Question

[G.Dr]

Giải phương trình:

a) \(3x^2-5x+2=0\)

b) \(7x^2-5x-2=0\)

c) \(\left(x^2+x\right)^2+5\left(x^2+x\right)+6=0\)

d) \(x-7\sqrt{x}-9=0\)

e) \(x-5\sqrt{x}+4=0\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(3x^2-5x+2=0\)

\(\Leftrightarrow3x^2-3x-2x+2=0\)

\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)

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

Vậy: Tập nghiệm \(S=\left\{1;\frac{2}{3}\right\}\)

b) Ta có: \(7x^2-5x-2=0\)

\(\Leftrightarrow7x^2-7x+2x-2=0\)

\(\Leftrightarrow7x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{-2}{7}\right\}\)

c) Ta có: \(\left(x^2+x\right)^2+5\left(x^2+x\right)+6=0\)

\(\Leftrightarrow\left(x^2+x\right)^2+2\left(x^2+x\right)+3\left(x^2+x\right)+6=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+2\right)+3\left(x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x^2+x+2\right)\left(x^2+x+3\right)=0\)(1)

Ta có: \(x^2+x+2\)

\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\)

Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\forall x\)

hay \(x^2+x+2\ne0\forall x\)(2)

Ta có: \(x^2+x+3\)

\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{11}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\)

Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

hay \(x^2+x+3\ne0\forall x\)(3)

Từ (1), (2) và (3) suy ra \(x\in\varnothing\)

Vậy: Tập nghiệm \(S=\varnothing\)

d) Ta có: \(x-7\sqrt{x}-9=0\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{7}{2}+\frac{49}{4}-\frac{49}{4}-\frac{36}{4}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\frac{7}{2}\right)^2=\frac{85}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\frac{7}{2}=\frac{\sqrt{85}}{2}\\\sqrt{x}-\frac{7}{2}=-\frac{\sqrt{85}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\frac{\sqrt{85}}{2}+\frac{7}{2}=\frac{\sqrt{85}+7}{2}\\\sqrt{x}=\frac{-\sqrt{85}}{2}+\frac{7}{2}=\frac{7-\sqrt{85}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left(\frac{\sqrt{85}+7}{2}\right)^2=\frac{67+7\sqrt{85}}{2}\\x=\left(\frac{7-\sqrt{85}}{2}\right)^2=\frac{67-7\sqrt{85}}{2}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{\frac{67+7\sqrt{85}}{2};\frac{67-7\sqrt{85}}{2}\right\}\)

e) Ta có: \(x-5\sqrt{x}+4=0\)

\(\Leftrightarrow x-\sqrt{x}-4\sqrt{x}+4=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\)

Vậy: Tập nghiệm S={1;16}