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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
b: \(3x^2y^3=\dfrac{1}{9}\)
\(\Leftrightarrow3x^2=\dfrac{1}{9}:\dfrac{1}{27}=3\)
=>x=1 hoặc x=-1
a: \(A=\dfrac{-2}{3}\cdot\left(-27\right)\cdot4\cdot\dfrac{1}{2}=18\cdot2=36\)
Bài 1:
Giải:
Ta có: \(\left\{{}\begin{matrix}3x=4y\\5y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{6}=\dfrac{z}{5}\end{matrix}\right.\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)
Mà \(xyz=30\)
\(\Rightarrow240k^3=30\)
\(\Rightarrow k^3=\dfrac{1}{8}\)
\(\Rightarrow k=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=3\\z=2,5\end{matrix}\right.\)
Vậy...
Bài 2: sai đề
Bài 3:
Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)
Ta có: \(x+2y+3z=38\)
\(\Rightarrow2k+1+8k-6+18k+15=38\)
\(\Rightarrow28k=28\)
\(\Rightarrow k=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=1\\z=11\end{matrix}\right.\)
Vậy...
1) Ta có :
\(3x=4y\Rightarrow\dfrac{3x}{12}=\dfrac{4y}{12}\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\) <=> \(\dfrac{x}{8}=\dfrac{y}{6}\)
\(5y=6z\Rightarrow\dfrac{5y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\)
=> \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)
Thay vào đẳng thức xyz = 30
=> 8k.6k.5k = 30
<=> 240k3 = 30
<=> k3 = 8
<=> k = 2
\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)
b) Câu này cũng tương tự câu 1 nha ! Đặt k luôn , còn không bình phương lên rồi dùng tính chất dãy tỉ số bằng nhau .
c) Đặt \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=k\)
=> \(\left\{{}\begin{matrix}x=2k+1\\y=4k-3\\z=6k+5\end{matrix}\right.\)
Thay vào đẳng thức , ta có :
x + 2y + 3z = 2k + 1 + 2(4k - 3) + 3(6k + 5) = 38
=> 28k = 38
=> k = \(\dfrac{19}{14}\)
Vậy .....
Ta có:\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2\left(x+y+x\right)}{x+y+z}=2\)(theo tính chất của DTSBN)
Suy ra:\(\dfrac{1}{x+y+z}=2\)=>x+y+z=\(\dfrac{1}{2}\)
=>y+z=\(\dfrac{1}{2}\)-x
Tương tự, ta có được:
x+z=\(\dfrac{1}{2}-y\)
x+y=\(\dfrac{1}{2}-z\)
Thay các kết quả vừa tìm được, ta có:
\(\dfrac{0,5-x+1}{x}=\dfrac{0,5-y+2}{y}\dfrac{0,5-z-3}{z}=2\)=>\(\dfrac{1,5-x}{x}=\dfrac{2,5-y}{y}=\dfrac{-2,5-z}{z}=2\)
=>x=\(\dfrac{1}{2},y=\dfrac{5}{6},z=\dfrac{-5}{6}\)
Thay x=\(\dfrac{1}{2},y=\dfrac{5}{6},z=\dfrac{-5}{6}\)vào biểu thức A, ta có:
A=2018.\(\dfrac{1}{2}\)+\(\left(\dfrac{5}{6}\right)^{2017}\)+\(\left(\dfrac{-5}{6}\right)^{2017}\)
=>A=1009+\(\left[\left(\dfrac{5}{6}\right)^{2017}+\left(\dfrac{-5}{6}\right)^{2017}\right]\)
=>A=1009+0
=>A=1009
Vậy giá trị của biểu thức A là 1009
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
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 .
Bạn thay x, y, z vào đơn thức là được mà! Mấy đơn thức này còn thu gọn rồi! Bạn tự làm đi
a , thay vào
=> 15 . 8 . -8 . 27 = -25920
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