\(2x^2+30xy=5\left(x+5y\right)\sqrt{5xy}-50y^2\)\(\left(đk:x;y\ge0\right)\)

\(\Leftrightarrow2x^2+30xy-5\left(x+5y\right)\sqrt{5xy}+50y^2=0\left(1\right)\)

\(đặt:\sqrt{5xy}=b\ge0\Rightarrow5xy=b^2\Rightarrow10xy=2b^2\)

\(x+5y=a\ge0\Rightarrow x^2+10xy+25y^2=â^2\)

\(\Rightarrow2a^2=2x^2+20xy+50y^2\)

\(\Leftrightarrow\left(1\right)\Leftrightarrow2a^2+2b^2-5ab=0\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}b=2a\left(2\right)\\a=2b\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow\sqrt{5xy}=2x+10y\Leftrightarrow4x^2+35xy+100y^2=0\left(4\right)\)

\(với:y=0\) \(ko\) \(là\) \(nghiệm\)

\(với:y\ne0\Rightarrow\left(4\right)\Leftrightarrow4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100=0\)\(\left(vô-lí\right)\)

\(do:4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100>0\)

\(\left(3\right)\Rightarrow x+5y=2\sqrt{5xy}\Leftrightarrow x^2+10xy+25y^2=20xy\Leftrightarrow x^2-10xy+25y^2=0\Leftrightarrow\left(x-5y\right)^2=0\Leftrightarrow x=5y\)

\(thay:x=5y\) \(vào:2x^2+y^2=51\Rightarrow2\left(5y\right)^2+y^2-51=0\Leftrightarrow51y^2-51=0\Leftrightarrow\left[{}\begin{matrix}y=1\left(tm\right)\Rightarrow x=5\left(tm\right)\\y=-1\left(loại\right)\end{matrix}\right.\)

`a)100x^2-20x+1`

`=(10x-1)^2`

Thay `x=1/10`

`=>100x^2-20x+1=(1-1)^2=0`

`b)49x^2-42x+10`

`=49*4/49-42*2/7+10`

`=4-12+10=2`

`c)25x^2+40x+16y^2`

`=(5x+4y)^2=(2+3)^2=25`

a. (2x - 5)²

b. x(x² - 1) = x(x - 1)(x + 1)

c. (x - 3)(x² + 3x + 9)

d. 5x(x - y) + (x - y) = (x - y)(5x + 1)

Lời giải:
$(4x-2)^2-16x+16x=(4x-2)^2$

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$(3x-4)(3x+4)-(20x^2y-15xy^2):(5xy)$

$=(3x-4)(3x+4)-(4x-3y)$ không phân tích được thành nhân tử.

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$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$

$=3(x-2)(x^2+2x+4)-240x$

$=3(x^3-2^3)-240x=3x^3-240x-24$

$=3(x^3-80x-8)$

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-5xy=-1\\2\left(x+y\right)^2-4xy+xy=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-5xy=-1\\2\left(x+y\right)^2-3xy=8\end{matrix}\right.\)....(1)

đặt : \(\left\{{}\begin{matrix}xy=u\\x+y=v\end{matrix}\right.\) \(\Rightarrow\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}v^2-5u=-1\\2v^2-3u=8\end{matrix}\right.\) giải phương trình này bằng phương pháp thế

sau khi tìm được \(u\) và \(v\) tiếp đến ta áp dụng định lí vi ét đảo để tìm \(x\) và \(y\)

A=\(\frac{16x^2-40xy}{8x^2-24xy}=\frac{8x(2x-5y)}{ 8x(x-3y)} =\frac{2x-5y}{x-3y} \)

\(\frac{x}{y}=\frac{10}{3}<=>10y=3x <=>y=\frac{3}{10}x \)

=>A=(\(2x-\frac{3}{2}x):(x-\frac{9}{10}x) \)

=\(\frac{1}{2}x:\frac{1}{10}x=\frac{1}{2}x.\frac{10}{x}=5 \)