Question

Tìm x

a)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)6

b)\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

 

Nhanh nha kick cho :)

Answer 1

câu a) bỏ 1 số 6 đi nha

Answer 2

Ta có (x - 2)² - (x - 3) (x + 3) = 6 

=> (x - 2)² - (x² - 3²) = 6

=> x² - 4x + 2² - x² + 3² = 6

=> x² - x² - 4x + 4 + 9 = 6

=> - 4x + 13 = 6 

=> -4x = -7

=> x = \(\frac{-7}{-4}=\frac{7}{4}\)

Answer 3

a) \(\Leftrightarrow\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9=6\)

\(\Leftrightarrow13-4x=6\)

\(\Leftrightarrow4x=7\)

\(\Leftrightarrow x=\frac{7}{4}\)

b) \(4(x−3)^2−(2x−1)(2x+1)=10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)−4x^2+1=10\)

\(\Leftrightarrow4x^2-24x+36−4x^2+1=10\)

\(\Leftrightarrow37-24x=10\)

\(\Leftrightarrow24x=27\)

\(\Leftrightarrow x=\frac{9}{8}\)