Bài 2:

1:

a: Thay m=2 vào hệ phương trình, ta được:

\(\left\{{}\begin{matrix}3x+y=2-1=1\\x-2y=5\cdot2+2=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x+2y=2\\x-2y=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=14\\x-2y=12\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\2y=x-12=2-12=-10\end{matrix}\right.\)

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

b: Vì \(\dfrac{3}{1}\ne\dfrac{1}{-2}\)

nên hệ luôn có nghiệm duy nhất

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

=>\(\left\{{}\begin{matrix}6x+2y=2m-2\\x-2y=5m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x+2y+x-2y=2m-2+5m+2\\x-2y=5m+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x=7m\\2y=x-5m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=m-5m-2=-4m-2\end{matrix}\right.\)

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

\(T=x^2+y+12\)

\(=m^2-2m-1+12\)

\(=m^2-2m+11=\left(m-1\right)^2+10>=10\forall m\)

Dấu '=' xảy ra khi m-1=0

=>m=1

2:

a: Thay m=2 vào phương trình, ta được:

\(x^2-2\left(2+2\right)x+2^2+7=0\)

=>\(x^2-8x+11=0\)

=>\(\left(x-4\right)^2=5\)

=>\(x-4=\pm\sqrt{5}\)

=>\(x=4\pm\sqrt{5}\)

b: \(\Delta=\left(-2m-4\right)^2-4\left(m^2+7\right)\)

\(=4m^2+16m+16-4m^2-28=16m-12\)

Để phương trình có hai nghiệm phân biệt thì 16m-12>0

=>16m>12

=>\(m>\dfrac{3}{4}\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+2\right)=2m+4\\x_1x_2=\dfrac{c}{a}=m^2+7\end{matrix}\right.\)

\(x_1^2+x_2^2=x_1x_2+12\)

=>\(\left(x_1+x_2\right)^2-3x_1x_2=12\)

=>\(\left(2m+4\right)^2-3\left(m^2+7\right)-12=0\)

=>\(4m^2+16m+16-3m^2-21-12=0\)

=>\(m^2+16m-17=0\)

=>(m+17)(m-1)=0

=>\(\left[{}\begin{matrix}m=-17\left(loại\right)\\m=1\left(nhận\right)\end{matrix}\right.\)

a: \(\Delta=\left[-2\left(m+1\right)\right]^2-4\cdot1\cdot\left(-2\right)\left(m+5\right)\)

\(=4\left(m^2+2m+1\right)+8\left(m+5\right)\)

\(=4m^2+8m+4+8m+20\)

\(=4m^2+16m+24=\left(2m+4\right)^2+8>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

b: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+1\right)\\x_1x_2=\dfrac{c}{a}=-2\left(m+5\right)\end{matrix}\right.\)

\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=1\)

=>\(\dfrac{x_1+x_2}{x_1x_2}=1\)

=>\(\dfrac{2\left(m+1\right)}{-2\left(m+5\right)}=1\)

=>\(\dfrac{-\left(m+1\right)}{m+5}=1\)

=>-m-1=m+5

=>-2m=6

=>m=-3

c: Thay m=1 vào (1), ta được:

\(x^2-2\left(1+1\right)x-2\left(1+5\right)=0\)

=>\(x^2-4x-12=0\)

=>(x-6)(x+2)=0

=>\(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

Thịnh lm đúng rồi đó bạn! 

a: Thay m=1 vào phương trình, ta được:

\(x^2+\left(1-1\right)x-2\cdot1=0\)

=>\(X^2-2=0\)

=>\(x^2=2\)

=>\(x=\pm\sqrt{2}\)

b: \(\text{Δ}=\left(m-1\right)^2-4\cdot1\cdot\left(-2m\right)\)

\(=m^2-2m+1+8m\)

\(=m^2+6m+1\)

Để phương trình có nghiệm thì Δ>=0

=>\(m^2+6m+1>=0\)

=>\(\left(m+3\right)^2>=8\)

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

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

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

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

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

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

\(ĐKXĐ:x\ge0.x\ne1\)

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

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

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

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

\(A=\dfrac{-1}{\sqrt{x}-1}.\dfrac{\sqrt{x}+1}{1}+1\)

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

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

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

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

Giải thích: 

CDHG nội tiếp \(\Rightarrow\Delta PDH\sim\Delta PGC\Rightarrow\dfrac{PD}{PG}=\dfrac{PH}{PC}\Rightarrow\dfrac{7}{14}=\dfrac{6}{CD+7}\)

\(\Rightarrow CD=5\)

ABFFE nội tiếp \(\Rightarrow\Delta PBF\sim\Delta PEA\Rightarrow\dfrac{PB}{PE}=\dfrac{PF}{PA}\Rightarrow\dfrac{20}{32}=\dfrac{22}{20+AB}\)

\(\Rightarrow AB=15,2\)

\(AB=15,2\)

\(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=-1\end{matrix}\right.\)

Do \(x_1\) là nghiệm \(\Rightarrow x_1^2-3x_1-1=0\Rightarrow x_1^2=3x_1+1\)

\(\Rightarrow x_1^3=3x_1^2+x_1\)

\(P=3x_1^2+x_1+3x_2^2+x_2+1988\)

\(=3\left(x_1+x_2\right)^2-6x_1x_2+x_1+x_2+1988\)

\(=3.3^2-6.\left(-1\right)+3+1988=...\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=3\\x_1x_2=\dfrac{c}{a}=-1\end{matrix}\right.\)

\(P=x_1^3+3x_2^2+x_2+1988\)

\(=x_1^3+x_2^2\left(x_1+x_2\right)+x_2+1988\)

\(=x_1^3+x_2^3+x_2\left(x_1x_2+1\right)+1988\)

\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+x_2\left(x_1x_2+1\right)+1988\)

\(=3^3-3\cdot3\cdot\left(-1\right)+1988\)

=27+9+1988

=2024

a: Xét tứ giác BDOM có \(\widehat{BDO}+\widehat{BMO}=90^0+90^0=180^0\)

nên BDOM là tứ giác nội tiếp

=>\(\widehat{DOM}+\widehat{CBE}=180^0\)

Xét (O) có

\(\widehat{CBE}\) là góc nội tiếp chắn cung CE

\(\widehat{CAE}\) là góc nội tiếp chắn cung CE

Do đó: \(\widehat{CBE}=\widehat{CAE}\)

=>\(\widehat{DOM}+\widehat{CAE}=180^0\)

ae giúp nhanh vs cần gấp