\(TH1:\left|x+1\right|=x+1\\ \Leftrightarrow2x-\left(x+1\right)=-\dfrac{1}{2}\\ \Leftrightarrow2x-x-1=-\dfrac{1}{2}\\ \Leftrightarrow x-1=-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{2}+1\\ \Leftrightarrow x=\dfrac{1}{2}\\ TH2:\left|x+1\right|=-x-1\\ \Leftrightarrow2x-\left(-x-1\right)=-\dfrac{1}{2}\\ \Leftrightarrow2x+x+1=-\dfrac{1}{2}\\ \Leftrightarrow3x=-\dfrac{1}{2}-1\\ \Leftrightarrow3x=-\dfrac{3}{2}\\ \Leftrightarrow x=\left(-\dfrac{3}{2}\right):3=-\dfrac{1}{2}\)

Thay lần lượt \(x=\dfrac{1}{2};x=-\dfrac{1}{2}\) vào pt

\(\Rightarrow x=\dfrac{1}{2}\left(thoaman\right)\)

Vậy \(x=\dfrac{1}{2}\)

\(=2,02\times8+2,02\times\dfrac{5}{2}+2,02\times0,5\\ =2,02\times\left(8+\dfrac{5}{2}+0,5\right)\\ =2,02\times\left(\dfrac{8.2+5+1}{2}\right)\\ =2,02\times\dfrac{22}{2}\\ =2,02\times11\\ =22,22\)

a, Tam giác \(ABC\) cân tại \(A\)

\(\Rightarrow AB=AC;\widehat{B}=\widehat{C}\)

Xét \(\Delta ABM;\Delta ACM\) có

\(AB=AC\left(cmt\right)\\ \widehat{B}=\widehat{C}\left(cmt\right)\\ MB=MC\)

\(\Rightarrow\Delta ABM=\Delta ACM\left(c-g-c\right)\)

b, \(\Delta ABM=\Delta ACM\left(cmt\right)\)

\(\Rightarrow\widehat{HAM}=\widehat{KAM}\)

Xét \(\Delta AHM;\Delta AKM\) có

\(\widehat{HAM}=\widehat{KAM}\left(cmt\right)\\ \widehat{AHM}=\widehat{AKM}=90^o\)

\(AM\) chung

\(\Rightarrow\Delta AHM=\Delta AKM\left(ch-gn\right)\)

\(\Rightarrow HM=KM\)

\(a,A=27\times36+73\times99+27\times14-49\times73\\ =73\times\left(99-49\right)+27\times\left(36+14\right)\\ =73\times50+27\times50\\ =50\times\left(73+27\right)\\ =50\times100\\ =5000\)

\(b,B=\left(4^5.10.5^6+25^5.2^8\right):\left(2^8.5^4+5^7.2^5\right)\\ =\left(\left(2^2\right)^5.2.5.5^6+\left(5^2\right)^5.2^8\right):\left(2^8.5^4+5^7.2^5\right)\\ =\left(2^{11}.5^7+5^{10}.2^8\right):\left(2^8.5^4+5^7.2^5\right)\\ =\left[\left(2^8.5^7\right)\left(2^3+5^3\right)\right]:\left[\left(2^5.5^4\right)\left(2^3+5^3\right)\right]\\ =2^8.5^7:2^55^4\\ =\left(2^8:2^5\right).\left(5^7:5^4\right)\\ =2^3.5^3\\ =8.125\\ =1000\)

Số tiền được giảm là : \(350000\times20\%=70000\left(đồng\right)\)

Số tiền lan phải trả là : \(350000-70000=280000\left(đồng\right)\)

=> Đúng 

Chiều dài là : \(25,6\times2,5=64\left(m\right)\)

a, Diện tích thửa ruộng là : \(25,6\times64=1638,4\left(m^2\right)\)

b, Diện tích đất để trồng khoai là : \(1638,4\times60\%=983,04\left(m^2\right)\)

              Đáp số : a, \(1638,4m^2\)

                            b, \(983,4\left(m^2\right)\)

#Rinz

\(đk:\left\{{}\begin{matrix}3x-2\ne0\\3x+2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{2}{3}\\x\ne-\dfrac{2}{3}\end{matrix}\right.\)

\(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\\ =\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x-6}{9x^2-4}\\ =\dfrac{3x+2-\left(3x-2\right)+3x-6}{9x^2-4}\\ =\dfrac{3x+2-3x+2+3x-6}{9x^2-4}\\ =\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}\\ =\dfrac{1}{3x+2}\)

1 là phân số viết được dưới dạng 1/1

=> đúng đóa 

\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(=2,85\times100+2,85\times8\\ =2,85\times\left(100+8\right)\\ =2,85\times108\\ =307,8\)