Ta có:

\(\begin{array}{l}{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {R^2}\\ \Leftrightarrow {x^2} - 2ax + {a^2} + {y^2} - 2by + {b^2} - {R^2} = 0\\ \Leftrightarrow {x^2} + {y^2} - 2ax - 2by + c = 0\left( {{a^2} + {b^2} - {R^2} = c} \right)\end{array}\)

Bài 2 :

a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)

\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)

\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)

\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)

a. ( a + b + c)² + a² + b² + c²

= a² + b² + c² + 2ab + 2ac + 2bc + a² + b² + c²

= (a+b)² + (b+c)² + (a+c)²

b. 2.(a-b).(c-b) + 2.(b-a).(c-a) + 2.(b-c).(a-c)

đặt a - b = x; b-c = y; c-a = z => x + y + z = 0 (1)

ta có: 2.x.(-y) + 2.(-x).z + 2.y.(-z)

= -2xy - 2xz - 2yz  = -2.(xy+xz+yz)  

ta có: (x+y+z)² = x² + y² + z² + 2xy + 2yz + 2xz

 0² = x² + y² + z² + 2.(xy+yz+xz)

=> x² + y² + z²  = -2.(xy+yz+xz) (2)

Từ (2) => 2.(a-b).(c-b) + 2.(b-a) .(c-a) + 2.(b-c).(a-c) = x² + y² + z²

= (a-b)² + (b-c)² + (c-a)²

ĐÂY MÀ LÀ toán 5 ạ??

Gọi A là vế trái của BĐT cần chứng minh. Không mất tính tổng quát, ta giả sử a + b + c = 3. Áp dụng BĐT AM - GM ta có:

\(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{8bc\left(4a+4b+c\right)}}+\frac{ab\left(4a+4b+c\right)}{27}\)\(\ge\frac{1}{2}\left(a+b\right)\)

Suy ra 

             \(\sqrt{\frac{\left(a+b\right)^3}{8ab\left(4a+4b+c\right)}}\)\(+\frac{ab\left(4a+4b+c\right)}{54}\ge\frac{1}{4}\left(a+b\right)\)

Tương tự

            \(\sqrt{\frac{\left(b+c\right)^3}{8bc\left(4b+4c+a\right)}}+\frac{bc\left(4b+4c+a\right)}{54}\ge\frac{1}{4}\left(b+c\right)\)

và       \(\sqrt{\frac{\left(c+a\right)^3}{8ca\left(4c+4a+b\right)}}+\frac{ca\left(4c+4a+b\right)}{54}\ge\frac{1}{4}\left(c+a\right)\)

Cộng ba BĐT trên ta có: 

           \(\frac{1}{2\sqrt{2}}A\ge B\)

Với \(A=\frac{1}{54}[ab\left(4a+4b+c\right)+bc\left(4b+4c+a\right)\)

\(+ca\left(4c+4a+b\right)]\)

\(=\frac{1}{54}\left[4ab\left(a+b\right)+4bc\left(b+c\right)+4ca\left(c+a\right)+3abc\right]\)

\(=\frac{1}{54}\left[4\left(a+b+c\right)\left(ab+bc+ca\right)-9abc\right]\)

\(\le\frac{1}{54}\left(a+b+c\right)^3=\frac{1}{2}\)

và \(B=\frac{1}{4}.2\left(a+b+c\right)=\frac{3}{2}\)

Suy ra \(\frac{1}{2\sqrt{2}}A\ge\frac{3}{2}-\frac{1}{2}=1\Rightarrow A\ge2\sqrt{2}\)

Vậy 

              \(\sqrt{\frac{\left(a+b\right)^3}{ab\left(4a+4b+c\right)}}+\sqrt{\frac{\left(a+b\right)^3}{bc\left(4a+4b+c\right)}}+\sqrt{\frac{\left(c+a\right)^3}{ca\left(4c+4a+b\right)}}\ge2\sqrt{2}\)(đpcm)

toán lớp 5 phiên bản hack não

a: \(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)

\(=a^2+2ab+b^2+b^2+2bc+c^2+a^2+2ac+c^2\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)

b: \(=2\left(a-b\right)\left(c-b\right)-2\left(a-b\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)

\(=2\left(a-b\right)\left(c-b-c+a\right)+2\left(b-c\right)\left(a-c\right)\)

\(=2\left(a-b\right)\left(a-b\right)+2\left(b-c\right)\left(a-c\right)\)

\(=2\left(a^2-2ab+b^2+ab-bc-ac+c^2\right)\)

\(=2\left(a^2+b^2-ab-bc-ac+c^2\right)\)

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

a)      \(\left( {x + 4} \right)\left( {{x^2} - 4x + 16} \right) = {x^3} + {4^3} = {x^3} + 64\)

b)      \(\left( {4{x^2} + 2xy + {y^2}} \right)\left( {2x - y} \right) = {\left( {2x} \right)^3} - {y^3} = 8{x^3} - {y^3}\)

a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)

b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)

\(=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)

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

c) \(-16+\left(x-3\right)^2=\left(x-3\right)^2-16=\left(x-3-4\right)+\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)

d) \(64+16y+y^2\)

\(=8^2+2.8.y+y^2\)

\(=\left(8+y\right)^2\)

1/ a, \(=4^2-a^2=\left(4-a\right)\left(4+a\right)\)

b, \(=\left(a+b\right)^2-\left(2c\right)^2=\left(a+b-2c\right)\left(a+b+2c\right)\)

2/ a, \(101^2=\left(100+1\right)^2=100^2+2.100.1+1^2=10000+200+1=10201\)

b, \(199^2=\left(200-1\right)^2=200^2-2.200.1+1^2=40000-400+1=39601\)

c, \(47.53=\left(50-3\right)\left(50+3\right)=50^2-3^2=2500-9=2491\)

Giải:

a) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)

\(=\left[\left(x^2+x-1\right)-\left(x^2+2x+3\right)\right]\left[\left(x^2+x-1\right)+\left(x^2+2x+3\right)\right]\)

\(=\left(x^2+x-1-x^2-2x-3\right)\left(x^2+x-1+x^2+2x+3\right)\)

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

Vậy ...

b) \(-16+\left(x-3\right)^2\)

\(=\left(x-3\right)^2-16\)

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

\(=\left(x-3-4\right)\left(x-3+4\right)\)

\(=\left(x-7\right)\left(x+1\right)\)

Vậy ...

c) \(64+16y+y^2\)

\(=8^2+2.8.y+y^2\)

\(=\left(8+y\right)^2\)

Vậy ...