cmr:
1/4.1+1/4.2+1/4.3+...+1/7.9+1/8.0>7/12
3,1+1/2+1/3+1/4+...+1/62+1/63>6
giúp mk nhé đang cần gấp
Rút gọn: \(A=\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^4+1}+\frac{4.3}{4.3^4+1}+...+\frac{4.k}{4.k^4+1}\)
\(\frac{4k}{4k^4+1}=\frac{4k}{4k^4+4k^2+1-4k^2}=\frac{4k}{\left(2k^2+1\right)^2-\left(2k\right)^2}=\frac{4k}{\left(2k^2+2k+1\right)\left(2k^2-2k+1\right)}=\frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1}\)
\(=\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2k\left(k-1\right)+1}-\frac{1}{2k\left(k+1\right)+1}\)
\(=1-\frac{1}{2k\left(k+1\right)+1}=...\)
tìm số nguyên dương n thỏa mãn: \(\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^4+1}+\frac{4.3}{4.3^4+1}+...+\frac{4n}{4n^4+1}=\frac{220}{221}\)
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:
\(\frac{4.1}{4.1^4+1}+\frac{4.2}{4.2^2+1}+...+\frac{4n}{4n^4+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+...+\frac{1}{2n^2-2n+1}-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow1-\frac{1}{2n^2+2n+1}=\frac{220}{221}\)
\(\Leftrightarrow\frac{2n^2+2n}{2n^2+2n+1}=\frac{220}{221}\Rightarrow n=10\)
CMR : 3< 1+1/2 +1/3 +...+1/63 < 6
GIúp mình với mình đang cần gấp !
Thực hiện phép tính:
a) -1 và 1/15 : 2 và 1/2 b)1/3-5/14.21/25 c)-5/7. 2/11+ âm 5/7.9/11+1 và 5/7
e)8 và 1/4-(2 và 5/9+3 và 1/4)
f) 1/4.12/13+1/4.1/13-25
LM NHANH GIÚP MK NHÉ!
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 :
\(\frac{5}{14}.\frac{21}{25}=\frac{5}{2.7}.\frac{3.7}{5.5}=\frac{1}{2}.\frac{3}{5}\)
Chứng minh rằng M < 6 biết:
M = \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{62}+\frac{1}{63}\)
Giúp mk nhé Mai.
\(M=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{62}+\frac{1}{63}\)
\(M=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)\)
\(M< 1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32\)
\(M< 1+1+1+1+1+1\)
\(M< 1.6=6\left(đpcm\right)\)
Tính nhanh :
a,1+2+3+.....+20 =
b.1+3+5+7+.....+21=
c. 2+4+6+.....+22=
Giúp mk nhé mk đang cần gấp
(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
Các bạn giúp mình nhé , mk đang cần gấp , bạn nào giải hộ , mk sẽ tick đều đặn. Help me
1.Cmr : A=9/10!+9/11!+9/12!+...+9/1000! < 1/9
2. CHo G = 5/3+8/3^2+11/3^3+...+302/3^100. CMR : 23/9<G<7/2
3.so sánh : L =(1-1/2)(1-1/3)...(1-1/20) với 1/21
4.C=1/101+1/102+...+1/200. CMR:
a/ C>7/12
b//C>5/8
5 cho C = 1/11+1/12+...+1/13+...+1/70
CMR : 4/3<C<2,5
6. Cho B = 4/3+10/9+28/27+...+399/398 . CMR B< 100
3) tìm x biết
a) \(\sqrt{x+9}=7\)
b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\)
c) \(\sqrt{x^2-6x+9}=2x+1\)
d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\)
lm nhanh giúp mk nhé mk đang cần gấp
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)
cho B =1+7+7^2+......+7^63.CMR B chia 7,57 du 1
Giúp mình với mình đang cần gấp
\(B=1+7+7^2+...+7^{63}\)
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