Question

[Lê]

Tìm x, biết:

a, x² - 10x + 9 = 0 (làm bằng 2 cách)

b, 8x² - 2x - 15 = 0 (làm bằng 2 cách)

c, 2x² + 8x - 7 = 0

d, 3x² - 15x + 3 = 0

e, 16x² - 24x - 4 = 0

f, -5x² + 6x + 3 = 0

i, 6x² - 9x + 40 = 0

Answer 1

[Nguyễn Lê Phước Thịnh]

a)

Cách 1:

Ta có: \(x^2-10x+9=0\)

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

\(\Leftrightarrow x\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)

Vậy: S={1;9}

Cách 2:

Ta có: \(x^2-10x+9=0\)

\(\Leftrightarrow x^2-10x+25-16=0\)

\(\Leftrightarrow\left(x-5\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

Vậy: S={9;1}

b)

Cách 1:

Ta có: \(8x^2-2x-15=0\)

\(\Leftrightarrow8x^2-12x+10x-15=0\)

\(\Leftrightarrow4x\left(2x-3\right)+5\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{3}{2};\frac{-5}{4}\right\}\)

Cách 2:

Ta có: \(8x^2-2x-15=0\)

\(\Leftrightarrow8\left(x^2-\frac{1}{4}x-\frac{15}{8}\right)=0\)

\(\Leftrightarrow x^2-\frac{1}{4}x-\frac{15}{8}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{8}+\frac{1}{64}-\frac{121}{64}=0\)

\(\Leftrightarrow\left(x-\frac{1}{8}\right)^2=\frac{121}{64}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{8}=\frac{11}{8}\\x-\frac{1}{8}=-\frac{11}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{12}{8}=\frac{3}{2}\\x=\frac{-11+1}{8}=\frac{-10}{8}=\frac{-5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{3}{2};\frac{-5}{4}\right\}\)

c) Ta có: \(2x^2+8x-7=0\)

\(\Leftrightarrow2\left(x^2+4x-\frac{7}{2}\right)=0\)

\(\Leftrightarrow x^2+4x+4-\frac{15}{2}=0\)

\(\Leftrightarrow\left(x+2\right)^2=\frac{15}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{\frac{15}{2}}\\x+2=-\sqrt{\frac{15}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\frac{15}{2}}-2\\x=-\sqrt{\frac{15}{2}}-2\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{\frac{15}{2}}-2;-\sqrt{\frac{15}{2}}-2\right\}\)

d) Ta có: \(3x^2-15x+3=0\)

\(\Leftrightarrow3\left(x^2-5x+1\right)=0\)

\(\Leftrightarrow x^2-5x+1=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{21}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{21}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5}{2}=\frac{\sqrt{21}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{21}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{21}+5}{2}\\x=\frac{-\sqrt{21}+5}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{\sqrt{21}+5}{2};\frac{-\sqrt{21}+5}{2}\right\}\)

e) Ta có: \(16x^2-24x-4=0\)

\(\Leftrightarrow4\left(4x^2-6x-1\right)=0\)

\(\Leftrightarrow4x^2-6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{13}{4}=0\)

\(\Leftrightarrow\left(2x-\frac{3}{2}\right)^2=\frac{13}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{3}{2}=\frac{\sqrt{13}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{13}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{3+\sqrt{13}}{2}\\2x=\frac{3-\sqrt{13}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{13}}{2}:2=\frac{3+\sqrt{13}}{4}\\x=\frac{3-\sqrt{13}}{2}:2=\frac{3-\sqrt{13}}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\frac{3+\sqrt{13}}{4};\frac{3-\sqrt{13}}{4}\right\}\)

f) Ta có: \(-5x^2+6x+3=0\)

\(\Leftrightarrow-5\left(x^2-\frac{6}{5}x-\frac{3}{5}\right)=0\)

\(\Leftrightarrow x^2-\frac{6}{5}x-\frac{3}{5}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{5}+\frac{9}{25}-\frac{24}{25}=0\)

\(\Leftrightarrow\left(x-\frac{3}{5}\right)^2=\frac{24}{25}\)

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

Vậy: \(S=\left\{\frac{3+2\sqrt{6}}{5};\frac{3-2\sqrt{6}}{5}\right\}\)

i) Ta có: \(6x^2-9x+40=0\)

\(\Leftrightarrow6\left(x^2-\frac{3}{2}x+\frac{20}{3}\right)=0\)

\(\Leftrightarrow x^2-\frac{3}{2}x+\frac{20}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}+\frac{293}{48}=0\)

\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2+\frac{293}{48}=0\)(vô lý)

Vậy: \(S=\varnothing\)