Решение
а) (2x + 7) + (-x + 12) = 14
2x + 7 - x + 12 = 14
x + 19 = 14
x = 14 - 19
x = - 5
б) (-5y + 1) - (3y + 2) = - 9
-5y + 1 - 3y - 2 = - 9
-8y - 1 = - 9
- 8y = - 9 + 1
-8y = -8
y = 1
Нужно вместо а поставить какое либо число например 2)
![\sqrt{x+6-4 \sqrt{x+2} } + \sqrt{11+x-6 \sqrt{x+2} }=1](https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%2B6-4%20%5Csqrt%7Bx%2B2%7D%20%7D%20%2B%20%5Csqrt%7B11%2Bx-6%20%5Csqrt%7Bx%2B2%7D%20%7D%3D1)
![\sqrt{x+6-4 \sqrt{x+2} }=\sqrt{(\sqrt{x+2}-2)^2}=\sqrt{x+2}-2](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B6-4%20%5Csqrt%7Bx%2B2%7D%20%7D%3D%5Csqrt%7B%28%5Csqrt%7Bx%2B2%7D-2%29%5E2%7D%3D%5Csqrt%7Bx%2B2%7D-2)
![\sqrt{11+x-6 \sqrt{x+2} }=\sqrt{(3-\sqrt{x+2})^2}=3-\sqrt{x+2}](https://tex.z-dn.net/?f=%5Csqrt%7B11%2Bx-6%20%5Csqrt%7Bx%2B2%7D%20%7D%3D%5Csqrt%7B%283-%5Csqrt%7Bx%2B2%7D%29%5E2%7D%3D3-%5Csqrt%7Bx%2B2%7D)
отсюда
![\sqrt{x+2}-2+3-\sqrt{x+2}=1 \\ 3-2=1](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B2%7D-2%2B3-%5Csqrt%7Bx%2B2%7D%3D1%20%5C%5C%203-2%3D1)
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![\sqrt{x+2}](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B2%7D)
-любое число, но х+2≥0 ⇒ х ≥ -2
A)-3,-2,-1,0,1,2,3,4,5
б)-2,-1,0,1,2
в)-2
г)-4,-3,-2,-1