Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
\(9x^2-6x+2=\left(3x-1\right)^2+1=t\ge1\)
\(Pt\Rightarrow\sqrt{t}+\sqrt{5t-1}=\sqrt{10-t}\)
\(\Leftrightarrow5t-1=10-t+t-2\sqrt{t\left(10t-1\right)}\)
\(\Leftrightarrow2\sqrt{t\left(10t-1\right)}+5t=11\)
\(\Rightarrow VT\ge VP\left(t\ge1\right)\Rightarrow t=1\Rightarrow x=\frac{1}{3}\)
Đk: `x >=-1`.
`5sqrt(x+1) + sqrt(4x+4) - sqrt(9x+9) = 2`.
`<=> 5sqrt(x+1) + 2 sqrt(x+1) - 3sqrt(x+1) = 2`.
`<=> 4 sqrt(x+1) =2.`
`<=> sqrt(x+1) = 1/2`
`<=> x + 1 = 1/4`
`<=> x = 3/4 (tm)`.
Vậy `x = 3/4`.
\(5\sqrt{x+1}+\sqrt{4x+4}-\sqrt{9x+9}=2\)
\(\Leftrightarrow5\sqrt{x+1}+2\sqrt{x+1}-3\sqrt{x+1}=2\) (1)
ĐKXĐ: \(x\ge-1\)
(1) \(\Leftrightarrow4\sqrt{x+1}=2\)
\(\Leftrightarrow\sqrt{x+1}=\dfrac{1}{2}\)
\(\Leftrightarrow x+1=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}-1\)
\(\Leftrightarrow x=-\dfrac{3}{4}\) (nhận)
Vậy \(x=-\dfrac{3}{4}\)
Lời giải:
ĐKXĐ: $9x^2+6x+1\geq 0$
$\Leftrightarrow (3x+1)^2\geq 0$
$\Leftrightarrow x\in\mathbb{R}$
--------------------------
$\sqrt{9x^2+6x+1}=2-x$
\(\Rightarrow \left\{\begin{matrix} 2-x\geq 0\\ 9x^2+6x+1=(2-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ 9x^2+6x+1=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ 8x^2+10x-3=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 2\\ (4x-1)(2x+3)=0\end{matrix}\right.\Leftrightarrow x=\frac{1}{4}\) hoặc $x=\frac{-3}{2}$
a, \(\sqrt{9x-9}-2\sqrt{x-1}=8\)ĐK : x >= 1
\(\Leftrightarrow3\sqrt{x-1}-2\sqrt{x-1}=8\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\Leftrightarrow x=65\)
b, mình chưa hiểu đề lắm
\(\sqrt{x^2-6x+9}=5\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=5\)
\(\Leftrightarrow\left|x-3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\)