Ai trả lời nhanh nhất mình tik cho

Xét tổng 1 + 3 +5 +... + 99:

- Số số hạng là: (99-1):2+1= 50 số

- Tổng là: (1 + 99) x 50 : 2 = 2500

Vậy ta có biểu thức:

 \(2500=\left(x-2\right)^2\\ \Leftrightarrow50^2=\left(x-2\right)^2\\\Leftrightarrow \left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)

Vậy...

\(1+3+5+...+99=\left(x-2\right)^2\)

\(\Rightarrow\left[\left(99-1\right):2+1\right]\left(1+99\right):2=\left(x-2\right)^2\)

\(\Rightarrow50.100:2=\left(x-2\right)^2\)

\(\Rightarrow\left(x-2\right)^2=2500=50^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=50\\x-2=-50\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=52\\x=-48\end{matrix}\right.\)

\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)

\(\Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^2-1\right]=0\)

\(\Rightarrow\left(3x-1\right)^{10}\left(3x-1+1\right)\left(3x-1-1\right)=0\)

\(\Rightarrow\left(3x-1\right)^{10}.3x.\left(3x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x=0\\3x-1=0\\3x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=1\\3x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

x = 0 hoặc 1/3 hoặc 2/3

Diện tích được chồng lên nhau :

\(19.19=361\left(cm^2\right)\)

Hiệu diện tích các phần không bị chồng lên nhau :

\(\left(20-19\right).20=20\left(cm^2\right)\)

20cm2

\(...=57.\left(12+15+63\right)=57.90=5130\)

\(12.57+57.15+63.57\\ \Rightarrow57.\left(12+15+63\right).\\ \Rightarrow57.90=5130.\)

`@` `\text {Ans}`

`\downarrow`

`22 \times x + 27 \times x + x = 25`

`x \times (22 + 27 + 1) = 25`

`x \times 50 = 25`

`x = 25 \div 50`

`x =`\(\dfrac{25}{50}=\dfrac{1}{2}\)

Vậy, \(x=\dfrac{1}{2}\)

\(22xX+27xX+X=25\\ \left(22+27+1\right)xX=25\\ 50xX=25\\ x=50:25=2.\)

Vậy \(X=2.\)

a) \(\left|2x+1\right|=\left|x+4\right|\Rightarrow\left[{}\begin{matrix}2x+1=x+4\\2x+1=-x-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(\left|2x-1\right|=x+4\Rightarrow\left[{}\begin{matrix}2x-1=x+4\\2x-1=-x-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Ta có:

\(8=2^3\\ 31=31\)

\(\Rightarrow BCNN\left(8,31\right)=2^3.31=248.\)

\(\Rightarrow BCNN\left(8,31\right)=B\left(248\right)=\left\{0;248;496;744;...\right\}\)

Vì 31 là số nguyên tố

=>BC(8,31)=1

\(S_{ABCD}=\left(AB+CD\right)x\dfrac{h}{2}\) (h là chiều cao, vuông góc AB và CD)

\(S_{ABCD}=\left(\dfrac{1}{3}xCD+CD\right)x\dfrac{h}{2}\)

\(S_{ABCD}=\dfrac{4}{3}xCDx\dfrac{h}{2}=4x\left(\dfrac{1}{3}xCDx\dfrac{h}{2}\right)\)

mà \(S_{ABD}=ABx\dfrac{h}{2}=\dfrac{1}{3}xCDx\dfrac{h}{2}\)

Nên \(S_{ABCD}=4xS_{ABD}\)

\(\Rightarrow S_{ABD}=S_{ABCD}:4=200:4=50\left(m^2\right)\)

Vậy diện tích tam giác ABD là \(50m^2\)

mình cũng hông bt á

em cũng ko biết