Question

Câu 1: Tìm x

1. A = x² + 4x - 2

2. B = 2x² - 4x + 3

3. C = x² + y² - 4x + 2y + 5

Câu 2: Tìm giá trị lớn nhất:

1. A = -x² + 6x +5

2. B = - 4x² - 9y² - 4x + 6y + 3

Câu 3: Tìm x, y:

1. x² + y² - 2x + 4y + 5 = 0

2. 5x² + 9y² - 12xy - 6x + 9 = 0

Câu 4: Tìm a và b biết:

a) A(x) = 2x³ + 7x²

B(x) = x² + x - 1

b) A(x) = a . x³ + b . x - 24

B(x) = x² + 4x + 3

c) A(x) = 6x⁴ - x³ + a . x² + 4

B(x) = x² - 4

Câu 5: Cho x = y + 1. CMR:

1. x³ - y³ - 3xy = 1

2. (x + y) (x² + y²) ( x⁴ + y⁴) (x⁸ + y⁸) = x¹⁸ - 1y¹⁶

Answer 1

Câu 1 : Tìm x :

1. \(A=x^2+4x-2\)

\(A=x^2+2.x.2+2^2-2^2-2\)

\(A=\left(x^2+4x+2^2\right)-4-2\)

\(A=\left(x+2\right)^2-6\)

\(\left(x+2\right)^2-6\ge-6\)

MIn A= -6 khi \(\left(x+2\right)^2=0\)

=> \(x+2=0hayx=-2\)

Vậy x=2

những câu tiếp theo làm tg tự như thế nhé

Answer 2

Câu 1:

a) Ta có: \(A=x^2+4x-2\)

\(=x^2+4x+4-6\)

\(=\left(x+2\right)^2-6\)

Ta có: \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2-6\ge-6\forall x\)

Dấu '=' xảy ra khi

\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

Vậy: x=-2

b) Ta có: \(B=2x^2-4x+3\)

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

\(=2\left(x^2-2\cdot x\cdot1+1+\frac{1}{2}\right)\)

\(=2\left[\left(x^2-2x\cdot1+1\right)+\frac{1}{2}\right]\)

\(=2\left[\left(x-1\right)^2+\frac{1}{2}\right]\)

\(=2\left(x-1\right)^2+1\)

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi

\(2\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy: x=1

c) Ta có: \(C=x^2+y^2-4x+2y+5\)

\(=x^2-4x+4+y^2+2y+1\)

\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)

\(=\left(x-2\right)^2+\left(y+1\right)^2\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\left(y+1\right)^2\ge0\forall y\)

Do đó: \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

Vậy: x=2 và y=-1

Câu 2:

a) Ta có: \(A=-x^2+6x+5\)

\(=-\left(x^2-6x-5\right)\)

\(=-\left(x^2-6x+9-14\right)\)

\(=-\left[\left(x^2-6x+9\right)-14\right]\)

\(=-\left[\left(x-3\right)^2-14\right]\)

\(=-\left(x-3\right)^2+14\)

Ta có: \(\left(x-3\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)

\(\Leftrightarrow-\left(x-3\right)^2+14\le14\forall x\)

Dấu '=' xảy ra khi

\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy: GTLN của đa thức \(A=-x^2+6x+5\) là 14 khi x=3

b) Ta có: \(B=-4x^2-9y^2-4x+6y+3\)

\(=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)-5\right]\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2-5\right]\)

\(=-\left(2x+1\right)^2-\left(3y-1\right)^2+5\)

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Rightarrow-\left(2x+1\right)^2\le0\forall x\)(1)

Ta có: \(\left(3y-1\right)^2\ge0\forall y\)

\(\Rightarrow-\left(3y-1\right)^2\le0\forall y\)(2)

Từ (1) và (2) suy ra

\(-\left(2x+1\right)^2-\left(3y-1\right)^2\le0\forall x,y\)

\(\Rightarrow-\left(2x+1\right)^2-\left(3y-1\right)^2+5\le5\forall x,y\)

Dấu '=' xảy ra khi

\(\left\{{}\begin{matrix}-\left(2x+1\right)^2=0\\-\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\3y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

Vậy: GTLN của đa thức \(B=-4x^2-9y^2-4x+6y+3\) là 5 khi và chỉ khi \(x=\frac{-1}{2}\) và \(y=\frac{1}{3}\)

Câu 3:

a) Ta có: \(x^2+y^2-2x+4y+5=0\)

\(\Rightarrow x^2-2x+1+y^2+4y+4=0\)

\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy: x=1 và y=-2

b) Ta có: \(5x^2+9y^2-12xy-6x+9=0\)

\(\Rightarrow x^2+4x^2+9y^2-12xy-6x+9=0\)

\(\Rightarrow\left(4x^2+12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

\(\Rightarrow\left(2x+3y\right)^2+\left(x-3\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)

Vậy: x=3 và y=-2