đề bài là j ạ?

đề bài là gì vậy bạn !!!

Vì \(x,y\ge0\Rightarrow\left|19x+5y\right|+1975=\left|19y+5x\right|+2014^x\)

\(\Leftrightarrow19x+5y+1975=19y+5x+2014^x\)

\(\Leftrightarrow24x+24y+2014^x=1975\)

\(\Leftrightarrow2\left(12x+12y+2014^{x-1}\cdot1007\right)=1975\)

Do \(VT⋮2\Rightarrow VF⋮2\) mà \(VF\) không chia hết cho 2.

Vậy không có số tự nhiên x;y thỏa mãn đề bài.

Ta có : 3(a+b)=4a+3c

<=> 3a+3b=4a+3c

<=> 3b-3c=4a-3a

<=> 3b-3c=a (đpcm)

\(\frac{MN}{MP}=\frac{24}{7}\Rightarrow\frac{MN}{24}=\frac{MP}{7}\)

\(\Leftrightarrow\left(\frac{MN}{24}\right)^2+\left(\frac{MP}{7}\right)^2=\frac{MN^2}{576}+\frac{MP^2}{49}=\frac{MN^2+MP^2}{576+49}=\frac{NP^2}{625}=\frac{75^2}{625}=9\)

\(\Rightarrow\frac{MN^2}{576}=9\Rightarrow MN=72\left(cm\right)\)

\(\frac{MP^2}{49}=9\Rightarrow MP=21\left(cm\right)\)

\(\frac{3x-2y}{2015}=\frac{2x-4x}{2016}=\frac{4y-3z}{2017}\)

\(\Rightarrow\frac{12x-8y}{8060}=\frac{6z-12x}{6048}=\frac{8y-6z}{4034}=\frac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{8060+6048+4034}=0\)

\(\Leftrightarrow\hept{\begin{cases}3x-2y=0\\2z-4x=0\\4y-3z=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2y\\2z=4x\\4y=3z\end{cases}}}\Leftrightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{x}{2}=\frac{z}{4}\\\frac{y}{3}=\frac{z}{4}\end{cases}}\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\left(k\ne0\right)\)

\(\Rightarrow x=2k;y=3k;z=4k\)

Thay vào P ta có

\(P=\frac{4k^2-2.2k.3k-16k^2}{4k^2+9k^2+16k^2}=\frac{k^2\left(4-12-16\right)}{k^2\left(4+9+16\right)}=-\frac{24}{29}\)

Ấn máy tính mode 5,1 

hoặc dùng delta

TA CÓ :\(\frac{1}{n+1}>\frac{1}{2n},\frac{1}{n+2}>\frac{1}{2n},....\)\(\Rightarrow\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}>\frac{1}{2n}+\frac{1}{2n}+...+\frac{1}{2n}\)(n số)

=\(\frac{n}{2n}=\frac{1}{2}\left(đcpm\right)\)

\(\left|2x-27\right|^{2007}+\left(3y+10\right)^{2018}=0\)

Ta  có \(\left|2x-27\right|^{2017}\ge0\forall x;\left(3y+10\right)^{2018}\ge0\forall y\)

\(\Rightarrow\left|2x-27\right|^{2017}+\left(3.y+10\right)^{2018}\ge0\forall x;y\)

\(\Rightarrow\left|2x-17\right|^{2017}+\left(3y+10\right)^{2018}=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-17=0\\3.y+10=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{17}{2}\\y=-\frac{10}{3}\end{cases}}\)