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a)\(A=x^3+x^2y-xy-y^2+3y+x-1\)
Ta có:\(x+y-2=0\Rightarrow x+y=2\)
\(A=x^2\left(x+y\right)-y\left(x+y\right)+3y+x-1\)
\(=2x^2-2y+3y+x-1\)
\(=2x^2+y+x-1\)
\(=2x^2+2-1\)
\(=2x^2+1\)
b) x - y = 0 => x = y
B = x( x^2 + y^2 ) - y ( x^2 + y^2 ) + 3
= x(x^2 + x^2 ) - x (x^2 + x^2 ) + 3
= 3
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) = \(\dfrac{2}{9}\)
\(x^2+2xy+x+2y\)
\(=x\left(x+1\right)+2y\left(x+1\right)\)
\(=\left(x+1\right)\left(2y+x\right)\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
a)x2+2xy+x+2y
=(2xy+x2)+(2y+x)
=x(2y+x)+(2y+x)
=(x+1)(2y+x)
b)7x2-7xy-5x+5y
=(5y-7xy)+(7x2-5x)
=y(5-7x)-x(5-7x)
=(5-7x)(y-x)
c)x2-6x+9-9y2
=(x2+3xy-3x)-(3xy+9y2-9y)-(3x+9y-9)
=x(x+3y-3)-3y(x+3y-3)-3(x+3y-3)
=(x-3y-3)(x+3y-3)
d)x3-3x2+3x-1+2(x2-x)
Ta thấy x=1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x-1
=(x-1)(x2+1)
e) (x+y)(y+z)(z+x)+xyz
đề sai
f)x(y2-z2)+y(z2-x2)
=(xy2+yz2)+(x2y+xz2)
=y(xy+z2)-x(xy+z2)
=(y-x)(xy+z2)
bài 1 :
B=15-3x-3y
a) x+y-5=0
=>x+y=-5
B=15-3x-3y <=> B=15-3(x+y)
Thay x+y=-5 vào biểu thức B ta được :
B=15-3(-5)
B=15+15
B=30
Vậy giá trị của biểu thức B=15-3x-3y tại x+y+5=0 là 30
b)Theo đề bài ; ta có :
B=15-3x-3.2=10
15-3x-6=10
15-3x=16
3x=-1
\(x=\frac{-1}{3}\)
Bài 2:
a)3x2-7=5
3x2=12
x2=4
x=\(\pm2\)
b)3x-2x2=0
=> 3x=2x2
=>\(\frac{3x}{x^2}=2\)
=>\(\frac{x}{x^2}=\frac{2}{3}\)
=>\(\frac{1}{x}=\frac{2}{3}\)
=>\(3=2x\)
=>\(\frac{3}{2}=x\)
c) 8x2 + 10x + 3 = 0
=>\(8x^2-2x+12x-3=0\)
\(\Rightarrow\left(2x+3\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\4x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{1}{4}\end{cases}}}\)
vậy \(x\in\left\{-\frac{3}{2};\frac{1}{4}\right\}\)
Bài 5 đề sai vì |1| không thể =2
a/Ta có :
\(x+y+1=0\Leftrightarrow x+y=-1\)
\(A=x^2\left(x+y\right)-y^2\left(x+y\right)+x^2-y^2+2\left(x+y\right)+3\)
Mà \(x+y=-1\)
\(\Leftrightarrow A=x^2.\left(-1\right)-y^2.\left(-1\right)+x^2-y^2+2.\left(-1\right)+3\)
\(\Leftrightarrow A=-x^2+y^2+x^2-y^2-2+3\)
\(\Leftrightarrow A=\left(-x^2+x\right)+\left(y^2-y^2\right)-\left(2-3\right)\)
\(\Leftrightarrow A=0+0-\left(-1\right)\)
\(\Leftrightarrow A=1\)
Vậy ..
thanks bạn nha