Xét tứ giác AMON có \(\widehat{AMO}+\widehat{ANO}=90^0+90^0=180^0\)

nên AMON là tứ giác nội tiếp

Xét tứ giác AMHO có \(\widehat{AMO}=\widehat{AHO}=90^0\)

nên AMHO là tứ giác nội tiếp

\(\dfrac{2+\dfrac{2}{3}y}{y^2-\dfrac{1}{3}}=\left(\dfrac{2y+6}{3}\right):\left(\dfrac{3y^2-1}{3}\right)\)

\(=\dfrac{2y+6}{3}\cdot\dfrac{3}{3y^2-1}\)

\(=\dfrac{2y+6}{3y^2-1}\)

Bài 1:

1: \(3\sqrt{16}-5\sqrt{36}=3\cdot4-5\cdot6=12-30=-18\)

2: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

3: Để hàm số y=(2m+1)x-6 đồng biến thì 2m+1>0

=>2m>-1

=>\(m>-\dfrac{1}{2}\)

Bài 2:

1: \(\left\{{}\begin{matrix}2x+y=5\\x-3y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x+3y=15\\x-3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=14\\2x+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=5-2\cdot x=5-2\cdot2=1\end{matrix}\right.\)

2: Theo Vi-et, ta có:

\(x_1+x_2=-\dfrac{b}{a}=-4;x_1x_2=\dfrac{c}{a}=-1\)

Sửa đề: \(P=\dfrac{\left|x_1^2x_2-x_2x_1^2\right|}{x_2^2-4x_1}\)

\(=\dfrac{\left|x_1x_2\left(x_1-x_2\right)\right|}{x_2^2+x_1\left(x_1+x_2\right)}=\dfrac{\left|-1\right|\cdot\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}}{\left(x_1^2+x_2^2\right)+x_1x_2}\)

\(=\dfrac{\sqrt{\left(-4\right)^2-4\cdot\left(-1\right)}}{\left(x_1+x_2\right)^2-x_1x_2}=\dfrac{\sqrt{16+4}}{\left(-4\right)^2-\left(-1\right)}=\dfrac{2\sqrt{5}}{17}\)

a: Xét tứ giác DEBC có \(\widehat{DEC}=\widehat{DBC}=90^0\)

nên DEBC là tứ giác nội tiếp

Xét tứ giác KEHB có \(\widehat{KEH}+\widehat{KBH}=90^0+90^0=180^0\)

nên KEHB là tứ giác nội tiếp

b; ta có: DEBC là tứ giác nội tiếp

=>\(\widehat{DEB}+\widehat{DCB}=180^0\)

mà \(\widehat{DEB}+\widehat{MED}=180^0\)(hai góc kề bù)

nên \(\widehat{MED}=\widehat{MCB}\)

Xét ΔMED và ΔMCB có

\(\widehat{MED}=\widehat{MCB}\)

\(\widehat{M}\) chung

Do đó: ΔMED~ΔMCB

=>\(\dfrac{ME}{MC}=\dfrac{MD}{MB}\)

=>\(ME\cdot MB=MC\cdot MD\)

a: ĐKXĐ: \(x\notin\left\{1;-1;\dfrac{1}{2}\right\}\)

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

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

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

\(=\dfrac{-x-1+3x-3-4x^2}{2}\cdot\dfrac{1}{-2x+1}\)

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

b: Để A là số nguyên thì \(2x^2-x+2⋮2x-1\)

=>\(2⋮2x-1\)

=>\(2x-1\in\left\{1;-1;2;-2\right\}\)

=>\(2x\in\left\{2;0;3;-1\right\}\)

=>\(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2}\right\}\)

mà x nguyên

và \(x\notin\left\{1;-1;\dfrac{1}{2}\right\}\)

nên x=0

1: \(m^2< =2m-1\)

=>\(m^2-2m+1< =0\)

=>\(\left(m-1\right)^2< =0\)

=>\(m-1=0\)

=>m=1

2: 

ĐKXĐ: m<>2

\(\dfrac{m+3}{m-2}>=0\)

TH1: \(\left\{{}\begin{matrix}m+3>=0\\m-2>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m>=-3\\m>2\end{matrix}\right.\)

=>m>2

TH2: \(\left\{{}\begin{matrix}m+3< =0\\m-2< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< =-3\\m< 2\end{matrix}\right.\)

=>m<=-3

Lời giải:

ĐKXĐ: $x>0$

$B=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}$

$=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}=\frac{x+2\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}$

$=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}$

Suy ra:

$P=A:B=\frac{\sqrt{x}+2}{\sqrt{x}}: \frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}}=1+\frac{1}{\sqrt{x}}$

Để $P$ max thì $\frac{1}{\sqrt{x}}$ max

$\Rightarrow \sqrt{x}>0$ và $\sqrt{x}$ min

Với $x$ là số tự nhiên khác $0$, $\sqrt{x}_{\min}=1$ khi $x=1$

Khi đó: $P_{\max}=1+\frac{1}{\sqrt{1}}=2$

1: Xét (O) có \(\widehat{BAC}\) là góc nội tiếp chắn cung BC

nên \(\widehat{BOC}=2\cdot\widehat{BAC}=2\cdot65^0=130^0\)

2: Phương trình hoành độ giao điểm là:

\(\left(m-1\right)x^2=2mx-m\)

=>\(\left(m-1\right)x^2-2mx+m=0\)

Để (d) cắt (P) tại hai điểm phân biệt thì \(\left\{{}\begin{matrix}m-1< >0\\\text{Δ}>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< >1\\\left(2m\right)^2-4m\left(m-1\right)>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m< >1\\4m^2-4m^2+4m>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m\ne1\\4m>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}m>0\\m\ne1\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-\dfrac{1}{2}\\y\ne1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{y-1}{2x+1}+\dfrac{2x+1}{y-1}=-2\left(1\right)\\x+y=5\left(2\right)\end{matrix}\right.\)

Đặt \(\dfrac{2x+1}{y-1}=a\left(a\ne0\right)\)

(1) trở thành \(a+\dfrac{1}{a}=-2\)

=>\(\dfrac{a^2+1}{a}=-2\)

=>\(a^2+1+2a=0\)

=>(a+1)²=0

=>a+1=0

=>a=-1

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

=>2x+1=-y+1

=>2x+y=0

mà x+y=5

nên \(\left\{{}\begin{matrix}2x+y=0\\x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y-x-y=0-5\\x+y=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-5\\y=5-x=5-\left(-5\right)=10\end{matrix}\right.\)(nhận)