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a/ \(C=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-1\right)\)
\(C=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-1\right)=x+y-1\) (do x+y-2=0)
Mà x+y-2=0 => x+y-1=1 => C=1
b/ Với x=2; y=2 Ta nhận thấy \(x^3-2y^2=2^3-2.2^2=2^3-2^3=0\) => D=0
2) Ta có: \(\frac{x_1}{y_2}=\frac{x_2}{y_1}\Rightarrow\frac{x_1^2}{y_2^2}=\frac{x_2^2}{y_1^2}=\frac{x_1^2+x_2^2}{y_1^2+y_2^2}=\frac{2^2+3^2}{52}=\frac{1}{4}\)
\(\Rightarrow\frac{x_1^2}{y_2^2}=\frac{1}{4}\Rightarrow y_2^2=16\Rightarrow\)\(\orbr{\begin{cases}y_2=-4\\y_2=4\end{cases}\Rightarrow}\)\(\orbr{\begin{cases}y_1=-6\\y_1=6\end{cases}}\)
=> KL....
I2x+3I=x+2
TH1: Nếu \(x\le-\frac{3}{2}\)(*), =>I2x+3I=-2x-3
PT: -2x-3=x+2 <=> x=\(-\frac{5}{3}\)(tm (*))
TH2: Nếu \(x>-\frac{3}{2}\)(**), => I2x+3I=2x+3
PT: 2x+3=x+2 => x=-1 (tm (**))
Vậy x=...
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, Thay x=-1 vào biểu thức A ta có:
\(A=2\left(-1\right)^2+\left(-1\right)+1\)
\(A=2.1+\left(-1\right)+1\)
\(A=2\)
Thay \(x=\dfrac{1}{4}\) vào biểu thức A ta có:
\(A=2\left(\dfrac{1}{4}\right)^2+\dfrac{1}{4}+1\)
\(A=2.\dfrac{1}{16}+\dfrac{1}{4}+1\)
\(A=\dfrac{1}{8}+\dfrac{1}{4}+1\)
\(A=\dfrac{1}{8}+\dfrac{2}{8}+1\)
\(A=\dfrac{11}{8}\)
b, Thay x=-1; y=3 vào biểu thức B ta có:
\(B=\left(-1\right)^2.3^2+\left(-1\right).3+\left(-1\right)^3+3^3\)
\(B=1.9-3-1+27\)
\(B=2+27\)
\(B=29\)
c, Thay x=-1 vào biểu thức C ta có:
\(C=\left(-1\right)^2+\left(-1\right)^4+\left(-1\right)^6+\left(-1\right)^8+...+\left(-1\right)^{100}\)
\(C=1^4+1^6+1^8+1^9+...+1^{100}\)
\(C=100\)
d, Thay x+y=3; xy=-5 vào biểu thức D ta có:
\(D=3.\left(x+1\right).\left(y+1\right)\)
\(D=3.\left[\left(x.y\right)+1\right]\)
\(D=3.\left[\left(-5\right)+1\right]\)
\(D=3.\left(-4\right)\)
\(D=-12\)
Tích mình nha!!!
a: \(B=xy\left(y^2-x\right)=1\cdot1\cdot\left(1^2-1\right)=0\)
b: \(B=xy\left(y^2-x\right)=-1\cdot\left(-1\right)\cdot\left[\left(-1\right)^2-\left(-1\right)\right]=1\cdot\left(1+1\right)=2\)
c: \(B=xy\left(y^2-x\right)=1\cdot\left(-1\right)\cdot\left[\left(-1\right)^2-1\right]=0\)
d: \(B=xy\left(y^2-x\right)=-1\cdot1\cdot\left[1^2-\left(-1\right)^2\right]=0\)
Bài 2:
\(B=xy^3-x^2y=xy\left(y^2-x\right)\)
a) \(x=1;y=1\Rightarrow B=1.1.\left(1^2-1\right)=0\)
b) \(x=-1;y=-1\Rightarrow B=\left(-1\right).\left(-1\right).\left[\left(-1\right)^2-\left(-1\right)\right]=2\)
c) \(x=1;y=-1\Rightarrow B=1.\left(-1\right).\left[\left(-1\right)^2-\left(-1\right)\right]=-2\)
d) \(x=-1;y=1\Rightarrow B=\left(-1\right).1.\left[1^2-\left(-1\right)\right]=-2\)