Question

giải các pt sau

i)\(\left(1-\sqrt{2}\right)x^2-2\left(1+\sqrt{2}\right)x+1+3\sqrt{2}=0\)

 

ii)\(^{\left(6x^2-7x\right)^2-2\left(6x^2-7x\right)-3=0}\)

Answer 1

i)

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

\(\left(x-1\right)^2+\sqrt{2}\left(x^2-2x+3\right)=0\)

\(\left(x-1\right)^2+\sqrt{2}\left(x-1\right)^2+2\sqrt{2}=0\)

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

=> Phương trình vô nghiệm

ii)

Đặt: \(6x^2-7x=a\)

Ta có: \(a^2-2a-3=0\)

\(\left(a-3\right)\left(a+1\right)=0\)

\(\left(6x^2-7x-3\right)\left(6x^2-7x+1\right)=0\)

\(x=\frac{3}{2};-\frac{1}{3};1;\frac{1}{6}\)

Answer 2

 Phương trình vô nghiệm

ii)

Đặt: $6x^2-7x=a$6x2−7x=a

Ta có: $a^2-2a-3=0$a2−2a−3=0

$\left(a-3\right)\left(a+1\right)=0$(a−3)(a+1)=0

$\left(6x^2-7x-3\right)\left(6x^2-7x+1\right)=0$(6x2−7x−3)(6x2−7x+1)=0

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