4:

a: Xét tứ giác ABDC có

M là trung điểm chung của AD và BC

góc BAC=90 độ

=>ABDC là hcn

=>ΔACD vuông tại C

b: Xét ΔKAB vuông tại A và ΔKCD vuông tại C có

KA=KC

AB=CD

=>ΔKAB=ΔKCD

=>KB=KD

c: Xét ΔACD có

DK,CM là trung tuyến

DK cắt CM tại I

=>I là trọng tâm

=>KI=1/3KD

Xét ΔCAB có

AM,BK là trung tuyến

AM cắt BK tại N

=>N là trọng tâm

=>KN=1/3KB=KI

\(ĐK:x\ne\dfrac{1}{2};x\ne1;x\ne\dfrac{3}{2};x\ne2;x\ne\dfrac{5}{2}\\ PT\Leftrightarrow\dfrac{1}{\left(2x-1\right)\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(3x-2\right)}+\dfrac{1}{\left(3x-2\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(5x-2\right)}=\dfrac{4}{21}\\ \Leftrightarrow2\left[\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{1}{2}\right)\left(x-1\right)}+\dfrac{\dfrac{1}{2}}{\left(x-1\right)\left(x-\dfrac{3}{2}\right)}+\dfrac{\dfrac{1}{2}}{\left(x-\dfrac{3}{2}\right)\left(x-2\right)}+\dfrac{\dfrac{1}{2}}{\left(x-2\right)\left(x-\dfrac{5}{2}\right)}\right]=\dfrac{4}{21}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{1}{2}}+\dfrac{1}{x-\dfrac{3}{2}}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-\dfrac{3}{2}}+\dfrac{1}{x-\dfrac{5}{2}}-\dfrac{1}{x-2}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{x-\dfrac{5}{2}-x+1}{\left(x-1\right)\left(x-\dfrac{5}{2}\right)}=\dfrac{2}{21}\\ \Leftrightarrow\dfrac{-\dfrac{3}{2}}{x^2-\dfrac{7}{2}x+\dfrac{5}{2}}=\dfrac{2}{21}\\ \Leftrightarrow x^2-\dfrac{7}{2}x+\dfrac{5}{2}=-\dfrac{63}{4}\\ \Leftrightarrow4x^2-14x+10=-63\\ \Leftrightarrow4x^2-14x+73=0\\ \Leftrightarrow x\in\varnothing\)

  tìm x, y, z biết 

2.9 rồi 2.12 đâu ra vậy bn

  tìm x, y, z biết 

\(2x=3y\\ =>\dfrac{x}{3}=\dfrac{y}{2}\\ 4y=5z\\ =>\dfrac{y}{5}=\dfrac{z}{4}\\ \dfrac{x}{3}=\dfrac{y}{2}\\ =>\dfrac{x}{3.5}=\dfrac{y}{2.5}\\ =>\dfrac{x}{15}=\dfrac{y}{10}\\ \dfrac{y}{5}=\dfrac{z}{4}\\ =>\dfrac{y}{5.2}=\dfrac{z}{4.2}\\ =>\dfrac{y}{10}=\dfrac{z}{8}\\ =>\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{x+y+z}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\\ =>\left\{{}\begin{matrix}x=\dfrac{1}{3}.15=5\\y=\dfrac{1}{3}.10=\dfrac{10}{3}\\z=\dfrac{1}{3}.8=\dfrac{8}{3}\end{matrix}\right.\)

  tìm x, y, z biết

\(a,\dfrac{3^{10}.11+9^5.5}{27^3.2^4}.x=-9\\ =>\dfrac{3^{10}.11+\left(3^2\right)^5.5}{\left(3^3\right)^3.2^4}.x=-9\\ =>\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.16}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.2^4}{3^9.2^4}.x=-9\\ =>3^1.x=-9\\ =>x=-9:3\\ =>x=-3\)

  tìm x

Giải

\(\sqrt{\dfrac{9}{4}}-\left|2x+1\right|=0,75\)

TH1: \(\left|2x+1\right|=2x+1\)

\(=>\sqrt{\dfrac{9}{4}}-\left(2x+1\right)=0,75\\ =>\dfrac{3}{2}-2x-1=\dfrac{3}{4}\\ =>\left(\dfrac{3}{2}-1\right)-2x=\dfrac{3}{4}\\ =>\dfrac{1}{2}-2x=\dfrac{3}{4}\\ =>2x=\dfrac{1}{2}-\dfrac{3}{4}\\ =>2x=-\dfrac{1}{4}\\ =>x=\left(-\dfrac{1}{4}\right):2\\ =>x=-\dfrac{1}{8}\)

\(TH2:\left|2x+1\right|=-2x-1\\ =>\sqrt{\dfrac{9}{4}}-\left(-2x-1\right)=\dfrac{3}{4}\\ =>\dfrac{3}{2}+2x+1=\dfrac{3}{4}\\ =>\left(\dfrac{3}{2}+1\right)+2x=\dfrac{3}{4}\\ =>\dfrac{5}{2}+2x=\dfrac{3}{4}\\ =>2x=\dfrac{3}{4}-\dfrac{5}{2}\\ =>2x=-\dfrac{7}{4}\\ =>x=\left(-\dfrac{7}{4}\right):2\\ =>x=-\dfrac{7}{8}\)

Giải

\(a,5-\left(x-2\right)^2=-4\\ =>\left(x-2\right)^2=5-\left(-4\right)\\ =>\left(x-2\right)^2=9\\ =>\left[{}\begin{matrix}\left(x-2\right)^2=3^2\\\left(x-2\right)^2=\left(-3\right)^2\end{matrix}\right.=>\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy \(x=5;x=-1\)