Question

[youjthanh]

a. (x-4)(x+4)-x(x+2)=10

b.\(\frac{\left(x+3\right)}{2}\)-\(\frac{\left(x-2\right)}{3}\)=2-\(\frac{\left(x+3\right)}{2}\)

mong mọi người giúp dỡ

Answer 1

[Phạm Ngọc Ánh]

a) \(\left(x-4\right)\left(x+4\right)-x\left(x+2\right)=10\)

<=> \(x^2-16-x^2-2x=10\)

<=> \(-16-2x-10=0\)

<=> \(x=-13\)

Vậy pt có tập nghiệm S\(\)={-13}

b) \(\frac{\left(x+3\right)}{2}-\frac{\left(x-2\right)}{3}=2-\frac{\left(x+3\right)}{2}\)

<=> \(3\left(x+3\right)-2\left(x-2\right)=2.6-3\left(x+3\right)\)

<=> \(3x+9-2x+4=12-3x-9\)

<=> \(3x+9-2x+4-12+3x+9=0\)

<=> \(4x+10=0\)

<=> \(x=\frac{-5}{2}\)

Vậy pt có tập nghiệm S={\(\frac{-5}{2}\)}

Answer 2

[Nguyễn Quang Định]

a) \(\left(x-4\right)\left(x+4\right)-x\left(x+2\right)=10\)

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

\(\Leftrightarrow-26=2x\Leftrightarrow x=\frac{-26}{2}=-13\)

b) \(\frac{\left(x+3\right)}{2}-\frac{\left(x-2\right)}{3}=2-\frac{\left(x+3\right)}{2}\)

\(\Leftrightarrow\left(\frac{3\left(x+3\right)-2\left(x-2\right)}{6}\right)=\frac{12-3\left(x+3\right)}{6}\)

\(\Leftrightarrow3x+9-2x+4=12-3x-9\)

\(\Leftrightarrow x+13=-3x+3\)

\(\Leftrightarrow x+3x=-13+3\)

\(\Leftrightarrow4x=-10\Leftrightarrow x=\frac{-10}{4}=-2,5\)