Question

Giải phương trình:

a) (x-3)^2=11+6√2

b) x^2 -10x + 25=27-10√2

c) 4x^2 + 4x= 27-10√3

d) x^2+2√5x=16-4√5

e) x^2 + 4√3x=1-4√3

f) 4x^2-12√2 x -33+10√2=0

g) 2x^2-12x+9+4√2=0

h) 3x^2 -30x+26+8√3=0

Các bn giúp mk vs. Tks

Answer 1

a/ \(\left(x-2\right)^2=11+6\sqrt{2}\)

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

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

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

b/ \(x^2-10x+25=27-10\sqrt{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=5-\sqrt{2}\\x-5=\sqrt{2}-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\)

c/ \(4x^2+4x+1=28-10\sqrt{3}\)

\(\Leftrightarrow\left(2x+1\right)^2=\left(5-\sqrt{3}\right)^2\)

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

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

Answer 2

d/ \(x^2+2\sqrt{5}x+5=21-4\sqrt{5}\)

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

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

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

e/ \(x^2+2\sqrt{12}x+12=13-4\sqrt{3}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1-4\sqrt{3}\end{matrix}\right.\)

f/ \(4x^2-12\sqrt{2}x+18=51-10\sqrt{2}\)

\(\Leftrightarrow\left(2x-3\sqrt{2}\right)^2=\left(5\sqrt{2}-1\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{10\sqrt{2}-1}{2}\\x=\frac{1-3\sqrt{2}}{2}\end{matrix}\right.\)

Answer 3

g/

\(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.3\sqrt{2}+18=9-4\sqrt{2}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{10-\sqrt{2}}{2}\\x=\frac{2+\sqrt{2}}{2}\end{matrix}\right.\)

h/

\(\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.5\sqrt{3}+75=49-8\sqrt{3}\)

\(\Leftrightarrow\left(\sqrt{3}x-5\sqrt{3}\right)^2=\left(8\sqrt{3}-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3}x-5\sqrt{3}=8\sqrt{3}-1\\\sqrt{3}x-5\sqrt{3}=1-8\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3}x=13\sqrt{3}-1\\\sqrt{3}x=1-3\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{39-\sqrt{3}}{3}\\x=\frac{\sqrt{3}-9}{3}\end{matrix}\right.\)