Tính:
a) \(\left(\dfrac{4}{9}\right)^{x+1}\) = \(\left(\dfrac{8}{27}\right)^6\)
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\(2\dfrac{2}{3}-\dfrac{5}{12}-\left(1\dfrac{3}{4}-2\dfrac{1}{2}\right)-\left|1-1\dfrac{1}{6}\right|+\left|\dfrac{5}{3}\right|\)
\(=\dfrac{8}{3}-\dfrac{5}{12}-\left(\dfrac{7}{4}-\dfrac{5}{2}\right)-\left|1-\dfrac{7}{6}\right|+\dfrac{5}{3}\)
\(=\dfrac{8}{3}-\dfrac{5}{12}-\left(\dfrac{7}{4}-\dfrac{10}{4}\right)-\left|\dfrac{6}{6}-\dfrac{7}{6}\right|+\dfrac{5}{3}\)
\(=\dfrac{8}{3}-\dfrac{5}{12}-\left(\dfrac{7-10}{4}\right)-\left|\dfrac{6-7}{6}\right|+\dfrac{5}{3}\)
\(=\dfrac{8}{3}-\dfrac{5}{12}-\dfrac{-3}{4}-\dfrac{1}{6}+\dfrac{5}{3}\)
\(=\dfrac{8}{3}-\dfrac{5}{12}+\dfrac{3}{4}-\dfrac{1}{6}+\dfrac{5}{3}\)
`=32/12 - 5/12 + 9/12 - 2/12 + 20/12`
\(=\dfrac{32-5+9-2+20}{12}\)
`=54/12`
`=9/2`
Lời giải:
Theo định lý về tổng 3 góc trong tam giác vuông: $A+B+C=180^0$
$6A=4B=3C$
$\Rightarrow \frac{6A}{12}=\frac{4B}{12}=\frac{3C}{12}$
hay $\frac{A}{2}=\frac{B}{3}=\frac{C}{4}$
Áp dụng TCDTSBN:
$\frac{A}{2}=\frac{B}{3}=\frac{C}{4}=\frac{A+B+C}{2+3+4}=\frac{180^0}{9}=20^0$
$\Rightarrow A=2.20^0=40^0; B=3.20^0=60^0; C=4.20^0=80^0$
\(\left(\dfrac{4}{9}\right)^{x-1}=\left(\dfrac{8}{27}\right)^6\)
`<=>` \(\left[\left(\dfrac{2}{3}\right)^2\right]^{x+1}=\left[\left(\dfrac{2}{3}\right)^3\right]^6\)
`=>` \(\left(\dfrac{2}{3}\right)^{2x+2}=\left(\dfrac{2}{3}\right)^{18}\)
`=> 2x + 2 = 18`
`=> 2x = 18 - 2`
`=> 2x =16`
`=> x =16 : 2`
`=> x =8`
Vậy `x=8`
giúp mình với =(((