a)p+(x² -2y²)=x² -y² +3y² -1

\(\Rightarrow\)p=(x² -2y² +3y² -1)-(x-2y).(x+2y)

\(\Rightarrow\)p=(x²-2y² +3y² -1)-(x² +2xy-2xy)

\(\Rightarrow\)p=x² -2y²+3y² -1-x² -2xy+2xy

\(\Rightarrow\) p=2y²-1 

Vậy P=2y²-1

b)Q-(5x²-xyz)=xy+2x²-3xyz+5

\(\Rightarrow\)Q=(xy+2x²-3xyz+5)+(5x²-xyz)

\(\Rightarrow\)Q=7x²-4xyz+xy+5

Vậy Q=7x²-4xyz+xy+5

Tìm đa thức p biết : P + (2xy + 4) = 2x^2y - 2xy + 1

a) P + (x² - 2y²) = x² - y² + 3y² - 1

⇔ P = (x² - y² + 3y² - 1) - (x² - 2y²)

⇔ P = x² + 2y² - 1 - x² + 2y²

⇔ P = 4y² - 1

b) Q - (5x² - xyz) = xy + 2x² - 3xyz + 5

⇔ Q = (xy + 2x² - 3xyz + 5) + (5x² - xyz)

⇔ Q = xy + 2x² - 3xyz + 5 + 5x² - xyz

⇔ Q = xy + 7x² - 4xyz + 5

a) P + (x² – 2y²) = x² – y² + 3y² – 1

P = (x² – y² + 3y² – 1) - (x² – 2y²)

P = x² – y² + 3y² – 1 - x² + 2y²

P = x² – x² – y² + 3y² + 2y² – 1

P = 4y² – 1.

Vậy P = 4y² – 1.

b) Q – (5x² – xyz) = xy + 2x² – 3xyz + 5

Q = (xy + 2x² – 3xyz + 5) + (5x² – xyz)

Q = xy + 2x² – 3xyz + 5 + 5x² – xyz

Q = 7x² – 4xyz + xy + 5

Vậy Q = 7x² – 4xyz + xy + 5.

a) P + (x2 – 2y2) = x2 – y2 + 3y2 – 1

P = (x2 – y2 + 3y2 – 1) – (x2 – 2y2)

P = x2 – y2 + 3y2 – 1 – x2 + 2y2

P = x2 – x2 – y2 + 3y2 + 2y2 – 1

P = 4y2 – 1

b) Q – (5x2 – xyz) = xy + 2x2 – 3xyz + 5

Q = (xy + 2x2 – 3xyz + 5) + (5x2 – xyz)

Q = xy + 2x2 – 3xyz + 5 + 5x2 – xyz

Q = 7x2– 4xyz + xy + 5

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)

bài 11

a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)

b)

\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)

c)

\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

bài 12

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x=26\\ x=-2\)

b)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)

Xét pt:

\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)

TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)

\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)

\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)

TH2: \(y=6-x\) thế vào...

\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)

\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)

a) \(B=-\frac{1}{2}x^3y\left(-2xy^2\right)^2\)

\(B=\left(-\frac{1}{2}.-2\right).\left(x^3.x\right)\left(y.y^2\right)^2\)

\(B=1x^4y^5\)

Hệ số: 1

Bậc: 9

Chưa định hình phần b) nó là như nào

a, Trừ vế theo vế hai phương trình ta được

\(x^2+6y-y^2-6x=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=6-y\end{matrix}\right.\)

Nếu \(x=y,pt\left(1\right)\Leftrightarrow x^2+x=5x+3\)

\(\Leftrightarrow x^2-4x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y=2+\sqrt{7}\\x=y=2-\sqrt{7}\end{matrix}\right.\)

Nếu \(x=6-y,pt\left(2\right)\Leftrightarrow y^2+6-y=5y+3\)

\(\Leftrightarrow y^2-6y+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=3+\sqrt{6}\\y=3-\sqrt{6}\end{matrix}\right.\)

\(y=3+\sqrt{6}\Rightarrow x=3-\sqrt{6}\)

\(y=3-\sqrt{6}\Rightarrow x=3+\sqrt{6}\)

b, Trừ vế theo vế hai phương trình

\(3x^3-3y^3=y^2-x^2\)

\(\Leftrightarrow3\left(x-y\right)\left(x^2+xy+y^2+x+y\right)=0\)

Từ \(pt\left(1\right)\) \(3x^3=y^2+2>0\Rightarrow x>0\)

Tương tự \(y>0\)

\(\Rightarrow x^2+xy+y^2+x+y>0,\forall x;y\)

\(\Rightarrow x=y\)

\(pt\left(1\right)\Leftrightarrow3x^3=x^2+2\)

\(\Leftrightarrow3x^3-x^2-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)

\(\Leftrightarrow x=y=1\left(\text{vì }3x^2+2x+2=2x^2+\left(x+1\right)^2+1>0\right)\)

2:

a: A(x)=0

=>5x-10-2x-6=0

=>3x-16=0

=>x=16/3

b: B(x)=0

=>5x^2-125=0

=>x^2-25=0

=>x=5 hoặc x=-5

c: C(x)=0

=>2x^2-x-3=0

=>2x^2-3x+2x-3=0

=>(2x-3)(x+1)=0

=>x=3/2 hoặc x=-1

\(\begin{array}{l}A + B = \left( {5{x^2}y + 5x - 3} \right) + \left( {xy - 4{x^2}y + 5x - 1} \right)\\ = 5{x^2}y + 5x - 3 + xy - 4{x^2}y + 5x - 1\\ = \left( {5{x^2}y - 4{x^2}y} \right) + xy + \left( {5x + 5x} \right) + \left( { - 3 - 1} \right)\\ = {x^2}y + xy + 10x - 4\end{array}\)