Em tách ra 1-2 bài/1 câu hỏi để mọi người hỗ trợ nhanh nhất nha!

6:

a: \(\Leftrightarrow\dfrac{\sqrt{4}-\sqrt{1}}{4-1}+\dfrac{\sqrt{7}-\sqrt{4}}{7-4}+...+\dfrac{\sqrt{3n+4}-\sqrt{3n+1}}{3}=8\)

=>\(-\sqrt{1}+\sqrt{4}-\sqrt{4}+\sqrt{7}-...-\sqrt{3n+1}+\sqrt{3n+4}=24\)

=>\(\sqrt{3n+4}=24+1=25\)

=>3n+4=625

=>3n=621

=>n=207

b: \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\cdot\sqrt{n+1}}=\dfrac{4}{5}\)

=>\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

=>\(1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

=>n+1=25

=>n=24

Câu 5: 

a: Ta có: \(A=\left(x-1\right)\left(x-3\right)+11\)

\(=x^2-4x+3+11\)

\(=x^2-4x+4+10\)

\(=\left(x-2\right)^2+10\ge10\forall x\)

Dấu '=' xảy ra khi x=2

b: Ta có: \(B=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Câu 5:

a) \(A=\left(x-1\right)\left(x-3\right)+11=x^2-4x+3+11\)

\(=x^2-4x+14\)

\(=\left(x^2-4x+4\right)+10=\left(x-2\right)^2+10\ge10\)

\(minA=10\Leftrightarrow x=2\)

b) \(B=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

mn giúp em vs ạ em cần gấp cảm ơn ạ 

Em đang cần gấp ạ

Câu 2: 

a: Ta có: \(25x^2-9=0\)

\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b: Ta có: \(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)

\(\Leftrightarrow x^2-8x+16-x^2+4=6\)

\(\Leftrightarrow-8x=-14\)

hay \(x=\dfrac{7}{4}\)

c: Ta có: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow x=-\dfrac{255}{2}\)

2A:

a: \(a^3+12a^2+48a+64\)

\(=a^3+3\cdot a^2\cdot4+3\cdot a\cdot4^2+4^3\)

\(=\left(a+4\right)^3\)

b: \(-b^3+6b^2-12b+8\)

\(=\left(-b\right)^3+3\cdot\left(-b\right)^2\cdot2+3\cdot\left(-b\right)\cdot2^2+2^3=\left(-b+2\right)^3\)

c: \(\left(m-n\right)^6-6\left(m-n\right)^4+12\left(m-n\right)^2-8\)

\(=\left\lbrack\left(m-n\right)^2\right\rbrack^3-3\cdot\left\lbrack\left(m-n\right)^2\right\rbrack^2\cdot2+3\cdot\left(m-n\right)^2\cdot2^2-2^3\)

\(=\left\lbrack\left(m-n\right)^2-2\right\rbrack^3\)

d: \(\frac{8}{27}a^3-\frac83a^2b+8b^2a-8b^3\)

\(=\left(\frac23a\right)^3-3\cdot\left(\frac23a\right)^2\cdot2b+3\cdot\left(\frac23a\right)\cdot\left(2b\right)^2-\left(2b\right)^3\)

\(=\left(\frac23a-2b\right)^3\)

Bài 2B:

a: \(\frac18x^3+\frac34x^2y^2+\frac32xy^4+y^6\)

\(=\left(\frac12x\right)^3+3\cdot\left(\frac12x\right)^2\cdot y^2+3\cdot\frac12x\cdot\left(y^2\right)^2+\left(y^2\right)^3\)

\(=\left(\frac12x+y^2\right)^3\)

b: \(m^3+9m^2n+27mn^2+27n^3\)

\(=m^3+3\cdot m^2\cdot\left(3n\right)+3\cdot m\cdot\left(3n\right)^2+\left(3n\right)^3\)

\(=\left(m+3n\right)^3\)

c: \(8u^3-48u^2v+96uv^2-64v^3\)

\(=\left(2u\right)^3-3\cdot\left(2u\right)^2\cdot4v+3\cdot\left(2u\right)\cdot\left(4v\right)^2-\left(4v\right)^3\)

\(=\left(2u-3v\right)^3\)

d: \(\left(z-t\right)^3+15\left(z-t\right)^2+75\left(z-t\right)+125\)

\(=\left(z-t\right)^3+3\cdot\left(z-t\right)^2\cdot5+3\cdot\left(z-t\right)\cdot5^2+5^3=\left(z-t+5\right)^3\)