a) x × 30 = 1320

x = 1320 : 30

x = 44

b) x : 24 = 65

x = 65 × 24

x = 1560

a) x *  30 = 1320

x = 1320 : 30

x = 44

b) x : 24 = 65

x = 65 * 24

x = 1560

k mk rồi mk sẽ kết bạn với cậu

a, X x 30 =1320

x = 1320: 30 = 44

Vậy x = 44

b, x : 24 = 65

x = 65 x 24

x = 1560

Bn k mik nha xong mik kp vs bn

\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)

\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)

\(x-2023=1\)

\(x=2024\)

Vậy..............

\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)

\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)

\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)

a)  x = 25 28

b)  x = 5 12

(x-24).(x-2012)=0

=>

TH1: x-24=0 => x= 24

TH2: x-2012=0=> x=2012

chúc bn học giỏi.

\(\left(x-24\right).\left(x-2012\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-24=0\\x-2012=0\end{cases}\Rightarrow\orbr{\begin{cases}x=24\\x=2012\end{cases}}}\)

x 20 = 14 20 + − 13 20 x 20 = 1 20       x   =   1

Ko tính đc bạn

bn biết cách lm ko

Ta có:

\(x^2+5y^2-4x-4xy+6y+5=0\\\Rightarrow[(x^2-4xy+4y^2)-(4x-8y)+4]+(y^2-2y+1)=0\\\Rightarrow[(x-2y)^2-4(x-2y)+4]+(y-1)^2=0\\\Rightarrow(x-2y-2)^2+(y-1)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-2y-2\right)^2\ge0\forall x,y\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-2y-2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

Mà: \(\left(x-2y-2\right)^2+\left(y-1\right)^2=0\)

nên: \(\left\{{}\begin{matrix}x-2y-2=0\\y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2y+2\\y=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot1+2=4\\y=1\end{matrix}\right.\)

Thay \(x=4;y=1\) vào \(P\), ta được:

\(P=\left(4-3\right)^{2023}+\left(1-2\right)^{2023}+\left(4+1-5\right)^{2023}\)

\(=1^{2023}+\left(-1\right)^{2023}+0^{2023}\)

\(=1-1=0\)

Vậy \(P=0\) khi \(x=4;y=1\).