`a, (x-y)(x-5y)`

`= x^2 - xy - 5xy + 5y^2`

`= x^2 - 6xy + 5y^2`

`b, (2x+y)(4x^2 -2xy + y^2)`

`= (2x)^3 + y^3`

`= 8x^3 + y^3`

a) \(\left(x-y\right)\left(x-5y\right)\)

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

\(=x^2-6xy+5y^2\)

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

\(=8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3\)

\(=8x^3+y^3\)

\(25{x^2} + 20xy + 4{y^2} = {\left( {5x} \right)^2} + 2.5x.2y + {\left( {2y} \right)^2} = {\left( {5x + 2y} \right)^2}\)

Chọn D.

a) \(=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=\frac{-7}{3}\)

b)\(=\frac{3x\left(x+y\right)}{y}\)

c) \(\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)

\(=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)

a) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=-\frac{7}{3}.\)

b) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)

c) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)

d) \(\frac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\frac{x-z}{2}\)

h) \(\frac{3x\left(1-x\right)}{2\left(x-1\right)}=-\frac{3x\left(x-1\right)}{2\left(x-1\right)}=\frac{-3x}{2}\)

j) \(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)

Câu b) bạn xem lại nhé.

Học tốt ^3^

a) \(x+2y+\left(x-y\right)\)

\(=x+2y+x-y\)

\(=2x+y\)

b) \(2x+y-\left(3x-5y\right)\)

\(=2x+y-3x+5y\)

\(=-x+6y\)

c) \(3x^2-4y^2+6xy+7+\left(-x^2+y^2-8xy+9x+1\right)\)

\(=3x^2-4y^2+6xy+7-x^2+y^2-8xy+9x+1\)

\(=2x^2-3y^2-2xy+9x+8\)

d) \(4x^2y-2xy^2+8-\left(3x^2y+9xy^2-12xy+6\right)\)

\(=4x^2y-2xy^2+8-3x^2y-9xy^2+12xy-6\)

\(=x^2y-11xy^2+2+12xy\)

\(ĐK:x\ge\dfrac{1}{5};y\ge\dfrac{3}{8}\)

\(PT\left(1\right)\Leftrightarrow\dfrac{3x^2-3y^2}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=3\left(x+y\right)\\ \Leftrightarrow3\left(x+y\right)\left(\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x-y=\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}\\ \Leftrightarrow\left(x-y\right)=\dfrac{3\left(x^2-y^2\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}\\ \Leftrightarrow\left(x-y\right)\left[\dfrac{3\left(x+y\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}-1\right]=0\)

\(\Leftrightarrow x=y\)

Với \(x+y=0\Leftrightarrow x=-y\), thay vào PT 2

\(\Leftrightarrow3\left(-y\right)\left(y-7\right)+10=\sqrt{10\left(-y\right)-2}+2\sqrt{8y-3}\\ \Leftrightarrow3y\left(7-y\right)+10=\sqrt{-10y-2}+2\sqrt{8y-3}\)

ĐK: \(\left\{{}\begin{matrix}-10y-2\ge0\\8y-3\ge0\end{matrix}\right.\Leftrightarrow y\in\varnothing\)

Với \(x-y=0\Leftrightarrow x=y\), thay vào PT 2

\(\Leftrightarrow3x^2-21x+10=\sqrt{10x-2}+2\sqrt{8x-3}\left(x\ge\dfrac{3}{8}\right)\\ \Leftrightarrow3x^2-24x+9=\sqrt{10x-2}-\left(x+1\right)+2\sqrt{8x-3}-2x\)

\(\Leftrightarrow3\left(x^2-8x+3\right)=\dfrac{-x^2+8x-3}{\sqrt{10x-2}+\left(x+1\right)}+\dfrac{2\left(-x^2+8x-3\right)}{\sqrt{8x-3}+x}\\ \Leftrightarrow\left(x^2-8x+3\right)\left(3+\dfrac{1}{\sqrt{10x-2}+x+1}+\dfrac{2}{\sqrt{8x-3}+x}\right)=0\)

