Ta có: \(4 H \left(\right. x \left.\right) = \left(\right. 2 x \left.\right)^{2} - 2.2 x . y + y^{2} + 3 y^{2} - 4 x + 4 y + 4\)

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

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

\(= \left(\right. 2 x - y - 1 \left.\right) + 3 \left(\left(\right. y + \frac{1}{2} \left.\right)\right)^{2} + \frac{8}{3} \geq \frac{8}{3}\).

Vậy giá trị nhỏ nhất của \(E\) là: \(\frac{8}{3} : \&\text{nbsp}; 4 = \frac{2}{3}\) tại \(x = \frac{2}{3} ; \&\text{nbsp}; y = - \frac{1}{3}\)

Ta có: \(4 H \left(\right. x \left.\right) = \left(\right. 2 x \left.\right)^{2} - 2.2 x . y + y^{2} + 3 y^{2} - 4 x + 4 y + 4\)

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

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

\(= \left(\right. 2 x - y - 1 \left.\right) + 3 \left(\left(\right. y + \frac{1}{2} \left.\right)\right)^{2} + \frac{8}{3} \geq \frac{8}{3}\).

Vậy giá trị nhỏ nhất của \(E\) là: \(\frac{8}{3} : \&\text{nbsp}; 4 = \frac{2}{3}\) tại \(x = \frac{2}{3} ; \&\text{nbsp}; y = - \frac{1}{3}\)

Ta có: \(4 H \left(\right. x \left.\right) = \left(\right. 2 x \left.\right)^{2} - 2.2 x . y + y^{2} + 3 y^{2} - 4 x + 4 y + 4\)

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

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

\(= \left(\right. 2 x - y - 1 \left.\right) + 3 \left(\left(\right. y + \frac{1}{2} \left.\right)\right)^{2} + \frac{8}{3} \geq \frac{8}{3}\).

Vậy giá trị nhỏ nhất của \(E\) là: \(\frac{8}{3} : \&\text{nbsp}; 4 = \frac{2}{3}\) tại \(x = \frac{2}{3} ; \&\text{nbsp}; y = - \frac{1}{3}\)

Ta có: \(4 H \left(\right. x \left.\right) = \left(\right. 2 x \left.\right)^{2} - 2.2 x . y + y^{2} + 3 y^{2} - 4 x + 4 y + 4\)

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

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

\(= \left(\right. 2 x - y - 1 \left.\right) + 3 \left(\left(\right. y + \frac{1}{2} \left.\right)\right)^{2} + \frac{8}{3} \geq \frac{8}{3}\).

Vậy giá trị nhỏ nhất của \(E\) là: \(\frac{8}{3} : \&\text{nbsp}; 4 = \frac{2}{3}\) tại \(x = \frac{2}{3} ; \&\text{nbsp}; y = - \frac{1}{3}\)

Từ \(x + y + z = 0\) suy ra \(x + y = - z\)

\(x^{2} + 2 x y + y^{2} = z^{2}\)

\(x^{2} + y^{2} - z^{2} = - 2 x y\).

Tương tự ta có: \(y^{2} + z^{2} - x^{2} = - 2 y z\) và \(z^{2} + x^{2} - y^{2} = - 2 z x\).

Do đó \(A = \frac{x y}{- 2 x y} + \frac{y z}{- 2 y z} + \frac{z x}{- 2 z x} = - \frac{1}{2} - \frac{1}{2} - \frac{1}{2} = - \frac{3}{2}\).

Vậy \(A = - \frac{3}{2}\).

a) \(\Delta A B C\) vuông tại \(A\) suy ra \(\hat{B A C} = 9 0^{\circ}\) suy ra \(\hat{D A E} = 9 0^{\circ}\).

Do \(H D ⊥ A B\) suy ra \(\hat{H D A} = 9 0^{\circ}\); \(H E ⊥ A C\) suy ra \(\hat{H E A} = 9 0^{\circ}\).

Tứ giác \(A D H E\) có \(\hat{D A E} = \hat{H D A} = \hat{H E A} = 9 0^{\circ}\) suy ra tứ giác \(A D H E\) là hình chữ nhật. 

b) Do \(\Delta A H D\) vuông tại \(D\), áp dụng định lí Pythagore suy ra:

\(A H^{2} = A D^{2} + D H^{2}\)

\(25 = 16 + D H^{2}\)

\(D H^{2} = 9\) nên \(D H = 3\) cm.

Do \(A D H E\) là hình chữ nhật suy ra \(S_{A D H E} = A D . D H = 4.3 = 12\) (cm\(^{2}\)).

Vì đồ thị hàm số \(y = a x + b\) đi qua điểm \(A \left(\right. - 1 ; 2 \left.\right)\) nên ta có:

   \(2 = - 1. a + b\) suy ra \(- a + b = 2\)

Vi đồ thị hàm số \(y = a x + b\) đi qua điểm \(B \left(\right. 1 ; 4 \left.\right)\) nên ta có:

   \(4 = 1. a + b\) suy ra \(a + b = 4 \left(\right. 2 \left.\right)\)

Từ (1) và (2) ta tìm được \(a = 1 ; b = 3\)

Vậy hàm số cần tìm là \(y = x + 3\).

a) Thay \(x = 2\) (thỏa mãn điều kiện xác định) vào \(Q = \frac{x + 1}{x^{2} - 9}\), ta được:

\(Q = \frac{x + 1}{x^{2} - 9} = \frac{2 + 1}{2^{2} - 9} = \frac{3}{- 5} = - \frac{3}{5}\).

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

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

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

\(P = \frac{x^{2} + 3 x}{x \left(\right. x + 1 \left.\right)} = \frac{x + 3}{x + 1}\).

c) Ta có \(M = P . Q = \frac{x + 3}{x + 1} . \frac{x + 1}{x^{2} - 9} = \frac{x + 3}{\left(\right. x - 3 \left.\right) \left(\right. x + 3 \left.\right)} = \frac{1}{x - 3}\)

\(M = \frac{- 1}{2}\) suy ra \(\frac{1}{x - 3} = \frac{- 1}{2}\)

\(x - 3 = - 2\)

\(x = 1\).

Vậy với \(x = 1\) thì\(M = \frac{- 1}{2}\).

a) 5(x+2y)−15x(x+2y)=5(x+2y).(1−3x).

=(x +2y)(5-15x)

b)4x² - 12x + 9 = (2x - 3)²

c) (3x−2)3−3(x−4)(x+4)+(x−3)3−(x+1)(x2−x+1)

\(= 27 x^{3} - 54 x^{2} + 36 x - 8 - 3 \left(\right. x^{2} - 16 \left.\right) + x^{3} - 9 x^{2} + 27 x - 27 - \left(\right. x^{3} + 1 \left.\right)\)

a) 5(x + 2y) - 15x(x + 2y)

\(= \left(\right. 27 x^{3} + x^{3} - x^{3} \left.\right) + \left(\right. - 54 x^{2} - 3 x^{2} - 9 x^{2} \left.\right) + \left(\right. 36 x + 27 x \left.\right) + \left(\right. - 8 + 48 - 27 - 1 \left.\right)\)

\(= 27 x^{3} - 66 x^{2} + 63 x + 12\).

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