Question

Tìm x

a)4(x+3)(3x-2)-3(x+1)(4x-1)=-27

b)(x+1)(3x²-x+1)+x²(4-3x)=5/2

c)2(x-2)(x+2)+4(x-2)(x+1)+(x+2)(8+5x)=0

d)(2x+1)(5x-1)=20x²-16x-1

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Answer 1

a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)

\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)

\(\Leftrightarrow19x-21=-27\)

=>19x=-6

hay x=-6/19

b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2=\dfrac{3}{2}\)

\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)

=>x=1/2 hoặc x=-1/2

c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)

\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)

\(\Leftrightarrow3x^2+22x+16=0\)

\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)

\(\Leftrightarrow10x^2-19x=0\)

=>x(10x-19)=0

=>x=0 hoặc x=19/10