Đoạn dấu \(\left|x-2008\right|+\left|8-x\right|\le\left|x-2008+8-x\right|\) nhầm rồi ạ. Phải là dấu \(\ge\)

G = |\(x\) - 2008| + |\(x\) - 8| 

Vì |\(x-8\)| = |8 - \(x\)| 

⇒ G = |\(x\) - 2008| + |\(x\) - 8| = |\(x\) - 2008| + |8 - \(x\)|

G = |\(x\) - 2008| + |8-\(x\)| \(\ge\) |\(x-2008\) + 8 - \(x\)| = 2000

Dấu bằng xảy ra ⇔ (\(x\) - 2008).(8 - \(x\)) ≥ 0

Lập bảng ta có:

Theo bảng trên ta có: G_min = 2000 ⇔ 8 ≤ \(x\) ≤ 2008

a) Đặt \(ƯCLN\left(5a+3,7a+4\right)=d\)

\(\Rightarrow\left\{{}\begin{matrix}5a+3⋮d\\7a+4⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}35a+21⋮d\\35a+20⋮d\end{matrix}\right.\)

\(\Rightarrow\left(35a+21\right)-\left(35a+20\right)⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d=1\)

 Vậy \(ƯCLN\left(5a+3,7a+4\right)=1\) hay phân số \(\dfrac{5a+3}{7a+4}\) là phân số tối giản. Thế thì phân số này không thể rút gọn cho nguyên nào khác 1.

b) \(A=\dfrac{5a+3}{7a+4}\)

\(A=\dfrac{\dfrac{5}{7}\left(7a+4\right)+\dfrac{1}{7}}{7a+4}\)

\(A=\dfrac{5}{7}+\dfrac{1}{7\left(7a+4\right)}\)

 Nếu \(a< 0\) thì \(A< \dfrac{5}{7}\) còn nếu \(a\ge0\) thì \(A>\dfrac{5}{7}\). Do đó ta chỉ cần tìm giá trị lớn nhất của A khi \(a>0\). Để A lớn nhất thì \(7a+4\) nhỏ nhất hay \(a=0\). Vậy để phân số A lớn nhất thì \(a=0\)

\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)

\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)

\(\dfrac{5}{12}x=\dfrac{-1}{6}\)

\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)

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

Giúp mik ik 

a) \(24=2^3.3\)

\(60=2^2.3.5\)

\(UCLN\left(a;b\right)=UCLN\left(24;60\right)=2^2.3=6\)

\(BCNN\left(a;b\right)=BCNN\left(24;60\right)=2^3.3.5=120\)

\(a.b=UCLN\left(a;b\right).BCNN\left(a;b\right)\)

\(\Rightarrow a.b=6.120=720\)

mà \(\dfrac{a}{b}=\dfrac{24}{60}\Rightarrow\dfrac{a}{24}=\dfrac{b}{60}=\dfrac{720}{24.60}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a=24.\dfrac{1}{2}=12\\b=60.\dfrac{1}{2}=30\end{matrix}\right.\)

Vậy Phân số cần tìm là \(\dfrac{12}{30}\) 

b) \(\left\{{}\begin{matrix}14=2.7\\21=3.7\end{matrix}\right.\)

\(\Rightarrow UCLN\left(a;b\right)=UCLN\left(14;21\right)=7\)

\(a.b=UCLN\left(14;21\right).BCNN\left(14;21\right)\)

\(\Rightarrow a.b=7.3456=24192\)

\(\dfrac{a}{b}=\dfrac{14}{21}\Rightarrow\dfrac{a}{14}=\dfrac{b}{21}=\dfrac{a.b}{14.21}=\dfrac{24192}{294}=\dfrac{576}{7}\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{576}{7}.14=1152\\b=\dfrac{576}{7}.21=1728\end{matrix}\right.\)

Vậy phân số cần tìm là \(\dfrac{1152}{1728}\)

a) \(\widehat{AOB}=\widehat{AOM}+\widehat{MOB}\)

\(\Rightarrow\widehat{MOB}=\widehat{AOB}-\widehat{AOM}\)

mà \(\widehat{AOM}=90^o\) (OM vuông góc OA)

\(\Rightarrow\widehat{MOB}=100^o-90^o\)

\(\Rightarrow\widehat{MOB}=10^o\)

Câu b bạn xem lại đề

b) \(\widehat{B'OA}+\widehat{AOB}=180^o\) (2 góc kề bù)

\(\Rightarrow\widehat{B'OA}=180^o-\widehat{AOB}\)

\(\Rightarrow\widehat{B'OA}=180^o-100^o\)

\(\Rightarrow\widehat{B'OA}=80^o\)

\(\widehat{MOB'}=\widehat{B'OA}+\widehat{AOM}\)

\(\Rightarrow\widehat{MOB'}=80^o+90^o\)

\(\Rightarrow\widehat{MOB'}=170^o\)

\(C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)

\(\Rightarrow C=-2\left|\dfrac{1}{3}x+4\right|+\dfrac{5}{3}\)

mà \(-2\left|\dfrac{1}{3}x+4\right|\le0,\forall x\)

\(\Rightarrow C=-2\left|\dfrac{1}{3}x+4\right|+\dfrac{5}{3}\le\dfrac{5}{3}\)

\(\Rightarrow GTLN\left(C\right)=\dfrac{5}{3}\left(tạix=-12\right)\)

Ta có :

\(\widehat{xOy}+\widehat{yOz}=180^o\) (2 góc kề bù)

\(\Rightarrow\widehat{yOz}=180^o-\widehat{xOy}\)

\(\Rightarrow\widehat{yOz}=180^o-80^o\)

\(\Rightarrow\widehat{yOz}=100^o\)

Ta lại có :

\(\widehat{tOz}=\widehat{tOy}+\widehat{yOz}\)

mà \(\widehat{tOy}=\dfrac{1}{2}\widehat{xOy}=\dfrac{1}{2}.80^o=40^o\) (Ot là phân giác góc xOy)

\(\Rightarrow\widehat{tOz}=40^o+100^o\)

\(\Rightarrow\widehat{tOz}=140^o\)

\(\widehat{xOt}=\dfrac{1}{2}\widehat{xOy}=\dfrac{1}{2}.80^o=40^o\) (Ot là phân giác góc xOy)

\(x=7\Rightarrow8=x+1\left(1\right)\)

Thay \(1\) vào \(F\) ta có:

\(F=x^{2006}-\left(x+1\right)^{2005}+\left(x+1\right)^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\)

\(F=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\)

\(F=-7-5\)

\(\Rightarrow F=-12\)

\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)\)

\(=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}\)

\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}\)

\(=\dfrac{10}{18}\cdot\dfrac{1}{4026042}\)

\(=\dfrac{5}{9}\cdot\dfrac{1}{4026042}\)

\(=\dfrac{5}{36234378}\)