a: Để hệ có nghiệm duy nhất thì \(\frac{m}{1}<>\frac{-1}{4\left(m+1\right)}\)

=>\(4m\left(m+1\right)<>-1\)

=>\(4m^2+4m+1<>0\)

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

=>2m+1<>0

=>m<>-1/2

\(\begin{cases}mx-y=1\\ x+4\left(m+1\right)y=4m\end{cases}\Rightarrow\begin{cases}y=mx-1\\ x+4\left(m+1\right)\left(mx-1\right)=4m\end{cases}\)

=>\(\begin{cases}y=mx-1\\ x+\left(4m+4\right)\left(mx-1\right)=4m\end{cases}\Rightarrow\begin{cases}y=mx-1\\ x+4m^2x-4m+4mx-4=4m\end{cases}\)

=>\(\begin{cases}y=mx-1\\ x\left(4m^2+4m+1\right)=4m+4m+4\end{cases}\Rightarrow\begin{cases}y=mx-1\\ x\left(2m+1\right)^2=8m+4=4\left(2m+1\right)\end{cases}\)

=>\(\begin{cases}x=\frac{4}{2m+1}\\ y=mx-1=\frac{4m}{2m+1}-1=\frac{4m-2m-1}{2m+1}=\frac{2m-1}{2m+1}\end{cases}\)

Để x,y nguyên thì 4⋮2m+1 và 2m-1⋮2m+1

=>4⋮2m+1 và 2m+1-2⋮2m+1

=>4⋮2m+1 và -2⋮2m+1

=>2m+1∈Ư(2)

mà 2m+1 lẻ

nên 2m+1∈{1;-1}

=>2m∈{0;-2}

=>m∈{0;-1}

b: Để hệ có nghiệm duy nhất thì \(\frac{m+1}{2}<>\frac{3m+1}{m+2}\)

=>\(\left(m+1\right)\left(m+2\right)<>2\left(3m+1\right)\)

=>\(m^2+3m+2-6m-2<>0\)

=>\(m^2-3m<>0\)

=>m(m-3)<>0

=>m∉{0;3}

\(\begin{cases}\left(m+1\right)x+\left(3m+1\right)y=2-m\\ 2x+\left(m+2\right)y=4\end{cases}\Rightarrow\begin{cases}\left(2m+2\right)x+\left(6m+2\right)y=4-2m\\ \left(2m+2\right)x+\left(m+2\right)\left(m+1\right)y=4\left(m+1\right)\end{cases}\)

=>\(\begin{cases}x\left(2m+2\right)+y\left(m^2+3m+2\right)-\left(2m+2\right)x-\left(6m+2\right)y=4\left(m+1\right)-4+2m\\ 2x+\left(m+2\right)y=4\end{cases}\)

=>\(\begin{cases}y\left(m^2-3m\right)=6m\\ 2x+\left(m+2\right)y=4\end{cases}\Rightarrow\begin{cases}y=\frac{6}{m-3}\\ 2x=4-\left(m+2\right)\cdot y=4-\frac{6\left(m+2\right)}{m-3}=\frac{4m-12-6m-12}{m-3}=\frac{-2m-24}{m-3}\end{cases}\)

=>\(\begin{cases}y=\frac{6}{m-3}\\ x=\frac{-m-12}{m-3}\end{cases}\)

Để x,y nguyên thì 6⋮m-3 và -m-12⋮m-3

=>6⋮m-3 và -m+3-15⋮m-3

=>6⋮m-3 và -15⋮m-3

=>m-3∈ƯC(6;-15)

=>m-3∈Ư(3)

=>m-3∈{1;-1;3;-3}

=>m∈{4;2;6;0}

mà m∉{0;3}

nên m∈{2;4;6}

Câu 1 :

Ta có :

\(\Delta=\left(m-1\right)^2-4.\left(2m-7\right)\)

\(=m^2-2m+1-8m+28\)

\(=m^2-10m+27>0\)

Do đó pt luôn có 2 nghiệm phân biệt

zzz

giải được bài 1

\(x^4+y^4+z^4=\dfrac{x^4+y^4}{2}+\dfrac{y^4+z^4}{2}+\dfrac{x^4+z^4}{2}\)

\(\ge x^2y^2+y^2z^2+x^2z^2=\dfrac{x^2y^2+y^2z^2}{2}+\dfrac{y^2z^2+x^2z^2}{2}+\dfrac{x^2y^2+x^2z^2}{2}\)

\(\ge xy^2z+xyz^2+x^2yz=xyz\left(x+y+z\right)=xyz\)

\(\Rightarrow x^4+y^4+z^4\ge xyz\)

Dấu " =" xảy ra \(\Leftrightarrow x=y=z\)

Thay vào PT (1) \(\Rightarrow x=y=z=\dfrac{1}{3}\)

1/ \(P=\frac{1}{x+y+x+z}+\frac{1}{x+y+y+z}+\frac{1}{x+z+y+z}\)

\(P\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)

\(P\le\frac{1}{2}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\le\frac{1}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{z}\right)\)

\(P\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)

Dấu "=" xảy ra khi \(x=y=z=\frac{3}{4}\)

2/ ĐKXĐ: ...

\(\Leftrightarrow4x^2-8x\sqrt{x+1}+3\left(x+1\right)\le0\)

\(\Leftrightarrow\left(2x-\sqrt{x+1}\right)\left(2x-3\sqrt{x+1}\right)\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\ge\sqrt{x+1}\\2x\le3\sqrt{x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2-x-1\ge0\\4x^2-9x-9\le0\end{matrix}\right.\) \(\Rightarrow\frac{-1+\sqrt{17}}{8}\le x\le3\)

\(\Rightarrow x=\left\{1;2;3\right\}\Rightarrow\sum x^2=1+4+9=14\)