a) Bạn áp dụng công thức: \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3\) vào lm nhé.

a) \(\left(2x-3\right)^3\)

\(=\left(2x\right)^3-3\left(2x\right)^2.3+3.2x.3^2-3^3\)

\(=8x^3-36x+54x-27\)

c) \(\left(3x-5\right)^5\)

\(=\left(3x\right)^3-3\left(3x\right)^2.5+3.3x.5^2-5^3\)

\(=27x^3-135x^2+225x-125\)

\(\left(2x^2-y\right)^3\)

\(=8x^6-12x^4y+6x^2y^2-y^3\)

Tổng các hệ số là :

\(8+\left(-12\right)+6+\left(-1\right)\)

\(=-4+6-1\)

\(=2-1=1\)

Ta có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

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

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2\) (vì a + b + c = 0)

\(\Leftrightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)  (1)

Lại có: \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\) (2)

Thay (1) vào (2) ta được:

\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4=2\left(a^4+b^4+c^4\right)\left(đpcm\right)\)

Ta có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\) (1)

Cần chứng minh: \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^4+b^4+c^4\right)\)

\(\Leftrightarrow2\left(a^2b^2+b^2c^2+a^2c^2\right)=a^4+b^4+c^4\)

\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2\) (Cộng hai vế cho 2(a²b²+b²c²+a²c²)

\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=\left[-2\left(ab+bc+ac\right)\right]^2\) (vì (1))

\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)+8\left(ab^2c+abc^2+a^2bc\right)\)

\(\Leftrightarrow8\left(ab^2c+abc^2+a^2bc\right)=0\)

<=> 8abc (a+b+c) = 0

<=> 0 = 0 (Vì a+b+c = 0 ) (luôn luôn đúng)

Vậy => đpcm

C1: Ta có: \(a+b+c=0\)

\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)        (1)

Ta có: \(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^3=0^3\)

\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\) (2)

Thay (1) vào (2) ta có:

\(a^3+b^3+c^3+3.\left(-a\right).\left(-b\right).\left(-c\right)=0\)

\(a^3+b^3+c^3-3abc=0\)

\(a^3+b^3+c^3=3abc\)

                        đpcm

C2: \(a+b+c=0\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(a^3+3a^2+3ab^2+b^2=-c^3\)

\(a^3+b^3+c^3+3ab\left(a+b\right)=0\)

Ta có: \(a+b=-c\)

\(\Rightarrow\)\(a^3+b^3+c^3+3ab\left(-c\right)=0\)

\(a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

                                   đpcm

Ta có:\(a+b+c=0\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^3=-c^3\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)

\(\Leftrightarrow a^3-3abc+b^3=-c^3\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)

\(2x^2+4x+15=2.\left(x^2+2x+1\right)+13=2.\left(x+1\right)^2+13\ge13,\forall x\inℝ\\ \)

Dấu "=" xảy ra <=> x=-1

Vậy \(Min\left(A\right)=13\Leftrightarrow x=-1\)

2x²+4x+15

=2(x²+2x+1)+13

=2(x+1)²+13

Có 2(x+1)²\(\ge\)0 \(\forall x\in R\)

=>2(x+1)²+13\(\ge13\forall x\in R\)

Vậy GTNN của phương trình trên là 13

\(2x^2+3\left(x^2-1\right)=5x\left(x+1\right)\)

\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\)

\(\Leftrightarrow-5x=3\)

\(x=-\frac{3}{5}\)

gọi biểu thức trên là A.

Ta có: \(A=x^2-2xy+2y^2-6y+9\)

\(\Rightarrow A=x^2-2xy+y^2+y^2-6y+9\)

\(\Rightarrow A=\left(x^2-2xy+y^2\right)+\left(y^2-6y+9\right)\)

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

Nhận xét: \(\left(x+y\right)^2\ge0\forall x,y\)

                 \(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow\left(x+y\right)^2+\left(y-3\right)^2\ge0\forall x,y\)

Vậy \(minA=0\) khi \(y-3=0\Rightarrow y=3\)

                                       \(x-y=0\Rightarrow x-3=0\Rightarrow x=3\)

KL: Vậy \(minA=0\) khi \(x=3;y=3\)

Đặt \(A=x^2-2xy+2y^2-6y+9=\left(x^2-2xy+y^2\right)+\left(y^2-6y+9\right)=\left(x-y\right)^2+\left(y-3\right)^2\)

Vì \(\left(x-y\right)^2\ge0;\left(y-3\right)^2\ge0\Rightarrow A=\left(x-y\right)^2+\left(y-3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-3=0\end{cases}\Leftrightarrow x=y=3}\)

Vậy Amin = 0 khi x = y = 3

d ) 

\(B=5\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)

\(\Rightarrow B=\frac{5}{3}\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)

\(\Rightarrow B=\frac{5}{3}\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)

\(\Rightarrow B=\frac{5}{3}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)

\(\Rightarrow B=\frac{5}{3}\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)

\(\Rightarrow B=\frac{5}{3}\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)

\(\Rightarrow B=\frac{5}{3}\left(2^{64}-1\right)\left(2^{64}+1\right)\)

\(\Rightarrow B=\frac{5}{3}\left(2^{128}-1\right)\)

Sửa lại dấu \(\Rightarrow\)dòng 3 : 

\(B=\frac{5}{3}\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\)