giải pt: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

làm thế này mà chả hiểu sao lại bị gạch, ai biết chỉ với, cảm ơn nak:

+ ĐK:\(\left\{{}\begin{matrix}x\ge1\\x+3-4\sqrt{x-1}\ge0\\x+8-6\sqrt{x-1}\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)

+ pt đã cho \(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\) (*)

Th1: \(\left\{{}\begin{matrix}\sqrt{x-1}-2< 0\\\sqrt{x-1}-3< 0\end{matrix}\right.\)

(*) \(\Leftrightarrow2-\sqrt{x-1}+3-\sqrt{x-1}=1\Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x=5\left(N\right)\)

Th2: \(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\ge0\end{matrix}\right.\)

(*) \(\Leftrightarrow\sqrt{x-1}-2+\sqrt{x-1}-3=1\Leftrightarrow2\sqrt{x-1}=6\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x=10\left(N\right)\)

Th3: \(\sqrt{x-1}-3< 0\le\sqrt{x-1}-2\)

(*) \(\Leftrightarrow\sqrt{x-1}-2+3-\sqrt{x-1}=1\Leftrightarrow1=1\left(đúng\right)\)

Kl: \(x\ge1\)

Giải hệ phương trình

\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)

Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?

Giải pt: { máy tính cho ra x=-1 , x=4 }

\(\left(x+1\right)\sqrt{16x+17}=8x^2-15x-23\) (1)

ĐK: \(16x+17\ge0\Leftrightarrow x\ge-\dfrac{17}{16}\)

(1) \(\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(N\right)\\\left\{{}\begin{matrix}16x+17=\left(x-\dfrac{23}{8}\right)^2\\x\ge\dfrac{23}{8}\end{matrix}\right.\end{matrix}\right.\)(2)

(2) \(\Leftrightarrow16x+17=x^2-\dfrac{23}{4}x+\dfrac{529}{64}\Leftrightarrow x^2-\dfrac{87}{4}-\dfrac{559}{64}=0\) (Xấu quéc!! Pt này không có nghiệm = 4---> sai ở đâu vậy ạ??)

Cảm ơn trước nak ^^!

Giải hệ pt

1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)

2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)

3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)

4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)

m.n giúp e mấy bài này vs ạ!!

Đề: tìm x biết : \(2.\left|2-x\right|+3.\left|x+1\right|-x+1=2x\)

giải

•nếu \(-1>x\) thì: \(\left|2-x\right|=2-x\\ \left|x+1\right|=-x-1\)

•nếu \(-1\le x< 2\) thì: \(\left|2-x\right|=2-x\\ \left|x+1\right|=x+1\)

•nếu\(x\ge2\) thì: \(\left|2-x\right|=x-2\\ \left|x+1\right|=x+1\)

◘ từ 3 ĐK trên, ta có:

\(\left[{}\begin{matrix}2.\left(2-x\right)+3.\left(-x-1\right)-x+1=2x\left(với\:-1>x\right)\\2.\left(2-x\right)+3.\left(x+1\right)-x+1=2x\left(với\:-1\le x< 2\right)\\2.\left(x-2\right)+3.\left(x+1\right)-x+1=2x\left(với\:x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4-2x-3x-3-x+1=2x\\4-2x+3x+3-x+1=2x\\2x-4+3x+3-x+1=2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-2\\-2x=-8\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(loại\right)\\x=4\left(loại\right)\\x=0\left(loại\right)\end{matrix}\right.\)

vậy phương trình đã cho vô nghiệm.

P/S: giải dùm cho 1 bạn nhờ, đừng ném đa hay gạch j nhé !!!

My name is ???

Giải hệ

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

b) \(\left\{{}\begin{matrix}\left(6x+5\right)\sqrt{2x+1}-2y-3y^3=0\\y+\sqrt{x}=\sqrt{2x^2+4x-23}\end{matrix}\right.\)

Giải bất pt

\(\dfrac{9}{\left|x-5\right|-3}\ge\left|x-2\right|\)