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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
a: \(=\left(15x^2y^3-12x^2y^3\right)+\left(7x^2-12x^2\right)+\left(-8x^3y^2+11x^3y^2\right)\)
\(=3x^2y^3-5x^2+3x^3y^2\)
bậc là 5
b: \(=\left(3x^5y-\dfrac{1}{2}x^5y\right)+\left(\dfrac{1}{3}xy^4+2xy^4\right)+\left(\dfrac{3}{4}x^2y^3-x^2y^3\right)\)
\(=\dfrac{5}{2}x^5y+\dfrac{7}{3}xy^4-\dfrac{1}{4}x^2y^3\)
Bậc là 6
c: \(=5xy-2xy+4xy-y^2+3x-2y\)
\(=-y^2+3x-2y+7xy\)
Bậc là 2
a)= \(4x^2y+2x^2y-5x^2y-3y^3-5y^3-6xy^2\)
=\(2x^2y-8y^3-6xy\)
b) =\(2xyz-8xyz-11xy^3+2xy^3+4xy-2xy-11\)
=\(-6xyz-9xy^3+2xy-11\)
mình ko viết đề bài đâu 
2 câu còn lại làm tương tự nhé
a. \(4x^2y-3y^3-6xy^2-5y^3+2x^2y-5x^2y\)
\(=-8y^3+x^2y-6xy^2\)
b. \(2xyz-11xy^3-8xyz+2xy^3+4xy-11-2xy\)
\(=-6xyz-9xy^3+2xy-11\)
c. \(x\left(x-5\right)-3x\left(x-1\right)+6\left(x-2\right)\)
\(=x^2-5x-3x^2-3x+6x-12\)
\(=-2x^2-2x-12\)
d. \(x^3\left(x-2\right)-2x^2\left(x^2-x\right)+5\left(2x^4-1\right)\)
\(=x^4-2x^3-2x^4-2x^3+10x^4-5\)
\(=9x^4-4x^3-5\)
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) = \(\dfrac{2}{9}\)
ko viết lại đề nữa nhé bạn .
a, = \(2xy^3.\dfrac{1}{9}x^4y^2z^2\) = \(\dfrac{2}{9}x^5y^5z^2\)
b,=\(9x^6y^3.\dfrac{1}{81}x^4x^6\)= \(\dfrac{1}{9}x^{16}y^3\) câu này có vẻ sai đề ý bạn nhưng mk vẫn làm theo đề bạn đưa .
c,\(=-\dfrac{1}{2}x^2y^3z.4x^4y^2z^4\)\(=-2x^6y^5z^5\)
d, câu d, bạn ghi ko rõ là ngoặc bình phương ở đâu nên mk ko làm . lần sau ghi đề ghi cẩn thận nha bạn .
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
I . Trắc Nghiệm 1B . 2D . 3C . 5A II . Tự luận 2,a,Ta có: A+(x22y-2xy22+5xy+1)=-2x22y+xy22-xy-1 ⇔⇔ A=(-2x22y+xy22-xy-1) - (x22y-2xy22+5xy+1) =-2x22y+xy22-xy-1 - x22y+2xy22-5xy-1 =(-2x22y - x22y) + (xy22+ 2xy22) + (-xy - 5xy ) + (-1 - 1) = -3x22y + 3xy22 - 6xy - 2 b, thay x=1,y=2 vào đa thức A Ta có A= -3x22y + 3xy22 - 6xy - 2 = -3 . 122 . 2 + 3 .1 . 222 - 6 . 1 . 2 -2 = -6 + 12 - 12 - 2 = -8 3,Sắp xếp f(x) =9-x55+4x-2x33+x22-7x44 =9-x55-7x44-2x33+x22+4x g(x) = x55-9+2x22+7x44+2x33-3x =-9+x55+7x44+2x33+2x22-3x b,f(x) + g(x)=(9-x55-7x44-2x33+x22+4x) + (-9+x55+7x44+2x33+2x22-3x) =9-x55-7x44-2x33+x22+4x-9+x55+7x44+2x33+2x22-3x =(9-9)+(-x55+x55)+(-7x44+7x44)+(-2x33+2x33)+(x22+2x22)+(4x-3x) = 3x22 + x g(x)-f(x)=(-9+x55+7x44+2x33+2x22-3x) - (9-x55-7x44-2x33+x22+4x) =-9+x55+7x44+2x33+2x22-3x-9+x55+7x44+2x 33-x22-4x =(-9-9)+(x55+x55)+(7x44+7x44)+(2x33+2x33)+(2x22-x22)+(3x-4x) = -18 + 2x55 + 14x44 + 4x33 + x22 - x
a: \(=-10bx^3y^2\)
b: \(\dfrac{-4}{5}ab^2c\cdot\left(-20\right)a^4bx=16a^5b^3c\cdot x\)
c: \(=8\cdot\dfrac{1}{4}\cdot a^3\cdot b^2c^4=2a^3b^2c^4\)
d: \(=2ab\cdot\dfrac{4}{3}a^2\cdot b^4\cdot7abc=\dfrac{56}{3}a^4b^6c\)
tìm nghiệm của đa thức sau:
a,\(\left(-\dfrac{5}{3}x^2+\dfrac{3}{5}\right)\left(x^2-2\right)\)
Xét \(\left(-\dfrac{5}{3}x^2+\dfrac{3}{5}\right)\left(x^2-2\right)\) \(=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{5}{3}x^2+\dfrac{3}{5}=0\\x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{5}{3x}x^2=-\dfrac{3}{5}\\x^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{9}{25}\\\left[{}\begin{matrix}x=-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{9}{25}\\x=-\dfrac{9}{25}\end{matrix}\right.\\\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy nghiệm của đa thức \(\left(-\dfrac{5}{3}x^2+\dfrac{3}{5}\right)\left(x^2-2\right)\) là \(\left\{\dfrac{9}{25};-\dfrac{9}{25};\sqrt{2};-\sqrt{2}\right\}\)
\(A-\left(x^2+y^2-4xy\right)=x^2+4xy+3x^2\)
\(\Leftrightarrow A=x^2+4xy+3x^2+x^2+y^2-4xy\)
\(\Leftrightarrow A=5x^2+y^2\)
\(B+\left(-x^4+x^2-2x^3-\dfrac{1}{3}\right)=3x^2-2x^3+x-\dfrac{2}{3}\)
\(\Leftrightarrow B=3x^2-2x^3+x-\dfrac{2}{3}+x^4-x^2+2x^3+\dfrac{1}{3}\)
\(\Leftrightarrow B=x^4+2x^2+x-\dfrac{1}{3}\)
A=5x2+y2.
B=x4+2x2+x-1/3.