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g(1)=16 - 6 x 15 + 6 x 14 - 6 x 13+ 6 x 12 - 6 x 1 +11
= 1 - 6 + 6 - 6 + 6 - 6 + 11
= 6
a) Ta có \(|5\left(2x+3\right)\ge0\)
\(|2\left(2x+3\right)|\ge0\)
\(|2x+3|\ge0\)
\(\Rightarrow|5\left(2x+3\right)|+|\left(2x+3\right)|+|2x+3|\ge0\)
\(\Rightarrow5\left(2x+3\right)+2\left(2x+3\right)+2x+3=16\)
\(\Rightarrow10x+15+4x+6+2x+3=16\)
\(\Rightarrow\left(10x+4x+2x\right)+\left(15+6+3\right)=16\)
\(\Rightarrow16x+24=16\)
\(\Rightarrow24=16x-16\)
\(\Rightarrow24=x\)
Vậy x=24
Ta có:
\(\left(\right. a - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (1)
Và: \(a + 1 = b + 2 = c + 3\)
\(\Rightarrow a = b + 2 - 1 = b + 1\)
Thay vào (1) ta có:
\(\left(\right. b + 1 - \frac{1}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\)
\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. c - 3 \left.\right) = 0\) (2)
Mà: \(b + 2 = c + 3\)
\(\Rightarrow c = b + 2 - 3 = b - 1\)
Thay vào (2) ta có:
\(\left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 1 - 3 \left.\right) = 0\)
\(\Rightarrow \left(\right. b + \frac{2}{3} \left.\right) \left(\right. b + \frac{1}{2} \left.\right) \left(\right. b - 4 \left.\right) = 0\)
\(\Rightarrow \left[\right. b = - \frac{2}{3} \\ b = - \frac{1}{2} \\ b = 4\)
TH1 khi b=\(- \frac{2}{3}\)
\(\Rightarrow a = b + 1 = - \frac{2}{3} + 1 = \frac{1}{3}\)
\(\Rightarrow c = b - 1 = - \frac{2}{3} - 1 = - \frac{5}{3}\)
TH2 khi \(b = - \frac{1}{2}\)
\(\Rightarrow a = b + 1 = - \frac{1}{2} + 1 = \frac{1}{2}\)
\(\Rightarrow c = b - 1 = - \frac{1}{2} - 1 = - \frac{3}{2}\)
TH3 khi \(b = 4\)
\(\Rightarrow a = b + 1 = 4 + 1 = 5\)
\(\Rightarrow c = b - 1 = 4 - 1 = 3\)
sai mình xin lỗi
\(\frac{5}{4} x + \frac{2}{3} - 9 x - 3 = \frac{4}{5} x - 2 x - 1\)
\(- \frac{31}{4} x - \frac{7}{3} = - \frac{6}{5} x - 1\)
Chuyển các số hạng:
\(\left(\right. - \frac{31}{4} + \frac{6}{5} \left.\right) x = \frac{4}{3}\)
\(- \frac{131}{20} x = \frac{4}{3}\)
Suy ra:
\(x=\frac{4}{3}\cdot\left(\right.-\frac{20}{131}\left.\right)=-\frac{80}{393}\)
\(\dfrac23\left(\dfrac65x+\dfrac94\right)-\dfrac23\left(6x+2\right)=\dfrac45x-\dfrac13\left(6x+3\right)\)
\(\Rightarrow\dfrac45x+\dfrac32-4x-\dfrac43=\dfrac45x-2x-1\)
\(\Rightarrow\dfrac45x-4x-\dfrac45x+2x=-\dfrac32+\dfrac43-1\)
\(\Rightarrow-2x=-\dfrac76\)
\(\Rightarrow x=-\dfrac76:-2\)
\(\Rightarrow x=\dfrac76\cdot\dfrac12\)
\(\Rightarrow x=\dfrac{7}{12}\)
Vậy \(x=\dfrac{7}{12}\)
Ta có: \(\frac23\left(\frac65x+\frac94\right)-\frac23\left(6x+2\right)=\frac45x-\frac13\left(6x+3\right)\)
=>\(\frac{12}{15}x+\frac{18}{12}-4x-\frac43=\frac45x-2x-1\)
=>\(x\left(\frac{12}{15}-4\right)+\frac{18}{12}-\frac{16}{12}=-\frac65x-1\)
=>\(x\left(\frac45-4\right)+\frac{2}{12}=-\frac65x-1\)
=>\(x\left(\frac45-4+\frac65\right)=-1-\frac16\)
=>\(x\left(2-4\right)=-\frac76\)
=>\(-2x=-\frac76\)
=>\(x=\frac76:2=\frac{7}{12}\)