Hôm nay sol vài bài trên olm rồi off tiếp

\(\sqrt{xy+y}=\sqrt{y\left(x+1\right)}\)

ĐKXĐ: \(x>-1,y>0\)

Đặt \(\sqrt{x+1}=a;\sqrt{y}=b\left(a,b>0\right)\)

HPT \(\Leftrightarrow\hept{\begin{cases}a^2-1+\frac{1}{a}=\frac{4}{a+b}-1\\b^2+\frac{1}{b}=2ab\end{cases}}\) 

\(\Leftrightarrow\hept{\begin{cases}a^4+a^3b-3a+b=0\\2ab^2-b^3-1=0\end{cases}}\)

PT(2) \(\Leftrightarrow2ab^2=\left(b+1\right)\left(b^2-b+1\right)\Rightarrow a=\frac{\left(b+1\right)\left(b^2-b+1\right)}{2b^2}\)

Thay ngược lên pt(1) tương đương  \(\left(3b^6+8b^3+1\right)\left(b^3-1\right)^2=0\)

\(\Rightarrow b=1\rightarrow a=1\)

HPT có nghiệm duy nhất a = b = 1

Khúc sau từ suy ra x, y nhé. Quên mất lỡ bấm gửi.

a) \(\hept{\begin{cases}\left(x-1\right)\left(2x+y\right)=0\\\left(y+1\right)\left(2y-x\right)=0\end{cases}}\)
\(\cdot x=1\Rightarrow\hept{\begin{cases}0=0\\\left(y+1\right)\left(2y-1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\y=-1;y=\frac{1}{2}\end{cases}}\)
\(\cdot y=-1\Rightarrow\hept{\begin{cases}\left(x-1\right)\left(2x-1\right)=0\\0=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1;x=\frac{1}{2}\\0=0\end{cases}}\)
\(\cdot x=2y\Rightarrow\hept{\begin{cases}\left(2y-1\right)5y=0\\0=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=0\\y=\frac{1}{2}\Rightarrow x=1\end{cases}}\)
\(y=-2x\Rightarrow\hept{\begin{cases}0=0\\\left(1-2x\right)5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\Rightarrow y=-1\\x=0\Rightarrow y=0\end{cases}}\)

b) \(\hept{\begin{cases}x+y=\frac{21}{8}\\\frac{x}{y}+\frac{y}{x}=\frac{37}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\\left(\frac{21}{8}-y\right)^2+y^2=\frac{37}{6}y\left(\frac{21}{8}-y\right)\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\2y^2-\frac{21}{4}y+\frac{441}{64}=-\frac{37}{6}y^2+\frac{259}{16}y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\1568y^2-4116y+1323=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{8}\\y=\frac{9}{4}\end{cases}}hay\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{3}{8}\end{cases}}\)

c) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x-y\right)^2=-4x^2y^2+2xy\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}8x^2y^2-4x^2y-4xy^2+x^2+y^2-2xy+2xy=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4x^2y^2-4x^2y+x^2+4x^2y^2-4xy^2+y^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x\right)^2+\left(2xy-y\right)^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=\frac{-1}{2}\end{cases}}\)
d) \(\hept{\begin{cases}xy+x+y=71\\x^2y+xy^2=880\end{cases}}\). Đặt \(\hept{\begin{cases}x+y=S\\xy=P\end{cases}}\), ta có: \(\hept{\begin{cases}S+P=71\\SP=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P\left(71-P\right)=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P^2-71P+880=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=16\\P=55\end{cases}}hay\hept{\begin{cases}S=55\\P=16\end{cases}}\)
\(\cdot\hept{\begin{cases}S=16\\P=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=16\\xy=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y\left(16-y\right)=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y^2-16y+55=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=11\end{cases}}hay\hept{\begin{cases}x=11\\y=5\end{cases}}\)

