-12.(x-5)+7(3-x)=5
giúp tớ vs ạ!!!
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tìm x biết:
2x:(1+\(\dfrac{1}{1+2}\)\(+\dfrac{1}{1+2+3}\)\(+.....\)\(+\dfrac{1}{1+2+3+...+x}\))=2023
\(2x:\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...x}\right)=2023\left(1\right)\)
Đặt \(A=\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...x}\right)\)
\(\Rightarrow A=\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}\right)\)
\(\Rightarrow\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}\right)\)
\(\Rightarrow\dfrac{1}{2}A=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)\)
\(\Rightarrow\dfrac{1}{2}A=1-\dfrac{1}{x+1}\)
\(\Rightarrow A=2\left(1-\dfrac{1}{x+1}\right)\Rightarrow A=\dfrac{2x}{x+1}\)
\(\left(1\right)\Rightarrow2x:\dfrac{2x}{x+1}=2023\)
\(\Rightarrow2x.\dfrac{x+1}{2x}=2023\left(x\ne0\right)\)
\(\Rightarrow x+1=2023\)
\(\Rightarrow x=2022\)
Để olm.vn giúp em, em nhé.
Kẻ AB kéo dài về phía B cắt đường thẳng y tại L
Kéo dài CB vế phía B cắt đường thẳng x tại M
Vì x//y ta có: \(\widehat{MAB}\) = \(\widehat{BLC}\) ( so le trong)
\(\widehat{ABC}\) = \(\widehat{BCL}\) + \(\widehat{BLC}\) (góc ngoài tam giác bằng tổng hai góc trong không kề với nó)
⇒ \(\widehat{ABC}\) = \(\widehat{BCL}\) + \(\widehat{MAB}\) (đpcm)
d, -(24 - 87) + (- 87 + 48 - 24)
= - 24 + 87 - 87 + 48 - 24
= - (24 + 24) + (87 - 87 ) + 48
= -48 + 0 + 48
= 0
e, (2027. 20262024 - 20262024) : 20262025
= 20262024.(2027 - 1):20262025
= 20262024.2026: 20262025
= 20262025:20262025
= 1
\(A=\dfrac{10}{7.12}+\dfrac{10}{12.17}+\dfrac{10}{17.22}+...+\dfrac{10}{502.507}\) (sửa 502+507 thành 503.507)
\(\Rightarrow A=10\left(\dfrac{1}{7.12}+\dfrac{1}{12.17}+\dfrac{1}{17.22}+...+\dfrac{1}{502.507}\right)\)
\(\Rightarrow A=10.\dfrac{1}{5}\left(\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}+...+\dfrac{1}{502}-\dfrac{1}{507}\right)\)
\(\Rightarrow A=2.\left(\dfrac{1}{7}-\dfrac{1}{507}\right)=2.\left(\dfrac{500}{3549}\right)=\dfrac{1000}{3549}\)
\(B=\dfrac{4}{8.13}+\dfrac{4}{13.18}+\dfrac{4}{18.23}+...+\dfrac{4}{253.258}\)
\(\Rightarrow B=4\left(\dfrac{1}{8.13}+\dfrac{1}{13.18}+\dfrac{1}{18.23}+...+\dfrac{1}{253.258}\right)\)
\(\Rightarrow B=4.\dfrac{1}{5}\left(\dfrac{1}{8}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{23}+...+\dfrac{1}{253}-\dfrac{1}{258}\right)\)
\(\Rightarrow B=\dfrac{4}{5}\left(\dfrac{1}{8}-\dfrac{1}{258}\right)=\dfrac{4}{5}\left(\dfrac{129}{1032}-\dfrac{8}{1032}\right)=\dfrac{4}{5}.\dfrac{121}{1032}=\dfrac{121}{1290}\)
\(x\left(x-y\right)=\dfrac{10}{9}\) (1)
\(y\left(x-y\right)=\dfrac{-2}{3}\) (2)
Trừ 1 và 2, ta được:
\(x\left(x-y\right)-y\left(x-y\right)=\dfrac{10}{9}-\left(\dfrac{-2}{3}\right)\)
\(\left(x-y\right)\times\left(x-y\right)=\dfrac{16}{9}\)
\(\left(x-y\right)^2=\left(\pm\dfrac{4}{3}\right)^2\)
=> \(x-y=\pm\dfrac{4}{3}\)
TH1:
Nếu \(x-y=\dfrac{4}{3}\) thay vào 1 và 2, Ta có:
\(x\times\dfrac{4}{3}=\dfrac{10}{9}\) => \(x=\dfrac{10}{9}\div\dfrac{4}{3}\) => \(x=\dfrac{5}{6}\)
\(y\times\dfrac{4}{3}=\dfrac{-2}{3}\) => \(y=\dfrac{-2}{3}\div\dfrac{4}{3}\) => \(y=-\dfrac{1}{2}\)
TH2:
+) Nếu \(x-y=-\dfrac{4}{3}\) thay vào 1 và 2, ta được:
\(x\times\dfrac{-4}{3}=\dfrac{10}{9}\) => \(x=\dfrac{10}{9}\div\dfrac{-4}{3}=\dfrac{-5}{6}\)
\(y\times\dfrac{-4}{3}=\dfrac{-2}{3}\) => \(y=\dfrac{-2}{3}\div\dfrac{-4}{3}=\dfrac{1}{2}\)
Vậy ta có 2 cặp số (x,y) thoả mãn là \(\left(\dfrac{5}{6},\dfrac{-1}{2}\right);\left(\dfrac{-5}{6},\dfrac{1}{2}\right)\)
a) \(8,5.2,3+3,7.4,2=9.2+4.4=18+16=34\)
b) \(2,6.\left(15,245+84,564\right)=3.\left(15+85\right)=2.100=300\)
c) \(5,37.12,8:24,56=5.13:25=\dfrac{13}{5}\)
các thầy cô giúp em với ạ, em cảm ơn nhiều ạ:33
\(-12\cdot\left(x-5\right)+7\cdot\left(3-x\right)=5\\ -12\cdot x+60+21-7x=5\\ x\left(-12-7\right)+\left(60+21\right)=5\\ x\cdot\left(-19\right)+81=5\\ x\left(-19\right)=-76\\ x=4\)