Ta có: \(\widehat{AMC}+\widehat{AMD}=180^o\)(2 góc kề bù) (1)
Mà \(\widehat{AMC}=2\widehat{AMD}\)(Đề cho) (Ngoặc ''}'' 2 điều lại)
=> \(2\widehat{AMD}+\widehat{AMD}=180^o\)
=>    \(\left(2+1\right)\widehat{AMD}=180^o\)
=>                    \(3\widehat{AMD}=180^o\)
=>                       \(\widehat{AMD}=180^o:3\)
=>                       \(\widehat{AMD}=60^o\)(2)
Từ (1) và (2) => \(\widehat{AMC}=180^o-60^o=120^o\)
Lại có: \(\widehat{AMC}=\widehat{BMD}\)(2 góc đối đỉnh) (Ngoặc ''}'' 2 điều lại)
=> \(\widehat{BMD}=120^o\)
Mặt khác: \(\widehat{AMD}=\widehat{BMC}\)(2 góc đối đỉnh)
Mà \(\widehat{AMD}=60^o\)(Theo (2)) (Ngoặc ''}'' 2 điều lại)
=> \(\widehat{BMC}=60^o\)
Vậy \(\widehat{AMC}=\widehat{BMD}=120^o\)
       \(\widehat{AMD}=\widehat{BMC}=60^o\)

Hình vẽ sai số đo nên tự chỉnh lại y như đáp án nhé

sub kênh mik đã 

Ender dragon Boy Vcl

a) a là số hữu tỉ dương khi n>3

b)a là số hữu tỉ âm khi a<3

c)...........................................................khi =3

a) \(\left|2x-3\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=5\\2x-3=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)

Vậy : \(x\in\left\{4,-1\right\}\)

b) \(\left|2x-1\right|=\left|2x-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=2x-3\\2x-1=3-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-2x=-3+1\\2x+2x=3+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0=-2\\4x=4\end{cases}}\)

\(\Rightarrow x=1\)

Vậy : \(x=1\)

c) \(\left|x-1\right|+3x=11\)

\(\Leftrightarrow\left|x-1\right|=11-3x\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=11-3x\\x-1=3x-11\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3x=11+1\\3x-x=-1+11\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=12\\2x=10\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

Vậy : \(x\in\left\{3,5\right\}\)

d) \(\left|5x-3\right|-x=7\)

\(\Leftrightarrow\left|5x-3\right|=7+x\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7+x\\5x-3=-7-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x-x=7+3\\5x+x=-7+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}4x=10\\6x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{4}=\frac{5}{2}\\x=-\frac{4}{6}=-\frac{2}{3}\end{cases}}\)

Vậy : \(x\in\left\{\frac{5}{2},-\frac{2}{3}\right\}\)

a,\(|2x-3|=5\)

\(\Rightarrow2x-3=5\)hoặc\(2x-3=-5\)

\(2x=5+3\)hoặc\(2x=-5+3\)

\(2x=8\)hoặc\(2x=-2\)

\(\Rightarrow x=\frac{8}{2}=4\)hoặc\(x=-\frac{2}{2}=-1\)

Vậy.......

b,Sai vế phải r bn ơi

c,\(|x-1|+3x=11\)

\(|x-1|=11-3x\)

\(\Rightarrow x-1=+\left(11-3x\right)\)hoặc\(\Rightarrow x-1=-\left(11-3x\right)\)

\(x-1=11-3x\)hoặc\(x-1=-11+3x\)

\(x+3x=11+1\)hoặc\(x-3x=-11+1\)

\(4x=12\)hoặc\(-2x=-10\)

\(\Rightarrow x=\frac{12}{4}=3\)hoặc\(x=\frac{-10}{-2}=5\)

Vậy.....

d,\(|5x-3|-x=7\)

\(|5x-3|=7+x\)

\(\Rightarrow5x-3=+\left(7+x\right)\)hoặc\(5x-3=-\left(7+x\right)\)

\(5x-3=7+x\)hoặc\(5x-3=-7-x\)

\(5x-x=7+3\)hoặc\(5x+x=-7+3\)

\(4x=10\)hoặc\(6x=-4\)

\(\Rightarrow x=\frac{10}{4}=2,5\)hoặc\(x=-\frac{4}{6}=-\frac{2}{3}\)

Vậy......

a, \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}\Rightarrow}\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

b. \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=-1\left(Voly\right)\\x=4\end{cases}\Rightarrow x=4}\)

c, \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

d, \(\left(\frac{4}{5}\right)^{5x}=\left(\frac{4}{5}\right)^7\)

\(\Rightarrow5x=7\)

\(\Rightarrow x=\frac{7}{5}\)

e, Ta có: \(A=\frac{x+5}{x-2}=\frac{\left(x-2\right)+7}{x-2}=1+\frac{7}{x-2}\)

Để A ∈ Z <=> (x - 2) ∈ Ư(7) = { ±1; ±7 }

 Vậy....

a) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=1\\2x=\frac{1}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

Vậy : ....

b) \(\left(x^2+1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-1\left(loại\right)\\x=4\end{cases}}\)

c) \(2x^2-\frac{1}{3}x=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-\frac{1}{3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{6}\end{cases}}\)

Vậy :...

a) \(\left|a\right|+a=a+a=2a\)

b) \(\left|a\right|-a=a-a=0\)

c) \(\left|a\right|.a=a.a=a^2\)

d) \(\left|a\right|:a=a:a=1\)( Với \(a\ne0\) )

e) \(3\left(x-1\right)-2\left|x+3\right|=3x-3-2.\left(x+3\right)\)

\(=3x-3-2x-6=x-9\)

g) \(2\left|x-3\right|-\left|4x-1\right|=2.\left(x-3\right)-\left(4x-1\right)\)

\(=2x-6-4x+1=-2x-5\)

a,\(|a|+a=a+a=2a\)

b,\(|a|-a=a-a=0\)

c,\(|a|.a=a.a=a^2\)

d,\(|a|:a=a:a=1\)

e,\(3\left(x-1\right)-2|x+3|=3x-3-2\left(x+3\right)\)

\(=3x-3-2x-6=x-9\)

g,\(2|x-3|-|4x-1|=2\left(x-3\right)-\left(4x-1\right)\)

\(=2x-6-4x+1=4x-5\)

+) x = -1

-1 = 7/2a - 1

-2a + 1 = 7

-2x = 7 - 1

-2x = 6

x = -3

để x = -1 thì a = -3.

+) x = 7

7 = 7/2a - 1

7(2a - 1) = 7

2a - 1 = 1

2a = 1 + 1

2a = 2

a = 1

để x = 7 thì a = 1.

=79305375744 nhé

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