\(\cdot\hept{\begin{cases}S=55\\P=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=55\\xy=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y\left(55-y\right)=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y^2-55y+16=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{55-3\sqrt{329}}{2}\\y=\frac{55+3\sqrt{329}}{2}\end{cases}}hay\hept{\begin{cases}x=\frac{55+3\sqrt{329}}{2}\\y=\frac{55-3\sqrt{329}}{2}\end{cases}}\)

e) \(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{cases}}\). Đặt \(\hept{\begin{cases}S=\sqrt{x}+\sqrt{y}\\P=\sqrt{xy}\end{cases}}\), ta có \(\hept{\begin{cases}SP=12\\P\left(S^2-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\P\left(\frac{144}{P^2}-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\2P^4+28P^2-144P=0\end{cases}}\)
Tự làm tiếp nhá! Đuối lắm luôn

Điều kiện xác định \(x,y>0\)

Hệ đã cho tương đương với

\(\hept{\begin{cases}\sqrt{x}-\sqrt{y}+\frac{3}{\sqrt{x}}-\frac{3}{\sqrt{y}}=0\left(1\right)\\2x-\sqrt{xy}=1\left(2\right)\end{cases}}\)

Giải (1) \(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)-3\left(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\right)=0\)

           \(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(1-\frac{3}{\sqrt{xy}}\right)=0\)

            \(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-\sqrt{y}=0\\1-\frac{3}{\sqrt{xy}}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\sqrt{x}=\sqrt{y}\\\frac{3}{\sqrt{xy}}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=y\\\sqrt{xy}=3\end{cases}.}\)

Với x=y ta thế vào (2) có \(2x-\sqrt{x^2}=1\Leftrightarrow x=1\left(TMĐK\right)\)

                                     \(\Rightarrow x=y=1\)

Với \(\sqrt{xy}=3\)thế vào (2) có \(2x-3=1\Leftrightarrow x=2\left(TMĐK\right)\)

                                      \(\Rightarrow\sqrt{2y}=3\Leftrightarrow y=\frac{9}{2}\left(TMĐK\right)\)

Vậy hệ có 2 nghiệm.......

Ta có: \(\sqrt{8x-y+5}+\sqrt{x+y-1}=3\sqrt{x}+2\)

\(\Leftrightarrow8x-y+5+x+y-1+2\sqrt{\left(8x-y+5\right)\left(x+y-1\right)}=9x+12\sqrt{x}+4\)

\(\Leftrightarrow9x+4+2\sqrt{8x^2-y^2+7xy-3x+6y-5}=9x+4+12\sqrt{x}\)

\(\Leftrightarrow\sqrt{8x^2-y^2+7xy-3x+6y-5}=6\sqrt{x}\)

\(\Leftrightarrow8x^2-y^2+7xy-3x+6y-5=36x\)

\(\Leftrightarrow8x^2-y^2+7xy-39x+6y-5=0\)

\(\Leftrightarrow\left(8x^2+8xy-40x\right)-y^2-xy-5+x+6y=0\)

\(\Leftrightarrow8x\left(x+y-5\right)-\left(y^2+xy-5y\right)+\left(x+y-5\right)=0\)

\(\Leftrightarrow\left(x+y-5\right)\left(8x-y+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=5-x\\y=8x+1\end{cases}}\)

Thay vào pt dưới ta có:

\(\sqrt{xy}+\frac{1}{\sqrt{x}}=\sqrt{8x-y+5}\left(1\right)\)

+) với y=5-x (1) thành:

\(\sqrt{x\left(5-x\right)}+\frac{1}{\sqrt{x}}=\sqrt{8x-\left(5-x\right)+5}\)

\(\Leftrightarrow\sqrt{5x-x^2}+\frac{1}{\sqrt{x}}=\sqrt{9x}\)\(\Leftrightarrow\sqrt{5x^2-x^3}+1=3x\)\(\Leftrightarrow\sqrt{5x^2-x^3}=3x-1\)

\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{3}\\5x^2-x^3=9x^2-6x+1\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{3}\\x^3+4x^2-6x+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge\frac{1}{3}\\x=1\left(tm\right)\end{cases}}}\)

Với x=1=>y=4

ĐKXĐ: \(x;y\)\(\ge\)0

Biến đổi phương trình thứ nhất ta có \(y-2x+\sqrt{y}-\sqrt{x}+\sqrt{xy}=0\Leftrightarrow y-x+\sqrt{y}-\sqrt{x}-x+\sqrt{xy}=0\)

\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)+\left(\sqrt{y}-\sqrt{x}\right)+\sqrt{xy}-\sqrt{x}=0\)

