![\sqrt{2x^{2}+8x+7}=x+2\\2x^{2}+8x+7=(x+2)^{2}\\2x^{2}+8x+7=x^{2}+4x+4\\2x^{2}+8x+7-x^{2}-4x-4=0\\x^{2}+4x+3=0\\x_{1}*x_{2}=3\\x_{1}+x_{2}=-4\\x_{1}=-1\\x_{2}=-3](https://tex.z-dn.net/?f=+%5Csqrt%7B2x%5E%7B2%7D%2B8x%2B7%7D%3Dx%2B2%5C%5C2x%5E%7B2%7D%2B8x%2B7%3D%28x%2B2%29%5E%7B2%7D%5C%5C2x%5E%7B2%7D%2B8x%2B7%3Dx%5E%7B2%7D%2B4x%2B4%5C%5C2x%5E%7B2%7D%2B8x%2B7-x%5E%7B2%7D-4x-4%3D0%5C%5Cx%5E%7B2%7D%2B4x%2B3%3D0%5C%5Cx_%7B1%7D%2Ax_%7B2%7D%3D3%5C%5Cx_%7B1%7D%2Bx_%7B2%7D%3D-4%5C%5Cx_%7B1%7D%3D-1%5C%5Cx_%7B2%7D%3D-3)
-3 не удовлетворяет из ОДЗ, поэтому ответ: только -1.
А) = 68.7-44-0.375=25.075
б) 90.4+65.4-90.8=65
в) = 504-47.9+58.7-49=465.8
г) = 17.654-37+22.9+0.345=3.899
Х+у=39, а х-у=5, из первого уравнения выразим Х и подставим во второе
х=39-у, (39-у)-у=5, -2у=5-39, -2у=-34, у=17, , тогда х=39-17,
х=22
Ответ 22 и 17
<span>2р.75 к.=.275к.
6р.90к.=.690к.
4р.5к.=405к.
9р.17к.=917к.</span>
7.8+x=14.4
x=14.4-7.8
x=6.6
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