2 + 4y + 4) - 4 = (x - 1)2 + (y + 2)2 - 4 \(\ge-4\)
=> Min C = -4
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy Min C = -4 <=> x = 1 ; y = -2
b) Đặt D = x2 + 2y2 - 2xy - 2x + 4y + 5
= (x2 - 2xy + y2) - 2(x - y) + 1 + (y2 + 2y + 1) + 3
= (x - y)2 - 2(x - y) + 1 + (y + 1)2 + 3
= (x - y - 1)2 + (y + 1)2 + 3 \(\ge3\)
=> Min D = 3
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y-1=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
Vậy Min D = 3 <=> x = 0 ; y = -1
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\(A=\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
\(A\cdot\left(x-y\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
\(A\cdot\left(x-y\right)=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
\(A\cdot\left(x-y\right)=\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
\(A\left(x-y\right)=\left(x^8-y^8\right)\left(x^8+y^8\right)\)
\(A\left(x-y\right)=x^{16}+y^{16}\)
Mà x - y = 1
\(\Rightarrow A=x^{16}+y^{16}\)