\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)+\left(\sqrt{y}-\sqrt{x}\right)+\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)=0\)

\(\Leftrightarrow\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+2\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{y}-\sqrt{x}=0\Leftrightarrow x=y\\\sqrt{y}+2\sqrt{x}+1=0\end{cases}}\)Mặt khác \(\sqrt{y}+2\sqrt{x}+1\ge1>0\forall x;y\)
\(\Rightarrow\)vô nghiệm

Thay x=y vào phương trình thứ hai rồi tự tính tiếp nha bạn coa nghiệm x=y=1

\(\hept{\begin{cases}\frac{x}{\sqrt{y}}+\frac{2\sqrt{y}}{x}=\frac{2}{x}+\frac{1}{\sqrt{y}}-3\left(1\right)\\x^2-xy-9x+12=0\left(2\right)\end{cases}}\)

Đặt \(\frac{2}{x}=a,\frac{1}{\sqrt{y}}=b\left(b>0\right)\)

\(\left(1\right)\Leftrightarrow\frac{2b}{a}+\frac{a}{b}=a+b-3\)

\(\Leftrightarrow2b^2+a^2+3ab=ab\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a+2b\right)=\left(a+b\right)ab\)

\(\Leftrightarrow\left(a+b\right)\left(a-ab+2b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-b\left(3\right)\\a-ab+2b=0\left(4\right)\end{cases}}\)

Giải (3)

\(\left(3\right)\Leftrightarrow\frac{2}{x}=-\frac{1}{\sqrt{y}}\Leftrightarrow\frac{4}{x^2}=\frac{1}{y}\)

\(\Leftrightarrow y=\frac{x^2}{4}\). Thay vào (2) tìm nghiệm (x,y)

Giải (4)

\(\left(4\right)\Leftrightarrow\frac{2}{x}-\frac{2}{\sqrt{y}}+\frac{2}{x\sqrt{y}}=0\)

\(\Leftrightarrow\sqrt{y}-x+2=0\)

Giải tiếp là ra

Học tốt!!!!!!!!!

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

\(ĐK:x,y>0\)

\(\left(1\right)\Leftrightarrow\frac{y-x}{y\sqrt{x}}=\left(x-y\right)\left(x+2y\right)\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{y\sqrt{x}}\right)=0\)

Vì x, y > 0 nên \(x+2y+\frac{1}{y\sqrt{x}}>0\)suy ra x - y = 0 hay x = y

Thay x = y vào (2), ta được: \(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\)

\(\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}\)\(\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}\)

\(\Leftrightarrow\sqrt{x+3}.\sqrt{x}-\sqrt{x+3}-\sqrt{x}+1=0\)\(\Leftrightarrow\left(\sqrt{x+3}-1\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\\sqrt{x}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\\x=1\left(tmđk\right)\end{cases}}\Rightarrow x=y=1\)

Vậy hệ có một nghiệm duy nhất \(\left(x;y\right)=\left(1;1\right)\)

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

ĐK: \(\hept{\begin{cases}x>0\\y>0\end{cases}}\)và \(\hept{\begin{cases}x+3\ge0\\x^2+3x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x>0\\y>0\end{cases}}}\)

\(\left(1\right)\Leftrightarrow\frac{y-x}{y\sqrt{x}}=\left(x-y\right)\left(x+2y\right)\Leftrightarrow\left(x+y\right)\left(x+2y+\frac{1}{y\sqrt{x}}\right)=0\Leftrightarrow x=y\)do \(x+2y+\frac{1}{y\sqrt{x}}>0\forall x,y>0\)

Thay y=x vào pt (2) ta được

\(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\Leftrightarrow1+\sqrt{x^2+3x}=\frac{3}{\sqrt{x+3}-\sqrt{x}}\)

\(\Leftrightarrow1+\sqrt{x^2+3x}=\sqrt{x+3}+\sqrt{x}\Leftrightarrow\sqrt{x+3}\cdot\sqrt{x}-\sqrt{x+3}-\sqrt{x+1}=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-1\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}=1\\\sqrt{x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\left(loai\right)\\x=1\left(tm\right)\end{cases}\Rightarrow}x=y=1}\)

Vậy hệ có nghiệm duy nhất (x;y)=(1;1)

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