Question

Tìm x:

a.) 1(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x².

b.) 2(4x-1)²-(3x+2)(3x-2)=(7x-1)(x+2)+(2x+1)²-(4x²+7)

Answer 1

a: \(\Leftrightarrow x^2-25-x^2-6x-9+3x^2-12x+12=x^2+2x+1-x^2+16+3x^2\)

\(\Leftrightarrow3x^2-18x-22=3x^2+2x+17\)

=>-18x-22=2x+17

=>-20x=39

hay x=-39/20

b: \(\Leftrightarrow2\left(16x^2-8x+1\right)-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)

\(\Leftrightarrow32x^2-16x+2-9x^2+4=7x^2+17x-8\)

\(\Leftrightarrow23x^2-16x+6-7x^2-17x+8=0\)

\(\Leftrightarrow16x^2-33x+14=0\)

\(\text{Δ}=\left(-33\right)^2-4\cdot16\cdot14=193>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{33-\sqrt{193}}{32}\\x_2=\dfrac{33+\sqrt{193}}{32}\end{matrix}\right.\)