a) \(\dfrac{3}{7}-x=\dfrac{5}{21}\)

\(x=\dfrac{3}{7}-\dfrac{5}{21}=\dfrac{4}{21}\)

b) \(\dfrac{4}{5}x+\dfrac{1}{3}=1,5=\dfrac{3}{2}\)

\(\dfrac{4}{5}x=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6}\)

\(x=\dfrac{7}{6}:\dfrac{4}{5}=\dfrac{7}{6}.\dfrac{5}{4}=\dfrac{35}{24}\)

c) \(3\left(x+\dfrac{1}{2}\right)-\dfrac{1}{2}\left(4x-\dfrac{2}{3}\right)=\dfrac{5}{6}\)

\(3x+\dfrac{3}{2}-2x+\dfrac{1}{3}=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}-\dfrac{1}{3}-\dfrac{3}{2}=-1\)

d) \(\left(x-\dfrac{3}{2}\right)^2=\dfrac{9}{16}=\left(\dfrac{3}{4}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{3}{4}\\x-\dfrac{3}{2}=-\dfrac{3}{4}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)

e) \(\left(2x-1\right)^3=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

\(2x-1=\dfrac{2}{3}\)

\(2x=\dfrac{5}{3}\)

\(x=\dfrac{5}{6}\)

f) \(\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}=\left(\dfrac{7}{5}\right)^2\)

\(x=2\)

g) \(\left(\dfrac{1}{2}\right)^x=\dfrac{1}{32}=\left(\dfrac{1}{2}\right)^5\)

\(x=5\)

\(B=7-7^4+7^7-7^{10}+...+7^{295}-7^{298}+7^{301}\)

\(=7\left(1-7^3\right)+7^7\left(1-7^3\right)+...+7^{295}\left(1-7^3\right)+7^{301}\)

\(=\left(1-7^3\right)\left(7+7^7+...+7^{295}\right)+7^{301}\)

\(=\left(1-7^3\right)\left(\dfrac{7^{296}-7}{6}\right)+7^{301}\)

\(=-57\left(7^{296}-7\right)+7^{301}\)

\(\dfrac{17}{51}\) ; \(\dfrac{4}{3}\) là phân số tối giản

\(\dfrac{11}{22}=\dfrac{1}{2};\dfrac{6}{9}=\dfrac{2}{3}\)

a) \(5x^5=x^3\)

\(x^3\left(5x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{\sqrt{5}}\left(l\right)\\x=-\dfrac{1}{\sqrt{5}}\left(l\right)\end{matrix}\right.\)

b) \(\left(2x-3\right)^3=64\)

\(2x-3=4\)

\(x=\dfrac{7}{2}\left(l\right)\)

c) \(\left(5x-1\right)^2=2023^0.7+3^2=16\)

\(\left[{}\begin{matrix}5x-1=4\\5x-1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{3}{5}\left(l\right)\end{matrix}\right.\)

Ta có:

\(Ư\left(10\right)=\left\{1;2;5;10\right\}\)

=> chọn B

a) Ta có:

AE//DH

AD//HE

=> tứ giác AEHD là hình bình hành

b) Ta có:

D trung điểm HM

=> DM=DH=AE(AEHD là h.b.h)

=> AMDE là hình bình hành

=> MA//DE(1)

c) Xét tam giác HMN có:

E, D lầ lượt là trung điểm HN, HM

=> ED là đường trung bình tam giác HMN

=> ED//MN(2) ; DE=MA(3)

Từ (1) và (2): Áp dụng tiên đề Ơ clít:

=> M, A, N thẳng hàng

cmtt: => ANED là h.b.h

=> DE=AN(4)

Từ (3) và (4)=> \(DE=\dfrac{1}{2}MN\)

a) Ta có:

\(\widehat{P}=180^o-90^o-60^o=30^o\)(tổng 3 góc trong tam giác)

b) Ta có:

MN=\(\dfrac{1}{2}\)NP=HN=HM(cạnh đối diện góc 30 độ)

=> tam giác MNH đều

Tứ giác MNHE nội tiếp

=> \(\widehat{HNE}=\widehat{HME}=30^o\)

\(\Rightarrow\widehat{MNE}=30^o\)

=> EN là phân giác góc MNP

c) Ta có:

KF//MN

=> \(\widehat{KFE}=\widehat{MNN}=30^o\)(so le trong)

mà \(\widehat{ENH}=30^o\)

=> tam giác NKF cân

=> NK=KF

\(2xy-6x+y=13\)

\(2x\left(y-3\right)+y-3=10\)

\(\left(2x+1\right)\left(y-3\right)=10\)

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1=10\\y-3=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=1\\y-3=10\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=2\\y-3=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=5\\y-3=2\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left\{\left(0,13\right);\left(2,5\right)\right\}\)

\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+140\right)=5041\)

\(x+x+...+x+2+4+...+140=5041\)

Có tất cả số hạng là:

\(\dfrac{\left(140-2\right)}{2}+1=70\left(số\right)\)

=> \(71x+\dfrac{\left(140+2\right).70}{2}=5041\)

=> \(71x=71\)

=> \(x=1\)

\(\left(6^{2024}-6^{2023}\right):6^{2023}\)

\(=\dfrac{6^{2024}}{6^{2023}}-\dfrac{6^{2023}}{6^{2023}}\)

\(=6^{2024-2023}-6^{2023-2023}\)

\(=6^1-6^0\)

\(=6-1=5\)