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

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

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

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

Vậy \(A\ne B\)

(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)

= x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5

= (x^4y-x^4y)+(x^3y^2-x^3y^2)+(x^2y^3)+(xy^4-xy^4)+x^5-y^5

= 0+0+0+0+x^5-y^5

= x^5-y^5

Vay (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4) = x^5-y^5

nho k nha

a)\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)

\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)

\(=x^5-y^5+\left(x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(x^2y^3-x^2y^3\right)+\left(xy^4-xy^4\right)\)

\(\Rightarrow\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)

b)\(\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)

\(=a^3+b^3+\left(-a^2b+a^2b\right)+\left(ab^2-ab^2\right)\)

\(\Rightarrow\)\(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)

a) (x - y)(x⁴ + x³y + x²y² + xy³ + y⁴)

= x(x⁴ + x³y + x²y² + xy³ + y⁴) - y(x⁴ + x³y + x²y² + xy³ + y⁴)

= x⁵ + x⁴y + x³y² + x²y³ + xy⁴ - x⁴y - x³y² - x²y³ - xy⁴ - y⁵

= x⁵ - y⁵

b) (a + b)(a² - ab + b²)

= a(a² - ab + b²) + b(a² - ab + b²)

= a³ - a²b + ab² + a²b - ab² + b³

= a³ + b³

\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)

\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)

Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))

2. CMR:

a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)

Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.

b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)

Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)

=> đpcm.

c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)

\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)

=> đpcm.

1.

b. \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)

\(\Leftrightarrow4\left(x^2+5x-x-5\right)-\left(x^2+5x+2x+10\right)=3\left(x^2+2x-x-2\right)\)

\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)

\(\Leftrightarrow4x^2+20x-4x-x^2-5x-2x-3x^2-6x+3x=20+10-6\)

\(\Leftrightarrow6x=24\)

\(\Leftrightarrow x=4\)

Vậy ....

\(a\text{)}.\: \left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow8x-5x^2+16-10x+4x^2-4x-8+2x^2-8=0\\ \Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

a

a

a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)

=3(x-y)^2+9

b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)

c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)

Ta có hpt \(\left\{{}\begin{matrix}xy+3y-5x-15=xy\\2xy+30x-y^2-15y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x=3y-15\\6\left(3y-15\right)-y^2-15y=0\end{matrix}\right.\)

Ta có pt (2) \(\Leftrightarrow3y-y^2-80=0\Leftrightarrow y^2-3y+80=0\left(VN\right)\)

=> hpy vô nghiệm

c) Ta có hpt \(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left(xy+x+y\right)=30\\xy\left(x+y\right)+xy+x+y=11\end{matrix}\right.\)

Đặt j\(xy\left(x+y\right)=a;xy+x+y=b\), ta có hpt

\(\left\{{}\begin{matrix}ab=30\\a+b=11\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=5;b=6\\a=6;b=5\end{matrix}\right.\)

với a=5;b=6, ta có \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}xy=1;x+y=5\\xy=5;x+y=1\end{matrix}\right.\)

đến đây thì thế y hoặc x ra pt bậc 2, còn TH còn lại bn tự giải nhé !

b) Ta có hpt <=> \(\left\{{}\begin{matrix}2\sqrt{x}-3y+2=-4z^2\\2\sqrt{3x}+4y-2=6z^2\\-3\sqrt{x}+y-4=-2z^2\end{matrix}\right.\)

cộng 3 vế của 3 pt, ta có \(\left(2\sqrt{3}-1\right)\sqrt{x}=4\Leftrightarrow\sqrt{x}=\dfrac{4}{2\sqrt{3}-1}\Leftrightarrow x=\dfrac{16}{\left(2\sqrt{3}-1\right)^2}\)

đến đây thay căn(x)=...vào và đặt z^2=m, ta sẽ ra 1 hệ mới chỉ có 2 ẩn y và m bậc 1 , lát thế vào sẽ ra bậc 2 thì dễ rồi !

Bài 1:

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

= \(\left(\frac{1}{5}-3\right)x^4y^3\)

= \(-\frac{14}{5}x^4y^3.\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

= \(\left(5-\frac{1}{4}\right)x^2y^5\)

= \(\frac{19}{4}x^2y^5.\)

Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.

Chúc bạn học tốt!