Ta có: \(\widehat{aOc}=\widehat{bOd}=a\left(\text{đối đỉnh}\right)\)

\(\Rightarrow\widehat{cOm}=\dfrac{a}{2}\)

\(\widehat{bOc}=180-\widehat{bOd}=180-a\)

Theo bài ra ta có:

\(\widehat{bOm}=155\)

\(\Leftrightarrow\widehat{cOm}+\widehat{bOc}=155\)

\(\Leftrightarrow\dfrac{a}{2}+180-a=155\)

\(\Leftrightarrow180-155==a-\dfrac{a}{2}\)

\(\Leftrightarrow a=50\)

a) \(\dfrac{A}{x-3}=\dfrac{y-x}{3-x}\left(Đk:x\ne3\right)\)

\(A=\dfrac{\left(x-3\right)\left(y-x\right)}{3-x}=x-y\)

b) \(\dfrac{5x}{x+1}=\dfrac{Ax\left(x-1\right)}{\left(1-x\right)\left(x+1\right)}\left(Đk:x\ne\pm1\right)\)

\(A=\dfrac{5x\left(1-x\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=-5\)

c) \(\dfrac{4x^2-5x+1}{A}=\dfrac{4x-1}{x+3}\left(Đk:x\ne-3;A\ne0\right)\)

\(A=\dfrac{\left(4x^2-5x+1\right)\left(x+3\right)}{4x-1}=\dfrac{\left(x-1\right)\left(4x-1\right)\left(x+3\right)}{4x-1}\)

    \(=\left(x-1\right)\left(x+3\right)=x^2+2x-3\)

c) \(\dfrac{1+8x}{1+2x}-\dfrac{2x}{2x-1}+\dfrac{12x^2-9}{1-4x^2}=0\left(Đk:x\ne\pm\dfrac{1}{2}\right)\)

\(\left(1+8x\right)\left(1-2x\right)+2x\left(1+2x\right)+12x^2-9=0\)

\(1-2x+8x-16x^2+2x+4x^2+12x^2-9=0\)

\(8x-8=0\)

\(x=1\)

d) \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-1\right)\left(x-3\right)}=\dfrac{1}{2}\left(Đk:x\ne1;x\ne3\right)\)

\(x-1+2\left(5-3x\right)=x^2-4x+3\)

\(9-5x=x^2-4x+3\)

\(x^2+x-6=0\)

\(\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

a) \(\dfrac{1}{\left(2x-3\right)^2}+\dfrac{3}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2}{\left(2x+3\right)^2}=0\left(Đk:x\ne\pm\dfrac{2}{3}\right)\)

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

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

\(\Leftrightarrow2x+3=3-2x\)

\(\Leftrightarrow x=0\)

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

\(\Leftrightarrow2x+3=2\left(3-2x\right)\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

b) \(1+\dfrac{14}{\left(x-4\right)^2}=-\dfrac{9}{x-4}\left(Đk:x\ne4\right)\)

\(\left(x-4\right)^2+14+9\left(x-4\right)=0\)

\(x^2-8x+16+14+9x-36=0\)

\(x^2+x-6=0\)

\(\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\overrightarrow{MN}=\overrightarrow{PC}\\\overrightarrow{MP}=\overrightarrow{NC}\\\overrightarrow{PN}=\overrightarrow{MB}\end{matrix}\right.\) 

Bạn xem lại nha, có thể đáp án A hoặc B sẽ có \(\overrightarrow{MN}=\overrightarrow{PC}\)

\(9x^2-1=0\)

\(x^2=\dfrac{1}{9}\)

\(\Rightarrow x=\pm\dfrac{1}{3}\Rightarrow x\in Q\)

Chọn A

\(\left|\overrightarrow{CH}+\overrightarrow{HA}\right|=\left|\overrightarrow{CA}\right|=a\)

Áp dụng công thức đường trung tuyến

\(m_a^2+m_b^2+m_c^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}+\dfrac{c^2+a^2}{2}-\dfrac{b^2}{4}+\dfrac{a^2+b^2}{2}-\dfrac{c^2}{4}\)

                          \(=\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)

Chọn A

Áp dụng hệ thức lượng vào tam giác vuông EGH có đường cao EH

\(\dfrac{1}{EH^2}=\dfrac{1}{EG^2}+\dfrac{1}{EF^2}\)

\(\dfrac{1}{30^2}=\dfrac{1}{\left(\dfrac{6EF}{5}\right)^2}+\dfrac{1}{EF^2}\)

\(\Rightarrow EF=5\sqrt{61}\)\(\Rightarrow EG=\dfrac{6.5\sqrt{61}}{5}=6\sqrt{61}\)

Áp dụng định lí Pytago vào tam giác GEF vuông tại E

\(\Rightarrow GF=\sqrt{\left(5\sqrt{61}\right)^2+\left(6\sqrt{61}\right)^2}=61\)

Áp dụng định lí Pytago vào tam giác EHG vuông tại H

\(GH=\sqrt{\left(6\sqrt{61}\right)^2-30^2}=36\)

\(\Rightarrow HF=61-36=25\)

a) \(9x^2-6x+3=\left(3x-1\right)^2+2\)

Vì \(\left(3x-1\right)^2\ge0\forall x\)

\(\Rightarrow9x^2-6x+3>0\)

b) \(3x-x^2-7=-\left(x^2-3x+7\right)=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}\)

Vì \(-\left(x-\dfrac{3}{2}\right)^2\le0\forall x\)

\(\Rightarrow3x-x^2-7< 0\)