Dễ thấy ngoặc lớn vô nghiệm với \(x\ge\dfrac{3}{8}>0\)

\(\Leftrightarrow x^2-8x+3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{13}\left(n\right)\\x=4-\sqrt{13}\left(n\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=4+\sqrt{13}\\y=4-\sqrt{13}\end{matrix}\right.\)

Vậy HPT có nghiệm \(\left(x;y\right)\in\left\{\left(4+\sqrt{13};4+\sqrt{13}\right);\left(4-\sqrt{13};4-\sqrt{13}\right)\right\}\)

ko ghi đề bài nha làm luôn

a) \(\frac{\left(2x+2y\right)+\left(5x+5y\right)}{\left(2x+2y\right)-\left(5x+5y\right)}=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}=\frac{\left(2+5\right)\left(x+y\right)}{\left(2-5\right)\left(x+y\right)}=\frac{-7}{3}\)

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

a)ĐK: \(x\ne-y;x,y\ne0\)

\(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(2+5\right)}{\left(x+y\right)\left(2-5\right)}=-\frac{7}{3}\)

b) ĐK: ...bạn tự xét...

\(\frac{4x^2-4xy}{5x^3-5x^2y}=\frac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\frac{4x}{5x^2}=\frac{4}{5x}\)

Vậy ...

\(\hept{\begin{cases}\left(x+2y-2\right)\left(2x+y\right)=2x\left(5y-2\right)-2y\\x^2-7y=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x^2-5xy+2y^2=0\left(1\right)\\x^2-7y=-3\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\left(y-2x\right)\left(2y-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=2x\\x=2y\end{cases}}\)

Thế ngược lại (2) giải tiếp sẽ được nghiệm nhé.

1+2-2+444=?

\(\hept{\begin{cases}\left(x+2y-2\right)\left(2x+y\right)=2x\left(5y-2\right)-2y\left(1\right)\\x^2-7y=-3\left(2\right)\end{cases}}\)

Ta có PT (1) <=> \(2x^2+xy+4xy+2y^2-4x-2y=10xy-4x-2y\)

\(\Leftrightarrow2x^2-5xy+2y^2=0\Leftrightarrow\left(2x^2-4xy\right)+\left(2y^2-xy\right)=0\Leftrightarrow2x\left(x-2y\right)-y\left(2y-x\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(2x-y\right)=0\Leftrightarrow\orbr{\begin{cases}x-2y=0\\2x-y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2y\\2x=y\end{cases}}}\)

TH1: x=2y kết hợp với pt (2) ta có hệ \(\hept{\begin{cases}x=2y\\x^2-7y=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2y\\4y^2-7y+3=0\end{cases}}\)

<=> x=2y và \(\orbr{\begin{cases}x=1\\x=\frac{3}{4}\end{cases}}\)

<=> \(\hept{\begin{cases}x=1\\y=2\end{cases}}\)hoặc \(\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{3}{2}\end{cases}}\)

TH2: y=2x kết hợp với pt (2) ta có hệ \(\hept{\begin{cases}y=2x\\x^2-7y=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=2x\\x^2-14x+3=0\end{cases}}\)

\(\Leftrightarrow\)x=2y và \(\orbr{\begin{cases}x=7+\sqrt{46}\\x=7-\sqrt{46}\end{cases}}\)

<=> \(\hept{\begin{cases}x=7+\sqrt{46}\\y=14+2\sqrt{46}\end{cases}}\)hoặc \(\hept{\begin{cases}x=7-\sqrt{46}\\y=14-2\sqrt{46}\end{cases}}\)

Vậy hệ phương trình có 4 nghiệm \(\hept{\begin{cases}x=1\\y=2\end{cases};\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{3}{2}\end{cases};\hept{\begin{cases}x=7+\sqrt{46}\\y=14+2\sqrt{46}\end{cases};\hept{\begin{cases}x=7-\sqrt{46}\\y=14-2\sqrt{46}\end{cases}}}}}\)