\(\frac{63^2-47^2}{215^2-105^2}=\)  \(\frac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)

                           \(=\frac{16.110}{110.320}=\frac{16}{320}\)\(=\frac{1}{20}\)

các câu kia làm tương tự nha

Thanks bạn nhiều nhiều nha!

(2n + 1) (n² - 3n - 1)

➡️2n³ - 6n² - 2n + n² - 3n - 1

➡️2n³ - 5n² - 5n - 1

Hok tốt~

2/

a,Ta có: a+b+c=0

<=>(a+b+c)²=0

<=>a²+b²+c²+2(ab+bc+ca)=0

<=>2+2(ab+bc+ca)=0

<=>ab+bc+ca=\(\frac{-2}{2}=-1\)

<=>(ab+bc+ca)²=1

<=>a²b²+b²c²+c²a²+2abc(a+b+c)=1

<=>a²b²+b²c²+c²a²=1 (vì a+b+c=0)

Lại có: a²+b²+c²=2

<=>(a²+b²+c²)²=4

<=>a⁴+b⁴+c⁴+2(a²b²+b²c²+c²a²)=4

<=>a⁴+b⁴+c⁴+2=4 (vì a²b²+b²c²+c²a²=1)

<=>a⁴+b⁴+c⁴=2

b, tương tự a

1/

b, \(B=9x^2-6x+2=9x^2-6x+1+1=\left(3x-1\right)^2+1\)

Vì \(\left(3x-1\right)^2\ge0\Rightarrow B=\left(3x-1\right)^2+1\ge1\)

Dấu "=" xảy ra khi x=1/3

Vậy Bmin = 1 khi x = 1/3

c,\(C=x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow C=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi x=-1/2

Vậy...

d, \(D=2x^2+2x+1=2\left(x^2+x+\frac{1}{2}\right)=2\left(x^2+x+\frac{1}{4}+\frac{1}{4}\right)=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\)

Vì \(2\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow D=2\left(x+\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

Dấu "=" xảy ra khi x=-1/2

Vậy...

\("="\Leftrightarrow x=1;y=3\)

Bạn thêm vào dòng cuối nhé :v Mình quên ghi :v

\(\frac{-7x^2+42x-64}{x^2-6x+10}\)

\(\Rightarrow7+\frac{6}{\left(x-3\right)^2+1}=y^2+2y\)

\(\Rightarrow\frac{6}{\left(x-3\right)^2+1}=\left(y-1\right)^2+6\)

\(\Rightarrow6=\left[\left(y-1\right)^2+6\right]\left[\left(x-3\right)^2+1\right]\)

\(\Rightarrow0=\left(y-1\right)^2\left(x-3\right)^2+6\left(x-3\right)^2+\left(y-1\right)^2\)

ta có 

áp dụng bất đẳng thức Bunhia : 

\(P^2=\left(x+y+2z\right)^2\le\left(1^2+1^2+2^2\right)\left(x^2+y^2+z^2\right)=6\times6=36\)

Do đó \(-6\le P\le6\text{ nên GTNN P= - 6, GTLN P =6}\)