\(\frac{4k}{4k^4+1}=\frac{4k}{4k^4+4k^2+1-4k^2}=\frac{4k}{\left(2k^2+1\right)^2-\left(2k\right)^2}=\frac{4k}{\left(2k^2+2k+1\right)\left(2k^2-2k+1\right)}=\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1}\)

\(=\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)

\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)

\(=1-\frac{1}{2k\left(k+1\right)+1}=...\)

Ta có: \(4n^4+1=\left(4n^4+4n^2+1\right)-4n^2=\left(2n^2+2n+1\right)\left(2n^2-2n+1\right)\)

\(\frac{4n}{4n^4+1}=\frac{\left(2n^2+2n+1\right)-\left(2n^2-2n+1\right)}{\left(2n^2-2n+1\right)\left(2n^2+2n+1\right)}=\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}\)

Thay vào ta có: 

\(\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^2+1}+...+\frac{4n}{4n^4+1}=\frac{220}{221}\)

\(\Leftrightarrow1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)

\(\Leftrightarrow1-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)

\(\Leftrightarrow\frac{2n^2+2n}{2n^2+2n+1}=\frac{220}{221}\Rightarrow n=10\)

a) \(-1\frac{1}{15}:2\frac{1}{2}=-\frac{16}{15}:\frac{5}{2}=-\frac{16}{15}\cdot\frac{2}{5}=-\frac{32}{75}\)

b) \(\frac{1}{3}-\frac{5}{14}\cdot\frac{21}{25}=\frac{1}{3}-\frac{1}{2}\cdot\frac{3}{5}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)

c) \(\frac{-5}{7}\cdot\frac{2}{11}+\frac{-5}{7}\cdot\frac{9}{11}+1\frac{5}{7}\)

\(=-\frac{5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)

\(=-\frac{5}{7}\cdot1+\frac{12}{7}=-\frac{5}{7}+\frac{12}{7}=\frac{7}{7}=1\)

d) \(8\frac{1}{4}-\left(2\frac{5}{9}+3\frac{1}{4}\right)=8\frac{1}{4}-2\frac{5}{9}-3\frac{1}{4}=\left(8\frac{1}{4}-3\frac{1}{4}\right)-2\frac{5}{9}\)

\(=5-2\frac{5}{9}=5-\frac{23}{9}=\frac{22}{9}\)

e) \(\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-25=\frac{1}{4}\left(\frac{12}{13}+\frac{1}{13}\right)-25=\frac{1}{4}\cdot1-25=\frac{1}{4}-25=-\frac{99}{4}\)

bn ơi sao câu b ở đâu ra 1/2 vậy

Trương Bùi Linh Đây :

\(\frac{5}{14}.\frac{21}{25}=\frac{5}{2.7}.\frac{3.7}{5.5}=\frac{1}{2}.\frac{3}{5}\)

\(M=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{62}+\frac{1}{63}\)

\(M=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)\)

\(M< 1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32\)

\(M< 1+1+1+1+1+1\)

\(M< 1.6=6\left(đpcm\right)\)

(1+19)+(2+18)+(3+17)+(4+16)+(5+15)+(6+14)+(7+13)+(8+12)+(9+11)+10+20=110
mấy câu còn lại k tính nhanh được
chúc bạn hk giỏi

a.190

b.121

c.132

câu kia mình làm sai rồi

a.210

b.121

c,132

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

     \(B=1+7+7^2+...+7^{63}\)

Nhận thấy từ số hạng thứ 2 của B đều chia hết cho 7,  còn 1 chia 7 dư 1

nên  B chia 7 dư 1

    \(B=1+7+7^2+....+7^{63}\)

\(=1+\left(7+7^2+7^3\right)+\left(7^4+7^5+7^6\right)+...+\left(7^{61}+7^{62}+7^{63}\right)\)

\(=1+7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{61}\left(1+7+7^2\right)\)

\(=1+\left(1+7+7^2\right)\left(7+7^4+...+7^{61}\right)\)

\(=1+57\left(7+7^4+...+7^{61}\right)\)

Ta thấy   \(57\left(7+7^4+...+7^{61}\right)⋮57\)

nên  B chia 57 dư 1