а)
6x {}^{2} + x - 7 = 06x2+x−7=0
D = 1 {}^{2} - 4 \times 6 \times ( - 7) = 1 + 168 = 169D=12−4×6×(−7)=1+168=169
\begin{gathered}x1 = \frac{ - 1 + \sqrt{169} }{2 \times 6} = \frac{ - 1 + 13}{12} = \frac{12}{12} = 1 \\ x2 = \frac{ - 1 - \sqrt{169} }{2 \times 6} = \frac{ - 1 - 13}{12} = \frac{ - 14}{12} = - \frac{7}{6} \end{gathered}
Объяснение:
x1=2×6−1+169=12−1+13=1212=1x2=2×6−1−169=12−1−13=12−14=−67
имеется 2 корня
otvet:x1 = 1x2 = - \frac{7}{6}otvet:x1=1 x2=−67
б)
x {}^{2} - 6x + 2 = 0x2−6x+2=0
D = (- 6) {}^{2} - 4 \times 1 \times 2 = 36 - 8 = 28D=(−6)2−4×1×2=36−8=28
x1 = \frac{ - ( - 6) + \sqrt{28} }{2 \times 1} = \frac{6 + 2 \sqrt{7} }{2} = \frac{2(3 + \sqrt{7)} }{2} = 3 + \sqrt{7}x1=2×1−(−6)+28=26+27=22(3+7)=3+7
x2 = \frac{ - ( - 6) - \sqrt{28} }{2 \times 1} = \frac{6 - 2 \sqrt{7} }{2} = \frac{2(3 - \sqrt{7)} }{2} = 3 - \sqrt{7}x2=2×1−(−6)−28=26−27=22(3−7)=3−7
имеется 2 корня
otvet:x1 = 3 + \sqrt{7} x2 = 3 - \sqrt{7}otvet:x1=3+7 x2=3−7
a²-c²-2ab+b²-(a²-2ab+b²-c²)=0
a²-c²-2ab+b²-a²+2ab-b²+c²=0
0=0- верно.
2): 3х-5(2х+1)=3(3-2х)
3х-10х-5=9-6х
3х-10х+6х=9+5
-х=14
х=-14
3)(2х2у)³=2³х³·2³у³=8х³·8у³=64х³у³