Question

[Thanh Dii]

Giải các phương trình sau:

a, \(\left(x-3\right)^2+x^4=-y^2+6y-4\)

b, \(\sqrt{2x-3}+\sqrt{5-2x}-x^2+4x-6=0\)

c, \(4+4x-x^2=|x-1|+|x-2|+|2x-3|+|4x-14|\)

d, \(x^2-2x+3=\sqrt{2x^2-x}+\sqrt{1+3x-3x^2}\)

Answer 1

[Akai Haruma]

Câu a:

Ta có:

\((x-3)^2+x^4=-y^2+6y-4\)

\(\Leftrightarrow (x-3)^2+x^4+y^2-6y+4=0\)

\(\Leftrightarrow x^4+x^2-6x+9+y^2-6y+4=0\)

\(\Leftrightarrow x^4+x^2-6x+4+(y^2-6y+9)=0\)

\(\Leftrightarrow (x^4-2x^2+1)+3(x^2-2x+1)+(y^2-6y+9)=0\)

\(\Leftrightarrow (x^2-1)^2+3(x-1)^2+(y-3)^2=0\)

\(\Rightarrow (x^2-1)^2=(x-1)^2=(y-3)^2=0\)

\(\Rightarrow \left\{\begin{matrix} x=1\\ y=3\end{matrix}\right.\)

Vậy..........

Answer 2

[Akai Haruma]

Câu b:

ĐKXĐ: \(\frac{3}{2}\leq x\leq \frac{5}{2}\)

\(\sqrt{2x-3}+\sqrt{5-2x}-x^2+4x-6=0\)

\(\Leftrightarrow \sqrt{2x-3}+\sqrt{5-2x}=x^2-4x+6\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VT}^2\leq (1+1)(2x-3+5-2x)=4\)

\(\Rightarrow \text{VT}\leq 2\)

Mà \(\text{VP}=x^2-4x+6=(x-2)^2+2\geq 2\)

Do đó để \(\text{VT}=\text{VP}\) thì \(\text{VT}=2=\text{VP}\)

Điều này xảy ra khi \(\left\{\begin{matrix} \sqrt{2x-3}=\sqrt{5-2x}\\ (x-2)^2=0\end{matrix}\right.\Rightarrow x=2\) (t/m)

Vậy pt có nghiệm duy nhất $x=2$