2 chữ số 0

có 2 chữ số 0 nha

`(-5/4)^8 : (-5/4)^6 - x = (-3)/2`

\(\left(-\dfrac{5}{4}\right)^{8-6}-x=\dfrac{-3}{2}\)

`(-5/4)^2 - x = (-3)/2`

`25/16 - x = (-3)/2`

`x=25/16- (-3)/2`

`x=25/16+3/2`

`x=49/16`

\((\dfrac{-5}{4})^8:(\dfrac{-5}{4})^6-x=\dfrac{-3}{2}\)

\((\dfrac{-5}{4})^2-x=\dfrac{-3}{2}\)

\(\dfrac{25}{16}-x=\dfrac{-3}{2}\)

\(x=\dfrac{25}{16}-\dfrac{-3}{2}\)

\(x=\dfrac{49}{16}\)

`(-4)^2-5/3:x=([-1]/2)^3`

`=>16-5/3:x=[-1]/8`

`=>5/3:x=16-[-1]/8=129/8`

`=>x=5/3:129/8`

`=>x=40/387`

\(\left(-4\right)^2-\dfrac{5}{3}:x=\left(-\dfrac{1}{2}\right)^3\\ 16-\dfrac{5}{3}:x=-\dfrac{1}{8}\\ \dfrac{5}{3}:x=\dfrac{129}{8}\\ x=\dfrac{40}{387}\)

`(-5/4)^8 : (-5/4)^6 - x = (-3)/2`

\(\left(-\dfrac{5}{4}\right)^{8-6}-x=\dfrac{-3}{2}\)

`(-5/4)^2 - x = (-3)/2`

`25/16 - x = (-3)/2`

`x=25/16- (-3)/2`

`x=25/16+3/2`

`x=49/16`

\(\left(-\dfrac{5}{4}\right)^8:\left(-\dfrac{5}{4}\right)^6-x=-\dfrac{3}{2}\\ \Rightarrow\left(-\dfrac{5}{4}\right)^2-x=-\dfrac{3}{2}\\ \Rightarrow\dfrac{25}{16}-x=-\dfrac{3}{2}\\ \Rightarrow x=\dfrac{25}{16}+\dfrac{3}{2}\\ \Rightarrow x=\dfrac{49}{16}\)

`x + 1 3/7 = (-2/7)^4 : (-2/7)^3`

`x + 1 3/7 =` \(\left(-\dfrac{2}{7}\right)^{4-3}\)

`x + 1 3/7 = (-2)/7`

`x + 10/7 = (-2)/7`

`x= (-2)/7 -10/7`

`x=(-12)/7`

\(x+1\dfrac{3}{7}=-\dfrac{2^4}{7}\div-\dfrac{2^3}{7}\)

\(x+\dfrac{10}{7}=-\dfrac{16}{7}\div-\dfrac{8}{7}\)

\(x+\dfrac{10}{7}=\dfrac{2}{7}\)

\(x=\dfrac{-8}{7}\)

\(\dfrac{2^{10}\left(1+2^5\right)}{1+2^5}=2^{10}\)

\(\left(2^{15}+2^{10}\right)\div\left(1+2^5\right)\)

\(\Rightarrow\left(2^{10}.33\right)\div33\)

\(\Rightarrow2^{10}.\left(33\div33\right)\)

\(\Rightarrow2^{10}.1\)

\(\Rightarrow2^{10}\)

Theo định lí Pytago tam giác AHB vuông tại H

\(AB=\sqrt{AH^2+HB^2}=13cm\)

Theo định lí Pytago tam giác AHC vuông tại H

\(HC=\sqrt{AC^2-AH^2}=16cm\)

BC = HB + HC = 21 cm 

$\Delta ABH$ vuông tại $H,$ ta có:

$A{H}^2+B{H}^2=A{B}^2$ (định lí Pi-ta-go)

hay ${12}^2+5^2=A{B}^2$

$\Rightarrow 144+25=A{B}^2$

$\Rightarrow 169=A{B}^2$

$\Rightarrow AB=\sqrt{169}=13$ $\left(cm\right)$

$\Delta AHC$ vuông tại $H,$ ta có:

$A{H}^2+H{C}^2=A{C}^2$

hay ${12}^2+H{C}^2={20}^2$

$\Rightarrow H{C}^2={20}^2-{12}^2$

$\Rightarrow H{C}^2=400-144=256$

$\Rightarrow HC=\sqrt{256}=16$ $\left(cm\right)$

Mà $BC=HB+HC$

Hay $BC=5+16=21$ $\left(cm\right)$