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Lời giải:
a. $3x-5y+1=3.\frac{1}{3}-5.\frac{-1}{5}+1=1+1+1=3$
b.
Với $x=1$ thì $3x^2-2x-5=3.1^2-2.1-5=-4$
Với $x=-1$ thì $3x^2-2x-5=3(-1)^2-2.(-1)-5=0$
Với $x=\frac{5}{3}$ thì $3x^2-2x-5=3(\frac{5}{3})^2-2.\frac{5}{3}-5=0$
c.
$x-2y^2+z^3=4-2.(-1)^2+(-1)^3=1$
d.
$xy-x^2-xy^3=(-1)(-1)-(-1)^2-(-1)(-1)^3=-1$
Theo định nghĩa đơn thức, các biểu thức sau là đơn thức
b) 9x2yz;
c) 15,5;
Các biểu thức a) \(\frac{2}{5}\) + x2y; d) 1 - \(\frac{5}{9}\)x3; không phải là đơn thức vì chúng có chứa phép cộng hoặc phép trừ.
ý b , c là các đơn thức
ý a , d ko phải là đơn thức vì chúng có dấu công và dấu trừ
Lời giải:
$2ax(-y)^3-x(\frac{-2}{3}y)^2+b(by)^2$
$=-2axy^3-\frac{4}{9}xy^2+b^2y^2$
$ay^2(-7xz)^2byz^3=ay^2.(49x^2z^2)byz^3$
$=49abx^2y^3z^5$
a) A + x2 - 4xy2 + 2xz - 3y2 = 0
=> A = -x2 + 4xy2 - 2xz + 3y2
b) B + 5x2 - 2xy = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - 5x2 + 2xy= x2 + 11xy - y2
c) 3xy - 4y2 - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - x2 + 7xy - 8y2 = -12y2 + 10xy - x2
Trả lời:
a, A + ( x2 - 4xy2 + 2xz - 3y2 ) = 0
=> A = - ( x2 - 4xy2 + 2xz - 3y2 ) = - x2 + 4xy2 - 2xz + 3y2
b, B + ( 5x2 - 2xy ) = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - ( 5x2 - 2xy ) = 6x2 + 9xy - y2 - 5x2 + 2xy = x2 + 11xy - y2
c, ( 3xy - 4y2 ) - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - ( x2 - 7xy + 8y2 ) = 3xy - 4y2 - x2 + 7xy - 8y2 = 10xy - 12y2 - x2
d, B + ( 4x2y + 5y2 - 3xz + z2 ) = x2 + 11xy - y2 + 4x2y + 5y2 - 3xz + z2 = x2 + 11xy + 4y2 + 4x2y - 3xz + z2
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) = \(\dfrac{38}{9}\)
|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) = \(\dfrac{2}{9}\)
bài 1:
|x| = \(\dfrac{1}{3}\) => x = \(\pm\)\(\dfrac{1}{3}\) |y| = 1 => y = \(\pm\)1
a
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\) +5 = 2.\(\dfrac{1}{9}\) - 1 + 5
= \(\dfrac{2}{9}\) - 1 + 5 = \(\dfrac{2-9+45}{9}\) = \(\dfrac{38}{9}\)
+) A = 2x\(^2\) - 3x + 5
= 2\(\left(\dfrac{-1}{3}\right)^2\) - 3\(\left(\dfrac{-1}{3}\right)\) + 5
= 2.\(\dfrac{1}{9}\) - (-1) + 5 = \(\dfrac{2}{9}\) + 1 +5
= \(\dfrac{2+9+45}{9}\) = \(\dfrac{56}{9}\)
b) +) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{1}{3}\right)^2\) - 3.\(\dfrac{1}{3}\).1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - 1 + 1 = \(\dfrac{2}{9}\) - 1 + 1
