Question

Tìm x biết :

(x+2)^2-2(x+2)(x+3)+(x+5)^2=7

Mn giúp mik vs ạ mik cần gấp mai mik phải ik hok rùi ạ

 

Answer 1

ta có \(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)

\(\Leftrightarrow x^2+4x+4-2\left(x^2+5x+6\right)+x^2+10x+25=7\)

\(\Leftrightarrow4x+10=0\Leftrightarrow x=-\frac{5}{2}\)

Answer 2

Bạn áp dụng hằng đẳng thức số 1, nhân phá ngoặc là Ok nhé

\(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)

\(\Leftrightarrow x^2+4x+4-2\left(x^2+3x+2x+6\right)+x^2+10x+25-7=0\)

\(\Leftrightarrow2x^2+14x+22-2x^2-6x-4x-12=0\)

\(\Leftrightarrow4x+10=0\)

\(\Leftrightarrow4x=-10\)

\(\Leftrightarrow x=\frac{-5}{2}\)

Answer 3

Trả lời:

( x + 2 )² - 2 ( x + 2 ) ( x + 3 ) + ( x + 5 )² = 7

<=> x² + 4x + 4 - 2 ( x² + 5x + 6 ) + x² + 10x + 25 = 7

<=> x² + 4x + 4 - 2x² - 10x - 12 + x² + 10x + 25 =  7

<=> 4x + 17 = 7

<=> 4x = - 10

<=> x = - 5/2

Vậy x = - 5/2 là nghiệm của pt.

Answer 4

(x+2)² - 2(x+2)(x+3)+(x+5)²=7

\(\Leftrightarrow\)x²+4x+4 - (2x+4)(x+3)+x² +10x +25 - 7=0

\(\Leftrightarrow\)2x² + 14x+22 -(2x²+6x+4x+12)=0

\(\Leftrightarrow\)2x²+14x+22- 2x²- 6x- 4x- 12=0

\(\Leftrightarrow\)4x= -10

\(\Leftrightarrow\)x= \(\frac{-10}{4}\)

Vậy x =\(\frac{-10}{4}\)