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\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right).\left(1+\sqrt{2}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right).\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right).\left(\sqrt{99}+\sqrt{100}\right)}\)

\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{100}-\sqrt{99}}{100-99}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-1=10-1=9\)

ai k dung mik giai cho

ai k dung mik giai cho

ai k dung mik giai cho

Giải đi

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

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

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...