`B = x^2- 2xy + y^2 + 2x - 10y + 17

`2B = 2x^2 - 4xy + 2y^2 + 4x - 20y + 34`

`= (x-y)^2 + (x+2)^2 + (y-5)^2 + 5 >= 5`.

ko biết

Sửa đề:

A=/x+5/+10

Ta có: /x+5/>= 0 với mọi x>=0

=> A=/x+5/+10 >= 10

=> Amin=10. Dấu "=" xảy ra <=> x+5=0<=> x=-5

Vậy...

\(\text{a) }A=\left|x+5\right|+10\)

\(\text{Vì }\left|x+5\right|\ge0\forall x\)

\(\Rightarrow A=\left|x+5\right|+10\ge10\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left|x+5\right|=0\)

\(\Rightarrow x=-5\)

\(\text{Vậy Min}_A=10\Leftrightarrow x=-5\)

\(\text{b) }\left|3-x\right|+5\)

\(\text{Vì }\left|3-x\right|\ge0\forall x\)

\(\Rightarrow\left|3-x\right|+5\ge5\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left|3-x\right|=0\)

\(\Rightarrow x=3\)

\(\text{Vậy Min}_B=5\Leftrightarrow x=3\)

\(\text{d) }D=\left(x+2\right)^2+15\)

\(\text{Vì ( x + 2 )}^2\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2+15\ge15\)

\(\text{Dấu ''='' xảy ra khi :}\)

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

\(\Rightarrow x+2=0\)

\(\Rightarrow x=-2\)

\(C=\left|x+1\right|+\left|y+4\right|\)

Ta có +) \(\left|x+1\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

+)\(\left|y+4\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow y+4=0\Leftrightarrow y=-4\)

\(\Rightarrow B=\left|x+1\right|+\left|y+2\right|\ge0\)

\(\Rightarrow minB=0\Leftrightarrow\hept{\begin{cases}x=-1\\y=-4\end{cases}}\)

P/s: ko chắc nha 

Ta có \(xy\le\dfrac{\left(x+y\right)^2}{4}\).

Do đó ta có: \(x+y+xy=x+y-2xy+3xy\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)

\(\Rightarrow x^2+y^2\le x+y-2xy+\dfrac{3}{4}\left(x+y\right)^2\)

\(\Leftrightarrow\dfrac{1}{4}\left(x+y\right)^2-\left(x+y\right)\le0\)

\(\Leftrightarrow\left(x+y\right)\left[\dfrac{1}{4}\left(x+y\right)-1\right]\le0\)

\(\Leftrightarrow0\le x+y\le4\).

Do đó m = 0, n = 4.

Vậy m² + n² = 16. Chọn A.

\(a,x=\dfrac{18}{z};y=10\Leftrightarrow x:y=\dfrac{18}{z}:10=\dfrac{9}{5z}=9:5z\)

\(b,y=x:\dfrac{9}{5z}=\dfrac{9}{5xz}\)

\(c,x=-2\Leftrightarrow z=-9\Leftrightarrow y=\dfrac{9}{5\cdot\left(-9\right)\cdot\left(-2\right)}=\dfrac{1}{10}\\ x=\dfrac{1}{5}\Leftrightarrow z=90\Leftrightarrow y=\dfrac{9}{5\cdot\dfrac{1}{5}\cdot90}=\dfrac{1}{10}\)

a: \(A=\dfrac{-1}{3}\cdot\dfrac{-2}{3}\cdot\dfrac{3}{2}\cdot x^2y\cdot xy^3\cdot xy^2=\dfrac{1}{3}x^4y^6\)

Hệ số là 1/3

Bậc là 10

b: \(P=\dfrac{1}{25}+3\cdot\dfrac{1}{5}\cdot\left(-1\right)+1=\dfrac{26}{25}-\dfrac{3}{5}=\dfrac{26}{25}-\dfrac{15}{25}=\dfrac{11}{25}\)

a)A= -\(\dfrac{1}{3}\)x²y. -\(\dfrac{2}{3}\)xy³. 1\(\dfrac{1}{2}\)xy²

A=(-\(\dfrac{1}{3}\).-\(\dfrac{2}{3}\).1\(\dfrac{1}{2}\)). (x².x.x). (y.y³.y²)
A=1.x⁴.y⁶

^{_Hệ số là:1
_Bậc của đơn thức A là 10}
 

b)Thay x=\(\dfrac{1}{5}\) , y=-1vào biểu thức P, ta có:

P=\(\dfrac{1}{5}\).2+3.\(\dfrac{1}{5}\).-1+-1.2
P=\(\dfrac{2}{5}\)+ -\(\dfrac{3}{5}\) +-2
P=-\(\dfrac{1}{5}\)+ -2=-\(\dfrac{11}{5}\)

b)

\(\left(x+1\right)^2+3\) 

ta có: \(\left(x+1\right)^2\) \(\geq\)0 với mọi x

=> \(\left(x+1\right)^2+3\) \(\geq\) \(3\) với mọi x

dấu bằng xảy ra<=>x+1=0

                          <=>x=1

vậy GTNN của biểu thức \(\left(x+1\right)^2+3\) là \(3\) <=> x= \(-1\)\

1) A. 999.

2) C. 9.

1: A

2: C

Câu 1: A

Câu 2: C