Tìm x:
2012 + 1320 : ( x + 24 ) = 2023
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Tìm x :
a) x × 30 = 1320 b) x : 24 = 65
a) x × 30 = 1320
x = 1320 : 30
x = 44
b) x : 24 = 65
x = 65 × 24
x = 1560
Tìm x :
a , X x 30 = 1320 b , x : 24 = 65
Kết bạn với mik nhé
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
2013 x 2012 x 2
---------------------------
2011 x 2011 x 2023
mình đang cần gấp SOS !
ai mà giải mình gọi là cụ ông trời hhihi
tìm x biết
\(\dfrac{x-2023}{6}\)\(+\dfrac{x-2023}{10}\)\(+\dfrac{x-2023}{15}\)\(+\dfrac{x-2023}{21}\)=\(\dfrac{8}{21}\)
\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)
\(\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\left(x-2023\right).\dfrac{8}{21}=\dfrac{8}{21}\)
\(x-2023=1\)
\(x=2024\)
Vậy..............
\(...\Rightarrow\left(x-2023\right)\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{35+21+14+1}{210}\right)=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}\)
\(\Rightarrow\left(x-2023\right).\dfrac{71}{210}=\dfrac{8}{21}.\dfrac{210}{71}=\dfrac{80}{71}\)
\(\Rightarrow x-2023=\dfrac{80}{71}\Rightarrow x=\dfrac{80}{71}+2023=\dfrac{143713}{71}\)
\(\dfrac{x-2023}{6}+\dfrac{x-2023}{10}+\dfrac{x-2023}{15}+\dfrac{x-2023}{21}=\dfrac{8}{21}\)
\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)=\dfrac{8}{21}\)
\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)=\dfrac{4}{21}\)
\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)=\dfrac{4}{21}\)
\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)=\dfrac{4}{21}\)
\(\Leftrightarrow\left(x-2023\right).\left(\dfrac{1}{3}-\dfrac{1}{7}\right)=\dfrac{4}{21}\)
\(\Leftrightarrow x-2023=1\Leftrightarrow x=2024\)
Tìm x, biết:
a) x − 3 4 = 1 7
b) − x + 3 5 + 13 20 = 5 6
Tìm x:
(x-24).(x-2012)=0 (Trình bày cách làm đầy đủ)
(x-24).(x-2012)=0
=>
TH1: x-24=0 => x= 24
TH2: x-2012=0=> x=2012
chúc bn học giỏi.
\(\left(x-24\right).\left(x-2012\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-24=0\\x-2012=0\end{cases}\Rightarrow\orbr{\begin{cases}x=24\\x=2012\end{cases}}}\)
Tìm x biết x 20 = 7 10 + − 13 20
x 20 = 14 20 + − 13 20 x 20 = 1 20 x = 1
tìm x nguyên 2023+2022+2021+2020+...+x=2023
X=-2022 nhà lúc nãy mik nhầm mong bạn thông cảm
Tìm x,y thỏa mãn x^2 +5y^2 -4x -4xy +6y +5 = 0. Tính P=(x-3)^2023 + (y-2)^2023 +(x+y-5)^2023
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 \(x=4;y=1\).