a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)

\(\left|x-1\right|^{2023}>=0\forall x\)

=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)

mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)

nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)

=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)

\(P=x^{2023}+\left(y-10\right)^{2023}\)

\(=1^{2023}+\left(9-10\right)^{2023}\)

=1-1

=0

c: \(\left|x-3\right|>=0\forall x\)

=>\(\left|x-3\right|+2>=2\forall x\)

=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)

mà \(\left|y+3\right|>=0\forall y\)

nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)

=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-3=0

=>x=3 và y=3

Ta có: \(x+y=2,5=\frac{5}{2}\)

\(\Rightarrow\left(x+y\right)^2=\frac{25}{4}\)

Ta  có: 

\(A=\frac{x}{y}+\frac{y}{x}\)

\(=\frac{x.x}{x.y}+\frac{y.y}{y.x}\)

\(=\frac{x^2}{xy}+\frac{y^2}{xy}\)

\(=\frac{x^2+y^2}{xy}\)

\(=\frac{\frac{25}{4}}{1}=\frac{25}{4}\)

x=25/4

a) Ta có : A = - 15 - |7 - x| = -(15 + |7 - x|) 

vì \(\left|7-x\right|\ge0\forall x\Rightarrow15+\left|7-x\right|\ge15\Rightarrow-\left(15+\left|7-x\right|\right)\le-15\)

Dấu"=" xảy ra <=> 7 - x = 0

=> x = 7

Vậy GTLN của A là - 15 khi x = 7

b) Ta có : \(\hept{\begin{cases}\left|x+2,5\right|\ge0\forall x\\\left(y-1\right)^4\ge0\forall y\end{cases}\Rightarrow\left|x+2,5\right|+\left(y-1\right)^4\ge0}\)

=> \(\left|x+2,5\right|+\left(y-1\right)^4-6\ge-6\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+2,5=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-2,5\\y=1\end{cases}}}\)

Vậy GTNN của B là - 6 khi \(\hept{\begin{cases}x=-2,5\\y=1\end{cases}}\)

a) Vì \(\left|7-x\right|\ge0\forall x\)\(\Rightarrow-15-\left|7-x\right|\le-15\forall x\)

hay \(A\le-15\)

Dấu " = " xảy ra \(\Leftrightarrow7-x=0\)\(\Leftrightarrow x=7\)

Vậy \(maxA=-15\Leftrightarrow x=7\)

b) Vì \(\hept{\begin{cases}\left|x+2,5\right|\ge0\forall x\\\left(y-1\right)^4\ge0\forall y\end{cases}}\)\(\Rightarrow\left|x+2,5\right|+\left(y-1\right)^4\ge0\forall x,y\)

\(\Rightarrow\left|x+2,5\right|+\left|y-1\right|^4-6\ge-6\forall x,y\)

hay \(B\ge-6\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2,5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2,5\\y=1\end{cases}}\)

Vậy \(minB=-6\Leftrightarrow\hept{\begin{cases}x=-2,5\\y=1\end{cases}}\)

A = x² + 2x + 1 - y²

= (x + 1)² - y²

= (x + 1 - y) (x + 1 + y)

Với x = 6,5, y = 2,5, ta có:

= (6,5 + 1 - 2,5) (6,5 + 1 + 2,5)

= 5 . 10 = 50

Vậy khi x = 6,5, y = 2,5 thì A = 50.

a,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)

\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)

Tính giá trị biểu thức:

a) x + (–  26), biết x = 6

Với x = 6 thì x + (–  26) = 6 + (–  26) = -20

b) – 15 +  y, biết y = – 25

Với y = -25 thì – 15 +  y = – 15 +  (-25) = - 40 

a.

\(2,5:y=5\)

\(\Leftrightarrow y=2,5:5\)

\(\Leftrightarrow y=0,5\)

b.

\(2,2\times4,8+2,2\times5,2\)

\(=2,2\times\left(4,8+5,2\right)\)

\(=2,2\times10\)

\(=22\)

a.

$2,5:y=5$

$\iff y=2,5:5$

$\iff y=0,5$

b.

$2,2\times 4,8+2,2\times 5,2$

$=2,2\times \left(4,8+5,2\right)$

$=2,2\times 10$

$=22$