gọi A=a2−4ab+5b2+10a−22b+28

=(a2−4ab+4b2)+(b2−2b+1)+(10a−20b)+27

=(a−2b)^2+10(a−2b)+25+(b−1^)2+2

 =(a−2b+5)^2+(b−1)^2+2

Vì (a−2b+5)^2≥0và(b−1)^2≥0

=>(a−2b+5)2+(b−1)2+2≥2

Để dấu =xảy ra thì:

     {(a−2b+5)^2=0và(b−1)^2=0

⇔{a=2b−5vàb=1

⇔{a=−3 và b=1

Vậy min=2 khi (a;b)=(−3;1).

\(A=a^2-4ab+5b^2+10a-22b+28\)

\(=\left(a^2-4ab+4b^2\right)+\left(b^2-2b+1\right)+\left(10a-20b\right)+27\)

\(=\left(a-2b\right)^2+10.\left(a-2b\right)+25+\left(b-1\right)^2+2\)

\(=\left(a-2b+5\right)^2+\left(b-1\right)^2+2\)

Ta có: \(\hept{\begin{cases}\left(a-2b+5\right)^2\ge0\\\left(b-1\right)^2\ge0\end{cases}}\Rightarrow\left(a-2b+5\right)^2+\left(b-1\right)^2+2\ge2\)

Dấu '' = '' xảy ra khi: 

\(\hept{\begin{cases}\left(a-2b+5\right)^2=0\\\left(b-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}a-2b+5=0\\b-1=0\end{cases}}\Rightarrow\hept{\begin{cases}a=-3\\b=1\end{cases}}\)

Vậy giá trị nhỏ nhất của \(A=2\) khi \(\hept{\begin{cases}a=-3\\b=1\end{cases}}\)

12 doc tieng anh

) (x+3)^2-(x+2)(x-2)=4x+17

⇔ X² + 6X +9 - ( X² -4) = 4X +17

⇔ X² + 6X +9 - X² + 4 - 4X -17 =0

⇔ 2X -4 =0

⇔2X =4

⇔ X = 2

c. 3X² +7X =10

⇔ 3X² +7X -10 =0

⇔ (X -1) .(3X +10 ) =0

⇔ X -1 = 0         hoặc   3X + 10 =0

⇔ X =1               hoặc      X =-10/3

\(5x.\left(3x-2\right)=4-9x^2\)

\(\Rightarrow5x.\left(3x-2\right)-\left(4-9x^2\right)=0\)

\(\Rightarrow5x.\left(3x-2\right)+\left(9x^2-4\right)=0\)

\(\Rightarrow5x.\left(3x-2\right)+\left(3x-2\right).\left(3x+2\right)=0\)

\(\Rightarrow\left(3x-2\right).\left(5x+3x+2\right)=0\)

\(\Rightarrow\left(3x-2\right).\left(8x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=2\\8x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}\)

cảm ơn bạn

\(\left(x^2-y^2+6x+9\right):\left(x+y+3\right)\)

\(=[\left(x^2+2x.3+3^2\right)-y^2]:\left(x+y+3\right)\)

\(=[\left(x+3\right)^2-y^2]:\left(x+y+3\right)\)

\(=\left(x+3+y\right).\left(x+3-y\right):\left(x+y+3\right)\)

\(=\left(x+3-y\right)\)

x=3,605 nha 

x^3 - 64 = 0

x^3 = 0+64

x^3 = 64

Suy ra x^3 = 4^3 vì 64 = 4.4.4

Mà 4.4.4 = 4^3. Vậy x= 4

\(x^3-64=0\)

\(\left(x-4\right)\left(x^2+4x+16\right)=0\)

Dễ thấy \(x^2+4x+16=\left(x+2\right)^2+12>0\)

Nên x = 4

a) Điều kiện: \(x\ne\left\{0;\pm2\right\}\)

\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=[\frac{x^2}{x.\left(x-2\right).\left(x+2\right)}-\frac{6}{3.\left(x-2\right)}+\frac{1}{x+2}]:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{x-2.\left(x+2\right)+x-2}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)

\(=\frac{6}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)

\(=-\frac{1}{x-2}\)

b) \(A\) \(Max\)

\(\Rightarrow-\frac{1}{x-2}Max\)

\(\Rightarrow\frac{1}{x-2}Min\)

\(\Rightarrow\left(x-2\right)\) \(Max\)

\(\Rightarrow x\) \(Max\)

\(\Rightarrow x\in\varnothing\)