phuong thang dung va mat nam ngang = 90^o

goc toi + goc pxa = 30+90 = 120^o

goc giua tia toi va guong = (180-120)/2 = 30⁰

=> goc giua mat pxa va phuong nam ngang = 30+30 = 60^o

Nguyễn Mạnh Linh. Ơ này trên học 24 và đó là câu trả lời của tui mà

Nguyễn Mạnh Linh Spam rồi

\(150+\left(50+x\right)=300\)

\(\Leftrightarrow50+x=300-150\)

\(\Leftrightarrow50+x=150\)

\(\Leftrightarrow x=150-50\)

\(\Leftrightarrow x=100\)

a)  (a-b)-(c-b)-a=a-b-c+b-a=(a-a)+(b-b)-c=0+0-c=-c

1)

a) (a-b) - (c-b) - a= a - b - c + b - a = -c

b) -(300 - 400) + (300 - 400) + 100 = 100

2) a) 4x - 20 =  50 - ( 30 -3x)

        4x - 20 = 50 - 30 + 3x

        4x - 3x = 50 - 30 + 20= 40

        x          = 40

b) 100 - (-5x) = 40 - (-4x +10)

    100 + 5x    = 40 + 4x -10

    5x - 4x       = 40 - 10 -100

    x                = -70

c) (-50) + (-7x) = 100 - 8x

     -7x + 8x      = 100 + 50

      x                = 150

Bài 1.

a) \(\left(a-b\right)-\left(c-b\right)-a\)

\(=a-b-c+b-a\)

\(=\left(a-a\right)+\left(-b+b\right)-c\)

\(=-c\)

b)\(-\left(300-400\right)+\left(300-400\right)+100\)

\(=-300+400+300-400+100\)

\(=\left(-300+300\right)+\left(400-400\right)+100\)

\(=100\)

Bài 2.

a)\(4x-20=50-\left(30-3x\right)\)

\(\Leftrightarrow4x-20=50-30+3x\)

\(\Leftrightarrow4x-20=20+3x\)

\(\Leftrightarrow x=40\)

Vậy x = 40

b)\(100-\left(-5x\right)=40-\left(-4x+10\right)\)

\(\Leftrightarrow100+5x=40+4x-10\)

\(\Leftrightarrow x=-70\)

Vậy x = -70

c)\(\left(-50\right)+\left(-7x\right)=100-8x\)

\(\Leftrightarrow-50-7x=100-8x\)

\(\Leftrightarrow x=150\)

Vậy x = 150

Linz

`@` `\text {Ans}`

`\downarrow`

`a)`

`13/50 + 9% + 41/100 + 0,24`

`= 0,26 + 0,09 + 0,41 + 0,24`

`= (0,26 + 0,24) + (0,09 + 0,41)`

`= 0,5 + 0,5`

`= 1`

`b)`

`2018 \times 2020 - 1/2017 + 2018 \times 2019`

`= 2018 \times (2020 + 2019) - 1/2017`

`= 2018 \times 4039 - 1/2017`

`= 8150702`

`c)`

`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`

`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)

`=`\(1-\dfrac{1}{7}\)

`= 6/7`

\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)

Cảm ơn các bạn đã trả lời giúp mình câu hỏi nha

Từ GT ; ta có :  \(\left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=224\)

\(\Rightarrow\left(x-1\right).\dfrac{2}{9}=224\)

\(\Rightarrow\left(x-1\right)=1008\)

\(\Rightarrow x=1009\)

Vậy ... 

\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)

Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)

Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

a) \(10^{20}\) và \(9^{10}\)

Vì 10 > 9 ; 20 > 10

nên \(10^{20}>9^{10}\)

Vậy \(10^{20}>9^{10}\)

b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)

Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

           \(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)

Vì 243 > 125 nên \(125^{10}< 243^{10}\)

Vậy \(\left(-5\right)^{30}< \left(-3\right)^{50}\)

c) \(64^8\) và \(16^{12}\)

Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)

          \(16^{12}=\left(4^2\right)^{12}=4^{24}\)

Vậy \(64^8=16^{12}\left(=4^{24}\right)\)

d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)

Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)

Vậy \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)

\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+.....+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+.....+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow A=\frac{1}{5}-\frac{1}{12}\)

\(\Rightarrow A=\frac{12}{60}-\frac{5}{60}\)

\(\Rightarrow A=\frac{7}{60}\)

k mk nha mk nhanh nhất và đúng 100% luôn 

mk đợi k của cậu 

A =\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{110}+\frac{1}{132}\)

A =\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

A =\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

A =\(\frac{1}{5}-\frac{1}{12}\)

A =\(\frac{7}{60}\)

Càm ơn các bạn nha ! Nhưng mà cả hai bạn đều làm đúng , mình không biết nên tk cho ai ?

Mình tính từng cái ra nha, từng cái sẽ ra được kết quả của phép tính:

\(1-\dfrac{1}{5}-\dfrac{1}{6}\)

\(=\left(1-\dfrac{1}{5}\right)-\dfrac{1}{6}\)

\(=\left(\dfrac{5}{5}-\dfrac{1}{5}\right)-\dfrac{1}{6}\)

\(=\dfrac{4}{5}-\dfrac{1}{6}\)

\(=\dfrac{24}{30}-\dfrac{5}{30}\)

\(=\dfrac{19}{30}\)

\(\dfrac{19}{30}\) nha

19/30 nhé