cho x-y-2=0. tính giá trị của biểu thức: A=(1-2/x).(1-x/y).(1+y/2)
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\(A=x^2+y^2+xy=\left(x+y\right)^2-2xy+xy\\ A=1-xy\)
Mà \(x+y=1\Leftrightarrow x=1-y\)
\(\Leftrightarrow A=1-\left(1-y\right)y=1-y+y^2=\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{3}{4}\\ A=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\\ A_{min}=\dfrac{3}{4}\Leftrightarrow x=y=\dfrac{1}{2}\)
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a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)
b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)
Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)
c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)
Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{2}{xy}:\left(\dfrac{y-x}{xy}\right)^2-\left(\dfrac{x^2+y^2}{\left(x-y\right)^2}\right)\)
\(=\dfrac{2}{xy}\cdot\dfrac{\left(xy\right)^2}{\left(x-y\right)^2}-\dfrac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\dfrac{2xy-x^2-y^2}{\left(x-y\right)^2}=-1\)
2:
\(P=\dfrac{\left(5x+3\right)^2}{3x-2}\cdot\dfrac{\left(3x-2\right)\left(3x+2\right)}{5x+3}=\left(5x+3\right)\left(3x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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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) = \(\dfrac{2}{9}\)
\(A=\left(1-\frac{2}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{2}\right)\)
\(=\left(\frac{x-2}{x}\right)\left(\frac{y-x}{y}\right)\left(\frac{2+y}{2}\right)\)
mà \(x-2=y;y-x=-2;2+y=x\)
\(=\frac{-2xy}{2xy}=-1\)
giúp mk vs