Question

Giải phương trình
a, \(\sqrt{x^2-x+9}=2x+1\)

b. \(\sqrt{5x+7}-\sqrt{x+3}=\sqrt{3x+1}\)

c. \(x^2-3x-10+3\sqrt{x.\left(x-3\right)}=0\)

d. \(\sqrt{2-x}+\sqrt{4-x}=x^2-6x+11\)

e. \(x+6\sqrt{x+8}+4\sqrt{6-2x}=27\)

Answer 1

Câu a:

ĐKXĐ:...........

\(\sqrt{x^2-x+9}=2x+1\)

\(\Rightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-x+9=(2x+1)^2=4x^2+4x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+5x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x(x-1)+8(x-1)=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (x-1)(3x+8)=0\end{matrix}\right.\Rightarrow x=1\)

Vậy.....

Answer 2

Câu b:

ĐKXĐ:.........

Ta có: \(\sqrt{5x+7}-\sqrt{x+3}=\sqrt{3x+1}\)

\(\Rightarrow (\sqrt{5x+7}-\sqrt{x+3})^2=3x+1\)

\(\Leftrightarrow 5x+7+x+3-2\sqrt{(5x+7)(x+3)}=3x+1\)

\(\Leftrightarrow 3(x+3)=2\sqrt{(5x+7)(x+3)}\)

\(\Leftrightarrow \sqrt{x+3}(3\sqrt{x+3}-2\sqrt{5x+7})=0\)

Vì \(x\geq -\frac{7}{5}\Rightarrow \sqrt{x+3}>0\). Do đó:

\(3\sqrt{x+3}-2\sqrt{5x+7}=0\)

\(\Rightarrow 9(x+3)=4(5x+7)\)

\(\Rightarrow 11x=-1\Rightarrow x=\frac{-1}{11}\) (thỏa mãn)

Vậy..........

Answer 3

Câu c:

Đặt \(\sqrt{x(x-3)}=a\). Ta có:

\(x^2-3x-10+3\sqrt{x(x-3)}=0\)

\(\Leftrightarrow x(x-3)-10+3\sqrt{x(x-3)}=0\)

\(\Leftrightarrow a^2-10+3a=0\)

\(\Leftrightarrow (a-2)(a+5)=0\Rightarrow a=2\) (do \(a\geq 0\) )

\(\Rightarrow \sqrt{x(x-3)}=2\Rightarrow x(x-3)=4\)

\(\Leftrightarrow (x-4)(x+1)=0\Rightarrow \left[\begin{matrix} x=4\\ x=-1\end{matrix}\right.\) (đều thỏa mãn)

Vậy......

Answer 4

Câu d:

ĐKXĐ: \(x\leq 2\)

\(\sqrt{2-x}+\sqrt{4-x}=x^2-6x+11\)

\(\Leftrightarrow x^2-6x+11-\sqrt{2-x}-\sqrt{4-x}=0\)

\(\Leftrightarrow x^2-4x+\frac{9}{2}+(2-x-\sqrt{2-x}+\frac{1}{4})+(4-x-\sqrt{4-x}+\frac{1}{4})=0\)

\(\Leftrightarrow (x-2)^2+\frac{1}{2}+(\sqrt{2-x}-\frac{1}{2})^2+(\sqrt{4-x}-\frac{1}{2})^2=0\)

Dễ thấy vế trái của pt luôn dương với mọi \(x\leq 2\), vế phải thì bằng $0$ (vô lý)

Do đó pt vô nghiệm.

Answer 5

Câu e:

ĐKXĐ:...........

\(x+6\sqrt{x+8}+4\sqrt{6-2x}=27\)

\(\Leftrightarrow 27-x-6\sqrt{x+8}-4\sqrt{6-2x}=0\)

\(\Leftrightarrow (x+8-6\sqrt{x+8}+9)+(6-2x-4\sqrt{6-2x}+4)=0\)

\(\Leftrightarrow (\sqrt{x+8}-3)^2+(\sqrt{6-2x}-2)^2=0\)

Vì \((\sqrt{x+8}-3)^2\geq 0; (\sqrt{6-2x}-2)^2\geq 0\) nên để tổng của chúng bằng $0$ thì:

\((\sqrt{x+8}-3)^2=(\sqrt{6-2x}-2)^2=0\Rightarrow x=1\)

(thỏa mãn)

Vậy.............