= \(\dfrac{2-9+9}{9}\) = \(\dfrac{2}{9}\)
+) B = 2x\(^2\) - 3xy + y\(^2\)
= 2\(\left(\dfrac{-1}{3}\right)\)\(^2\) - 3\(\left(\dfrac{-1}{3}\right)\). 1 + 1\(^2\)
= 2.\(\dfrac{1}{9}\) - (-1) + 1 = \(\dfrac{2}{9}\) + 1 + 1
= \(\dfrac{2+9+9}{9}\) = \(\dfrac{20}{9}\)
bài 3
x.y.z = 2 và x + y + z = 0
A = ( x + y )( y +z )( z + x )
= x + y . y + z . z + x = ( x + y + z ) + ( x . y . z )
= 0 + 2 = 2
bài 4
a) | 2x - \(\dfrac{1}{3}\) | - \(\dfrac{1}{3}\) = 0 => | 2x - \(\dfrac{1}{3}\) | = \(\dfrac{1}{3}\)
=> 2x - \(\dfrac{1}{3}\) = \(\pm\) \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\)= \(\dfrac{1}{3}\)
=> 2x = \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
x = \(\dfrac{2}{3}\) : 2 = \(\dfrac{2}{3}\) . \(\dfrac{1}{2}\) = \(\dfrac{1}{3}\)
+) 2x - \(\dfrac{1}{3}\) = \(\dfrac{-1}{3}\)
2x = \(\dfrac{-1}{3}\) + \(\dfrac{1}{3}\) = 0
x = 0 : 2 = 2
I . Trắc Nghiệm
1B . 2D . 3C . 5A
II . Tự luận
2,a,Ta có: A+(x\(^2\)y-2xy\(^2\)+5xy+1)=-2x\(^2\)y+xy\(^2\)-xy-1
\(\Leftrightarrow\) A=(-2x\(^2\)y+xy\(^2\)-xy-1) - (x\(^2\)y-2xy\(^2\)+5xy+1)
=-2x\(^2\)y+xy\(^2\)-xy-1 - x\(^2\)y+2xy\(^2\)-5xy-1
=(-2x\(^2\)y - x\(^2\)y) + (xy\(^2\)+ 2xy\(^2\)) + (-xy - 5xy ) + (-1 - 1)
= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
b, thay x=1,y=2 vào đa thức A
Ta có A= -3x\(^2\)y + 3xy\(^2\) - 6xy - 2
= -3 . 1\(^2\) . 2 + 3 .1 . 2\(^2\) - 6 . 1 . 2 -2
= -6 + 12 - 12 - 2
= -8
3,Sắp xếp
f(x) =9-x\(^5\)+4x-2x\(^3\)+x\(^2\)-7x\(^4\)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x
g(x) = x\(^5\)-9+2x\(^2\)+7x\(^4\)+2x\(^3\)-3x
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
b,f(x) + g(x)=(9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x) + (-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x)
=9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x
=(9-9)+(-x\(^5\)+x\(^5\))+(-7x\(^4\)+7x\(^4\))+(-2x\(^3\)+2x\(^3\))+(x\(^2\)+2x\(^2\))+(4x-3x)
= 3x\(^2\) + x
g(x)-f(x)=(-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x) - (9-x\(^5\)-7x\(^4\)-2x\(^3\)+x\(^2\)+4x)
=-9+x\(^5\)+7x\(^4\)+2x\(^3\)+2x\(^2\)-3x-9+x\(^5\)+7x\(^4\)+2x \(^3\)-x\(^2\)-4x
=(-9-9)+(x\(^5\)+x\(^5\))+(7x\(^4\)+7x\(^4\))+(2x\(^3\)+2x\(^3\))+(2x\(^2\)-x\(^2\))+(3x-4x)
= -18 + 2x\(^5\) + 14x\(^4\) + 4x\(^3\) + x\(^2\) - x
Câu 2. Trong các biểu thức sau, biểu thức nào là biểu thức đại số:
A. (34 – 5) : 8 B. (x + y)2 C. x2 + 2x + 1 D. Cả 3 đáp án trên
cảm ơn nhoa