giải pt x\(^2\)- 1 = 2\(\sqrt{2x+1}\)
a) Giải pt: \(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)
b)Giải hệ pt \(\left\{{}\begin{matrix}xy-y^2+2y-x-1=\sqrt{y-1}-\sqrt{x}\\3\sqrt{6-y}+3\sqrt{2x+3y-7}=2x+7\end{matrix}\right.\)
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
giải pt: \(\sqrt{x-\sqrt{2x-1}}+\sqrt{x+\sqrt{2x-1}}=\sqrt{2}x\)
Từ pt suy ra \(x\ge0\).
PT \(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}=2x\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|+\left|\sqrt{2x-1}+1\right|=2x\). (*)
+) \(\sqrt{2x-1}-1\ge0\Leftrightarrow x\ge1\): Khi đó (*) tương đương \(2\sqrt{2x-1}=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow x=1\) (thoả mãn)
+) \(\sqrt{2x-1}-1< 0\Leftrightarrow x< 1\): Khi đó (*) tương đương \(2=2x\Leftrightarrow x=1\), vô lí.
Vậy x = 1
Giải PT: \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)
Giải PT: \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)
Giải PT: \(\sqrt{2x+3\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)
Sửa lại đề bài cho mk là: \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)
Giải pt:
\(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\)
\(ĐK:x\in R\)
\(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\) (*)
Đặt \(x^2+x+1=a;a\ge0\)
\(\rightarrow\left\{{}\begin{matrix}x^2+x+4=a+3\\2x^2+2x+9=2a+7\end{matrix}\right.\)
(*) \(\Rightarrow\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow\left(\sqrt{a+3}+\sqrt{a}\right)^2=\left(\sqrt{2a+7}\right)^2\)
\(\Leftrightarrow a+3+a+2\sqrt{a\left(a+3\right)}=2a+7\)
\(\Leftrightarrow2\sqrt{a\left(a+3\right)}=4\)
\(\Leftrightarrow\sqrt{a\left(a+3\right)}=2\)
\(\Leftrightarrow a\left(a+3\right)=4\)
\(\Leftrightarrow a^2+3a-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) \((tm)\)
Vậy \(S=\left\{0;-1\right\}\)
GIẢI CÁC PT SAU:
\(\sqrt{x^2+5x+1}=\sqrt{x+1}\)
\(\sqrt{x^2+2x+4}=\sqrt{2-x}\)
\(\sqrt{2x+4}-\sqrt{2-x}=0\)
Lời giải:
1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$
PT $\Leftrightarrow x^2+5x+1=x+1$
$\Leftrightarrow x^2+4x=0$
$\Leftrightarrow x(x+4)=0$
$\Rightarrow x=0$ hoặc $x=-4$
Kết hợp đkxđ suy ra $x=0$
2. ĐKXĐ: $x\leq 2$
PT $\Leftrightarrow x^2+2x+4=2-x$
$\Leftrightarrow x^2+3x+2=0$
$\Leftrightarrow (x+1)(x+2)=0$
$\Leftrightarrow x+1=0$ hoặc $x+2=0$
$\Leftrightarrow x=-1$ hoặc $x=-2$
3.
ĐKXĐ: $-2\leq x\leq 2$
PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$
$\Leftrightarrow 2x+4=2-x$
$\Leftrightarrow 3x=-2$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
\(\sqrt{x^2-1}+\sqrt{x^2-2x+1}=0\)
Giải PT
ĐK: `x<=-1 ; x>= 1`
`\sqrt(x^2-1)+\sqrt(x^2-2x+1)=0`
`<=> \sqrt((x-1)(x+1)) + \sqrt((x-1)^2)=0`
`<=> \sqrt(x-1) (\sqrt(x+1) + \sqrt(x-1))=0`
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}+\sqrt{x-1}=0\left(VN\right)\end{matrix}\right.\\ \Leftrightarrow x=1\)
Vậy `S={1}`.
ĐKXĐ : \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
\(\sqrt{x^2-1}+\sqrt{x^2-2x+1}=0\)\(\)
\(\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x^2-2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\\left(x-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\x=1\end{matrix}\right.\)\(\)
\(\Leftrightarrow x=1\)
Vậy S = {1}
1. Giải pt:
\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
2. Giải pt:
\(\sqrt{x+2\sqrt{x-1}}=3\sqrt{x-1}-5\)
1. đk: pt luôn xác định với mọi x
\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)
\(\Leftrightarrow\left|x-1\right|-\left|x-3\right|=10\)
Bạn mở dấu giá trị tuyệt đối như lớp 7 là ok rồi!
2. đk: \(x\geq 1\)
\(\sqrt{x+2\sqrt{x-1}}=3\sqrt{x-1}-5\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=3\sqrt{x-1}-5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}-3\sqrt{x-1}+5=0\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|-3\sqrt{x-1}+5=0\)
Đến đây thì ổn rồi! bạn cứ xét khoảng rồi mở trị và bình phương 1 chút là ok cái bài!
Giải PT: \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
ĐK: `{(2x^2+8x+6>=0),(x^2-1>=0),(2x+2>=0):} <=> {(x=-1),(x>=1):}`
`\sqrt(2x^2+8x+6)+\sqrt(x^2-1)=2x+2`
`<=>(2x^2+8x+6)+(x^2-1)+2\sqrt((2x^2+8x+6)(x^2-1))=(2x+2)^2`
`<=>2(x+3)(x+1)+(x-1)(x+2)+2\sqrt((x+1)^2 (x+3)(x-1))=4(x+1)^2`
`<=> (x+1)[2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0`
`<=> [(x=-1\ (TM)),([2(x+3)+(x-1)+2\sqrt((x+3)(x-1))-4(x+1)]=0\ (1)):}`
(1) `<=> x-1=2\sqrt((x+3)(x-1))`
`<=>x^2-2x+1=4(x+3)(x-1)`
`<=>x=1\ `(TM)
Vậy `S={\pm 1}`.
\(ĐK:x\le-3;x\ge-1\)
\(PT\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\left(x+3\right)+\left(x-1\right)+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\\ \Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\\ \Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{25}{7}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)
Vậy \(S=\left\{-1;1\right\}\)