Question

Bài 1: Tính giá trị các biểu thức sau tại: |x| = \(\dfrac {1}{3}\); |y| = 1
a) A= 2x² - 3x + 5     b) B= 2x² - 3xy + y2
Bài 2: Tính giá trị các biểu thức A sau biết x + y +1 = 0:
A= x (x + y) - y2 (x + y) + x² - y² + 2 (x + y) + 3
Bài 3: Cho x.y.z = 2 và x + y + z = 0. Tính giá trị biểu thức:
A= (x + y)(y + z)(z + x)
Bài 4: Tìm các giá trị của các biến để các biểu thức sau có giá trị bằng 0:
a) |2x - \(\dfrac {1}{3}\)| - \(\dfrac {1}{3}\)      b) |2x - \(\dfrac {1}{3} \)| - \(\dfrac {1}{3}\)       c) |3x + 2\(\dfrac {1}{3} \)| + |y + 2| = 0       d) (x - 2)² + (2x - y + 1)² = 0

Answer 1

Bài 1:

|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}

A(-1) = 2(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)) + 5

A(-1) = \(\dfrac{2}{9}\) + 1 + 5

A (-1) = \(\dfrac{56}{9}\)

A(1) = 2.(\(\dfrac{1}{3}\) )²- \(\dfrac{1}{3}\).3 + 5

A(1) = \(\dfrac{2}{9}\) - 1 + 5

A(1) = \(\dfrac{38}{9}\)

 

Answer 2

|y| = 1 ⇒ y \(\in\) {-1; 1} 

⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))

B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)²

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)

B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).1 + 1²

B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1

B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\) 

B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)²

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)

B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).1 + (1)²

B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1

B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)

 

Answer 3

Bài 2:

\(x+y+1=0\Rightarrow x+y=-1\)

A = \(x\)(\(x+y\)) - y².(\(x+y\)) + \(x^2\) - y² + 2(\(x+y\)) + 3

Thay \(x\) + y  = -1 vào biểu thức A ta có:

A = \(x\).( -1) - y² .(-1) + \(x^2\)  - y² + 2(-1) + 3

A = -\(x\) + y² + \(x^2\) - y² - 2 + 3

A = \(x^2\) - \(x\) + 1

Answer 4

Bài 4:

a; |2\(x\) - \(\dfrac{1}{3}\)| - \(\dfrac{1}{3}\) = 0

    |2\(x-\dfrac{1}{3}\)| = \(\dfrac{1}{3}\)

     \(\left[{}\begin{matrix}2x-\dfrac{1}{3}=-\dfrac{1}{3}\\2x-\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)

      \(\left[{}\begin{matrix}2x=0\\2x=\dfrac{2}{3}\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x\in\) {0; \(\dfrac{1}{3}\)}

 

Answer 5

Bài 4 

c; |3\(x\)\(\) +2\(\dfrac{1}{3}\)| +|y + 2| = 0

       \(\left\{{}\begin{matrix}3x+2\dfrac{1}{3}=0\\y+2=0\end{matrix}\right.\)

          \(\left\{{}\begin{matrix}3x=-\dfrac{7}{3}\\y=-2\end{matrix}\right.\)

          \(\left\{{}\begin{matrix}x=-\dfrac{7}{9}\\y=-2\end{matrix}\right.\)

Vậy (\(x;y\)) = (-\(\dfrac{7}{9}\); -2)

    

Answer 6

Bài 4 d

(\(x-2\))² + (2\(x-y+1\))² = 0

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

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

\(\left\{{}\begin{matrix}x=2\\2.2-y+1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=2\\5-y=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

Vậy (\(x;y\)) = (2; 5)