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a, \(19\dfrac{1}{3}\) . \(\dfrac{3}{7}\) - \(33\dfrac{1}{3}\)
= \(\dfrac{58}{3}\) . \(\dfrac{3}{7}\) - \(\dfrac{100}{3}\)
= \(\dfrac{58.1}{1.7}\) - \(\dfrac{100}{3}\)
= \(\dfrac{58}{7}\) - \(\dfrac{100}{3}\) = \(\dfrac{174}{21}-\dfrac{700}{21}\)
= \(\dfrac{-526}{21}=-25\dfrac{1}{21}\)
1. f(-2) = 3.(-2)2-1 = 3.4-1 = 11
f(1/4) = 3.(1/4)2-1=-13/16
2. f(x) = 47
=> 3x2 - 1 = 47
=> 3x2 = 48
=> x2 = 16
=> x = 4 hoặc x = -4
3. f(x) = f(-x)
<=> 3x2 - 1 = 3.(-x)2 - 1
Mà x2 = (-x)2
=> 3x2 - 1 = 3.(-x)2 - 1
=> f(x) = f(-x) (đpcm)
Lê Duy Khương TH 1 bn làm đúng r. Còn TH 2 là là mạch điện hở nếu chiều dòng diện là bn làm sai
A= \(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{3}}\)
A= \(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{\frac{11}{3}\left(0,75+0,6-\frac{3}{7}-\frac{3}{13}\right)}\)
A=\(\frac{1}{\frac{11}{3}}\)
A= \(\frac{3}{11}\)
Ta có:\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
\(\Rightarrow1+\frac{x+1}{11}+1+\frac{x+2}{10}=1+\frac{x+3}{9}+1+\frac{x+4}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}=\frac{x+12}{9}+\frac{x+12}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
\(\Rightarrow\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
Mà \(\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)>0\)
\(\Rightarrow x+12=0\Rightarrow x=-12\)
\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
<=> \(\frac{x+1}{11}+\frac{x+2}{10}-\frac{x+3}{9}-\frac{x+4}{8}=0\)
<=> \(\left(\frac{x+1}{11}+1\right)+\left(\frac{x+2}{10}+1\right)-\left(\frac{x+3}{9}+1\right)-\left(\frac{x+4}{8}+1\right)=0\)<=> \(\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
<=> \(\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
<=> x + 12 = 0.Vì \(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)
<=> x = -12
Câu 1 :
\(\dfrac{3}{35}\) - ( \(\dfrac{3}{5}\)+x ) = \(\dfrac{1}{2}\)
=> \(\dfrac{3}{5}\)+x = - \(\dfrac{29}{70}\)
=> x = \(\dfrac{-71}{70}\)
Câu 2 :
\(\dfrac{1}{2}\)x-\(\dfrac{1}{3}\)= \(\dfrac{1}{4}\)x -\(\dfrac{1}{6}\)
=> \(\dfrac{1}{2}\)x -\(\dfrac{1}{4}\)x = \(\dfrac{-1}{6}\)+\(\dfrac{1}{3}\)
=> x.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)) = \(\dfrac{1}{6}\)
=> x.\(\dfrac{1}{4}\)=\(\dfrac{1}{6}\)
=> x =\(\dfrac{2}{3}\)
Câu 3 :
1\(\dfrac{3}{4}\) x +\(\dfrac{11}{2}\)=\(\dfrac{-4}{5}\)
=>\(\dfrac{7}{4}\)x = \(\dfrac{-63}{10}\)
=>x =\(\dfrac{-18}{5}\)
Câu 4:
(2x-1)2-5 = 20
=>(2x-1)2 =25
=> (2x-1)2= 52
=> 2x-1 = 5
=> 2x= 6
=> x= 6:2
=> x =3
Câu 5 :-\(\dfrac{3}{4}\)-|\(\dfrac{4}{5}\)-x | =-1
=> | \(\dfrac{4}{5}\) - x | = \(\dfrac{1}{4}\)
=> \(\dfrac{4}{5}\)-x = \(\dfrac{1}{4}\) hoặc \(\dfrac{4}{5}\)-x =\(\dfrac{-1}{4}\)
+) \(\dfrac{4}{5}\)-x = \(\dfrac{1}{4}\)=>x=\(\dfrac{11}{20}\)
+) \(\dfrac{4}{5}\) -x = \(\dfrac{-1}{4}\)=> x=\(\dfrac{21}{20}\)
Vậy .......................
A=\(\frac{9^{61}-9}{8}\)=3^122-3^2/8
vi3^122-3^2/8 <3^121 nên A<3^121