Cho At // Bz , góc zBc = 130 độ , góc bCa = 60 độ. Tính cAt?
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\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(3x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(3x-\frac{1}{2}-2x-\frac{1}{2}+5=0\)
\(3x-2x-\frac{1}{2}-\frac{1}{2}+5=0\)
\(x-0+5=0\)
\(x+5=0\)
\(x=0-5=-5\)
\(\Leftrightarrow\left(\frac{3}{4}x-\frac{9}{16}\right)\left(\frac{1}{3}-\frac{3}{5}.\frac{1}{x}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{4};\frac{9}{5}\right\}\)
\(\frac{x+3}{5}\cdot\left(x+1\right)< 0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)< 0\)
\(TH1:\orbr{\begin{cases}x+3< 0\\x+1>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< -3\\x>-1\end{cases}}\Leftrightarrow-1< x< 3\)
\(TH2:\orbr{\begin{cases}x+3>0\\x+1< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-3\\x< -1\end{cases}}\Leftrightarrow-3< x< 1\)
\(\frac{x+3}{5}\cdot\left(x+1\right)< 0\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x+1\right)}{5}< 0\)
\(\Leftrightarrow\left(x+3\right)\left(x+1\right)< 0\)
Nên x+3 và x+1 phải trái dấu,mà x+1<x+3
Do đó x+1 sẽ <0
\(\Leftrightarrow\hept{\begin{cases}x+1< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-3\end{cases}\Leftrightarrow}}-3< x< -1\)
\(\frac{x-1}{3}\cdot\frac{x+2}{5}>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)>0\)
\(TH1:\orbr{\begin{cases}x-1>0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x>-2\end{cases}}\Leftrightarrow x>1\left(1\right)\)
\(TH2:\orbr{\begin{cases}x-1< 0\\x+2< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 1\\x< -2\end{cases}}\Leftrightarrow x< -2\left(2\right)\)
Tu ( 1 ) va ( 2 ) suy ra \(-2< x< 1\)
Cho mik sua 1 chut nha.
\(TH1:\Leftrightarrow x>-2\)
\(TH2:\Leftrightarrow x< 1\)
\(\Leftrightarrow\orbr{\begin{cases}x+1+x+2+x+3+x+4=20\\x+1+x+2+x+3+x+4=-20\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+x+x+x\right)+1+2+3+4=20\\\left(x+x+x+x\right)+1+2+3+4=-20\text{}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+10=20\\x+10=-20\end{cases}}\Rightarrow\orbr{\begin{cases}x=20-10\\x=-20-10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-30\end{cases}}\)
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|=20\)
\(\Leftrightarrow\orbr{\begin{cases}x+1+x+2+x+3+x+4=20\\x+1+x+2+x+3+x+4=-20\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left[x+x+x+x\right]+\left[1+2+3+4\right]=20\\\left[x+x+x+x\right]+\left[1+2+3+4\right]=-20\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x+10=20\\4x+10=-20\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=10\\4x=-30\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\2x=-15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{15}{2}\end{cases}}\)
Vũ Bách Quang sai từ dòng thứ ba đến cuối . Xem kĩ lại nhé
\(\left(x-\frac{1}{3}\right)\left(x+\frac{2}{5}\right)>0\)
\(\orbr{\begin{cases}x=\frac{1}{3}\\x=-\frac{2}{5}\end{cases}}\)
\(-\frac{2}{5}< x< \frac{1}{3}\)
\(\left(\frac{3}{4}x-\frac{9}{16}\right)-\left(\frac{1}{3}+\frac{-3}{5}x\right)=0.\)
\(\frac{3}{4}x-\frac{9}{16}-\frac{1}{3}-\frac{-3}{5}x=0\)
\(\frac{27}{20}x-\frac{43}{48}=0\)
\(\frac{27}{20}x=\frac{43}{48}\)
\(x=\frac{215}{324}\)
mong ủng hộ mk
(3 phần 4 nhân X - 9 phần 16 ) - (1 phần 3 + -3 phần 5 chia X ) = 0
=>\(\frac{3}{4}.x-\frac{9}{16}=\frac{1}{3}+\frac{-3}{5}:x\)
=\(\frac{3}{4}.x-\frac{9}{16}=\frac{-4}{15}:x\)
=>\(\frac{3}{4}.x-\frac{-4}{15}:x=\frac{9}{16}\)
Đến đây thì mk chịu =)
Dù sao thi cg ti.ck giúp mk tí
Bài 1
a
Ta có:
\(\widehat{O_1}=\widehat{O_2}=60^0\left(đ.đ\right)\)
\(\widehat{O_1}+\widehat{O_2}=180^0\Rightarrow\widehat{0_2}=180^0-\widehat{O_1}=180-60^0=120^0\)
\(\widehat{O_2}=\widehat{O_4}=120^0\left(đ.đ\right)\)
b
Ta có:
\(\widehat{x'Oy}=\widehat{y'Ox}\Rightarrow\frac{1}{2}\widehat{x'Oy}=\frac{1}{2}\widehat{y'Ox}\Rightarrow\widehat{yOn}=\widehat{xOm}\)
\(\widehat{x'Oy}+\widehat{yOx}=180^0\)
\(\Rightarrow2\cdot\widehat{yOn}+\widehat{yOx}=180^0\)
\(\Rightarrow\widehat{yOn}+\widehat{yOx}+\widehat{xOm}=180^0\)
\(\Rightarrowđpcm\)
Bài 2
a
Ta có:
\(\widehat{BOD}=\widehat{AOC}=90^0\Rightarrow\widehat{BOC}+\widehat{COD}=\widehat{AOD}+\widehat{COD}\Rightarrow\widehat{BOC}=\widehat{AOD}\)
b
Ta có:
\(\widehat{BOM}=\widehat{BOC}+\widehat{COM}=\widehat{AOD}+\widehat{MOD}=\widehat{MOA}\)
Hiển nhiên OM nằm giữa \(\widehat{AOB}\) nên suy ra đpcm