Question

Tìm x biết.

a) √x^2-6x+9 = 3-x

b)√1-12x+36x^2 = 5

c) √x+2√x-1 = 2

d) √x^2+2x+1 = √x+1

e) √x^2-9 + √x^2-6x+9 = 0

f) √2x-2-2√2x-3 = 1

g) √x^2-4x+5 + √x^2-4x+8 + √x^2-4x+9 = 3+√5

Các dấu căn kéo dài hết ạ

Giúp mình với !!

Answer 1

e) ĐKXĐ: \(x^2-9\ge0\Leftrightarrow\left(x-3\right).\left(x+3\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Leftrightarrow\sqrt{\left(x-3\right).\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

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

\(TH1:x-3=0\Leftrightarrow x=3\)

\(TH2:\sqrt{x-3}=-\sqrt{x+3}\Leftrightarrow x=3\text{ và }x=-3\left(loai\right)\)

Vậy giá trị x cần tìm là 3

Answer 2

ĐKXĐ: \(3-x\ge0\Leftrightarrow x\le3\)

\(\sqrt{x^2-6x+9}=3-x\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=3-x\)

\(\Leftrightarrow\left|x-3\right|=3-x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=3-x\\3-x=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\\text{vô số x tm}\left(x\le3\right)\end{matrix}\right.\)

Vậy giá trị x cần tìm là \(x\le3\)

Answer 3

b) \(\sqrt{36^2-12x+1}=5\)

\(\Leftrightarrow\sqrt{\left(6x-1\right)^2}=5\)

\(\Leftrightarrow\left|6x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{2}{3}\end{matrix}\right.\)

Vậy giá trị x thỏa mãn là \(\left[{}\begin{matrix}x=1\\x=-\frac{2}{3}\end{matrix}\right.\)

Answer 4

c) ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ge0\\x-1\ge0\end{matrix}\right.\Rightarrow x\ge1\)

\(\sqrt{x+2}.\sqrt{x-1}=2\)

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

\(\Leftrightarrow x^2+x-2=4\)

\(\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)=4+\frac{9}{4}=\frac{25}{4}\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{5}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{5}{2}\\x+\frac{1}{2}=-\frac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMDKXD\right)\\x=-3\left(KTMDKXD\right)\end{matrix}\right.\)

Vậy giá trị x thỏa mãn là: x=2

Answer 5

d) ĐKXĐ: \(x+1\ge0\Leftrightarrow x\ge-1\)

\(\sqrt{x^2+2x+1}=\sqrt{x+1}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(TMĐKXĐ\right)\\x=0\left(TMĐKXĐ\right)\end{matrix}\right.\)

Vậy giá trị x thỏa mãn là \(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)

Answer 6

Cảm ơn bạn!