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

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

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

2)Phần này tg tự

3)\(\left(x+y+z\right)\left(x+y-z\right)=\left(x+y\right)^2-z^2=x^2+2xy+y^2-z^2\)

B1 Xét (7x+1)\(^2\)-(x+7)\(^2\)-48(x\(^2\)-1)

=49\(x^2\)+14x+1-x\(^2\)-14x-49-48x\(^2\)+48

=0

Vậy \(\left(7x+1\right)^2-\left(x+7\right)^2=48\left(x^2-1\right)\)

B2 \(16x^2-\left(4x-5\right)^2=15\)

(4x)\(^2\)-(4x-5)\(^2\)-15=0

(4x-4x+5)(4x+4x-5)-15=09x-5)=0

5(8x-5)-15=0

40x-25-15=0

40x-40=0

x        =1

câu B3 mình không bik làm 

chúc bạn học tốt ~~~

Bài 3:

\(A=x^2+2x+3\)

\(=\left(x+1\right)^2+2\ge2\)

Vậy  MIN  \(A=2\)   khi    \(x=-1\)

p/s: chúc bạn học tốt

\(27x^3-27x^2+9x-1=\left(3x\right)^3-3.\left(3x^2\right).1+3.3x.1+1^3=\left(3x-1\right)^3\)

b) không hiểu đề

Tham khảo~

BÀI 1:

Ta có:   \(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)

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

                    \(=\left(8x+8\right)\left(6x-6\right)\)

                   \(=8\left(x+1\right).6\left(x-1\right)\)

                  \(=48\left(x^2-1\right)=VP\)  (đpcm)

Bài 2:

         \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow\)\(16x^2-16x^2+40x-25=15\)

\(\Leftrightarrow\)\(40x=40\)

\(\Leftrightarrow\)\(x=1\)

Vậy...

Bài 3:

\(A=x^2+2x+3=\left(x+1\right)^2+2\ge2\)

Vậy MIN A = 2  khi  x = -1

\(4=\left(a+b+c+d\right)^2\ge4\left(a+b+c\right).d\)

\(\Rightarrow1\ge\left(a+b+c\right).d\)

\(\Rightarrow a+b+c\ge\left(a+b+c\right)^2d\ge4\left(a+b\right).c.d\)

\(\Rightarrow A=\frac{\left(a+b+c\right)\left(a+b\right)}{abcd}\ge\frac{4\left(a+b\right)^2.cd}{abcd}\ge\frac{16ab.cd}{abcd}=16\)

Nên GTNN của A là 16 đạt được khi  \(a=b=\frac{1}{4};c=\frac{1}{2};d=1\)