Đầu kiện: \(x\ge0;y\ge0\)

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\\sqrt{x}-\sqrt{y}=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\3\sqrt{x}-3\sqrt{y}=13,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\3\sqrt{x}-3\sqrt{x}+2\sqrt{y}-(-3\sqrt{y})=6-13,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\5\sqrt{y}=-7,5\end{matrix}\right.\)

Vì \(5\sqrt{y}\ge0\Rightarrow5\sqrt{y}=-7,5< 0\Rightarrow5\sqrt{y}=-7,5\)vô nghiệm.

Vậy hệ phương trình đã cho vô nghiệm 

ĐKXĐ: \(x\ge0;y\ge0\)

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\\sqrt{x}-\sqrt{y}=4,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3\left(\sqrt{y}+4,5\right)+2\sqrt{y}=6\\\sqrt{x}=\sqrt{y}+4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y}=-7,5\\\sqrt{x}=\sqrt{y}+4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=-1,5< 0\left(ktm\right)\\\sqrt{x}=\sqrt{y}+4,5\end{matrix}\right.\)

Vậy hệ đã cho vô nghiệm

ĐKXĐ: \(x\ge0;y\ge0\)

\(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\2\sqrt{x}+3\left(3\sqrt{x}-5\right)=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\11\sqrt{x}=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\\sqrt{x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}10y^2=2\\x^2=3y^2+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y^2=\dfrac{1}{5}\\x^2=\dfrac{8}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{1}{\sqrt{5}}\\y=\pm\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}2x^2+3y^2=36\\3x^2+7y^2=37\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2+14y^2=74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2-6x^2+9y^2-14y^2=108-74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\-5y^2=34\end{matrix}\right.\)

Vì \(-5y^2=34\Rightarrow y^2=\dfrac{34}{-5}< 0\) vô nghiệm

Vậy hệ phương trình đã cho vô nghiệm

\(\left\{{}\begin{matrix}2x^2+3y^2=36\\3x^2+7y^2=37\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2+14y^2=74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\5y^2=-34\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\y^2=-\dfrac{34}{5}< 0\left(vô-lý\right)\end{matrix}\right.\)

Vậy hệ pt vô nghiệm

\(\left\{{}\begin{matrix}7x^2+13y=-39\\5x^2-11y=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}77x^2+143y=-429\\65x^2-143y=429\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}143x^2=0\\65x^2-143y=429\end{matrix}\right.\)

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

ĐKXĐ: \(x\ne3;y\ne-1\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-3}=u\\\dfrac{1}{y+1}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4u+5v=2\\5u+v=\dfrac{29}{20}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4u+5\left(\dfrac{29}{20}-5u\right)=2\\v=\dfrac{29}{20}-5u\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-21u=-\dfrac{21}{4}\\v=\dfrac{29}{20}-5u\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{1}{4}\\v=\dfrac{1}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-3}=\dfrac{1}{4}\\\dfrac{1}{y+1}=\dfrac{1}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=7\\y=4\end{matrix}\right.\)

a)Q=\(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{a+\sqrt{a}}:\dfrac{\sqrt{a}-1}{a+2\sqrt{a}+1}\left(a>0\right)\)

Q=\(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}:\dfrac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)^2}\)

Q= \(\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\) =\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) 

vậy...

b)ta có Q=4

<=>\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) =4 <->\(\sqrt{a}+1=4.\sqrt{a}\)

<->\(\sqrt{a}+1=4\sqrt{a}\)

<->-3\(\sqrt{a}\) =1<=>\(\sqrt{a}\) =\(\dfrac{-1}{3}\) 

<-> a=1/9 

vậy ..........

giúp em với ạ, em đang cần gấp ạ.

\(\Leftrightarrow x^2+y^2=y^4+2y^2+1\)

\(\Leftrightarrow x^2=y^4+y^2+1\)

Ta có: \(y^4+y^2+1>y^4=\left(y^2\right)^2\)

Và \(y^4+y^2+1\le y^4+2y^2+1=\left(y^2+1\right)^2\)

\(\Rightarrow\left(y^2\right)^2< x^2\le\left(y^2+1\right)^2\)

\(\Rightarrow x^2=\left(y^2+1\right)^2\) theo định lý kẹp

\(\Rightarrow y^4+y^2+1=\left(y^2+1\right)^2\)

\(\Rightarrow y^2=0\Rightarrow y=0\)

\(\Rightarrow x^2=1\Rightarrow x=\pm1\)

Cho tâm giác vuông ABC vuông tại A có AB < AC, đường cao AH, trung tuyến AM. Gọi D, E theo thứ tự là hình chiếu của H trên AB và AC. Gọi K là giao điểm của AM và DE.

a, Chứng minh tứ giác ADHE là hình chữ nhật.

b,Chứng minh rằng AB²/AC²= BH/CH

c, chứng minh AD²= AH . DK

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