Question

1/ Tìm x:

10x+3-5x=4x+2

2/ Tìm giá trị biểu thức:

Q = \(\left(x+3\right)^2+\left(x+3\right).\left(x-3\right)-2.\left(x+2\right).\left(x-4\right)\) Biết x = \(\dfrac{1}{2}\)

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Answer 1

Bài 1: \(10x+3-5x=4x+2\)

\(\Leftrightarrow5x+3=4x+2\) \(\Leftrightarrow5x-4x=2-3\Leftrightarrow x=-1\)

Vậy x = - 1

Bài 2: \(Q=\left(x+3\right)^2+\left(x+3\right)\left(x-3\right)-2\left(x+2\right)\left(x-4\right)\)

\(=x^2+6x+9+x^2-9-2\left(x^2-2x-8\right)\)

\(=x^2+6x+9+x^2-9-2x^2+4x+16\)

\(=6x+4x+16=10x+16\)

Thay \(x=\dfrac{1}{2}\) ta được \(Q=10\cdot\dfrac{1}{2}+16=5+16=21\)