\(\left(123+164\right).75+164.925+25.123\)

\(=123.75+164.75+164.925+25.123\)

\(=123.\left(75+25\right)+164.\left(75+925\right)\)

\(=123.100+164.1000\)

\(=12300+164000\)

\(=176300\)

\(16:\left\{400:\left[200-\left(37+46.3\right)\right]\right\}\)

\(=16:\left\{400:\left[200-\left(37+138\right)\right]\right\}\)

\(=16:\left\{400:\left[200-175\right]\right\}\)

\(=16:\left\{400:25\right\}\)

\(=16:16\)

\(=1\)

\(\left(123+164\right).75+164.925+25.123\)

\(=123.75+164.75+164.925+25.123\)

\(=\left(123.75+25.123\right)+\left(164.75+164.925\right)\)

\(=123.\left(75+25\right)+164.\left(75+925\right)\)

\(=123.100+164.1000\)

\(=12300+164000\)

\(=176300\)

=====================

\(16:\left\{400:\left[200-\left(37+46.3\right)\right]\right\}\)

\(=16:\left\{400:\left[200-175\right]\right\}\)

\(=16:\left\{400:25\right\}\)

\(=16:16\)

\(=1\)

\(\left(-5\right)^5=\left(-5\right)^4\cdot\left(-5\right)=5^4\cdot\left(-5\right)=625\cdot\left(-5\right)=-3125\)

`  25.x:17-6=19`

`=>   25.x:17 =19 + 6`

`=>   25.x:17= 25`

`=> 25.x = 25.17`

`=> x =25 . 17 : 25`

`=> x = 17`

Vậy `x = 17`

``
` 2021-10.(x-5)=2021`

`=> 10.(x-5) = 2021 - 2021`

`=> 10 (x-5) = 0`

`=> x - 5 = 0 : 10`

`=> x - 5 = 0`

`=> x = 0+5`

`=> x = 5`

Vậy `x = 5`

\(25x:17-6=19\)

\(25x:17=19+6\)

\(25x:17=25\)

\(x:17=25:25\)

\(x:17=1\)

\(x=1.17\)

\(x=17\)

Vậy `x = 17`

\(2021-10.\left(x-5\right)=2021\)

\(10.\left(x-5\right)=2021-2021\)

\(10.\left(x-5\right)=0\)

\(x-5=0:10\)

\(x-5=0\)

\(x=0+5\)

\(x=5\)

Vậy `x = 5`

a: Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

=>\(\widehat{C}+\widehat{D}=360^0-110^0-70^0=180^0\)

=>\(\dfrac{1}{3}\cdot\widehat{D}+\widehat{D}=180^0\)

=>\(\dfrac{4}{3}\cdot\widehat{D}=180^0\)

=>\(\widehat{D}=135^0\)

\(\widehat{C}=\dfrac{1}{3}\cdot135^0=45^0\)

b:

Sửa đề: Cho tứ giác ABCD.

Đặt \(\widehat{B}=x;\widehat{C}=y;\widehat{D}=z\)

\(\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}}{4}\)

=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

=>\(x+y+z=360^0-90^0=270^0\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{270}{9}=30^0\)

=>\(x=2\cdot30^0=60^0;y=3\cdot30^0=90^0;z=4\cdot30^0=120^0\)

Vậy: \(\widehat{B}=x=60^0;\widehat{C}=y=90^0;\widehat{D}=z=120^0\)

a: \(x=\left(x^3\right)^{\dfrac{1}{3}}\)

b: \(x=\left(x^5\right)^{\dfrac{1}{5}}\)

\(\dfrac{1}{3^6}=\dfrac{1}{3^4\cdot3^2}=\dfrac{1}{81\cdot9}=\dfrac{1}{729}\)

  \(\dfrac{1}{3^6}\) = \(\dfrac{1}{3^4.3^2}\) = \(\dfrac{1}{81.9}\) = \(\dfrac{1}{729}\) 

\(\widehat{C}=\widehat{B}+10^0=\widehat{A}+10^0+10^0=\widehat{A}+20^0\)

\(\widehat{D}=\widehat{C}+10^0=\widehat{A}+20^0+10^0=\widehat{A}+30^0\)

Xét tứ giác ABCD có \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

=>\(\widehat{A}+\widehat{A}+10^0+\widehat{A}+20^0+\widehat{A}+30^0=360^0\)

=>\(4\cdot\widehat{A}=300^0\)

=>\(\widehat{A}=75^0\)

\(\widehat{B}=75^0+10^0=85^0\)

\(\widehat{C}=75^0+20^0=95^0\)

\(\widehat{D}=75^0+30^0=105^0\)

Chưa rõ cụ thể em hỏi cái gì?

nhờ mn giúp mình viết 1 dãy gồm 10000 số pi với ạ mình cảm ơn