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ADTCDTSBN, ta có:
\(\frac{\left(x-1\right)}{2}=\frac{2\left(y-2\right)}{6}=\frac{3\left(z-3\right)}{12}\)\(=\frac{\left(x-2y+3z-6\right)}{20}=\frac{\left(14-6\right)}{8}\)=1
=>\(\hept{\begin{cases}x-1=2\\2y-4=6\\3z-9=12\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=3\\y=5\\z=7\end{cases}}\)
Ta có :
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
\(\Rightarrow\)\(\frac{x-1}{2}=\frac{2y-4}{6}\)\(=\frac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\frac{x-1}{2}=\frac{2y-4}{6}\)\(=\frac{3z-9}{12}\)\(=\)\(\frac{\left(x-1\right)-\left(2y-4\right)+3z-9}{2-6+12}\)
\(=\frac{\left(x-y+z\right)+\left(-1+4-9\right)}{8}\)
\(=\frac{14-6}{8}=1\)
\(\Rightarrow\)x - 1 = 2
y - 2 = 3
z - 3 = 4
\(\Rightarrow\)x = 3
y = 5
z = 7
\(0,1\left(2\right)=\frac{12-1}{90}=\frac{11}{90}\)
\(0,\left(27\right)=\frac{27}{99}=\frac{3}{11}\)
\(3,\left(42\right)=3+\frac{42}{99}=\frac{113}{33}\)
\(3,\left(45\right)=3+\frac{45}{99}=\frac{38}{11}\)
a, Xét tam giác AMB và tam giác AMC ta có :
MA _ chung
BA = AC ( gt )
MB = MC ( gt )
Vậy tam giác AMB = tam giác AMC ( c.c.c )
b, Xét tam giác NMB và tam giác NMC ta có :
MN _ chung
NB = NC ( N là trung điểm BC )
BM = MC ( gt )
Vậy tam giác NMB = tam giác NMC ( c.c.c )
a) Xét \(\Delta AMB\) và \(\Delta AMC\) có
AM chung
MB = MC ( giả thiết )
AB = AC ( giả thiết )
Nên \(\Delta AMB=\Delta AMC\left(c.c.c\right)\)
b) Xét \(\Delta NMB\) và \(\Delta NMC\) có
NM chung
NB = NC ( vì N là trung điểm của BC )
MB = MC ( giả thiết )
Nên \(\Delta NMB=\Delta NMC\left(c.c.c\right)\)
Đặt \(x=3k;y=7k\)
Ta có : \(xy=42\Rightarrow21k^2=42\Leftrightarrow k^2=2\Leftrightarrow k=\pm\sqrt{2}\)
Với \(k=\sqrt{2}\)thì \(x=3\sqrt{2};y=7\sqrt{2}\)
Với \(k=-\sqrt{2}\)thì \(x=-3\sqrt{2};y=-7\sqrt{2}\)
\(\frac{41.42...81}{1.3...79}=\frac{42.44....81}{1.3....39}=\frac{2^{20}.21.22...40.81}{1.3...39}=\frac{2^{30}.11.12...20.81}{1.3...19}=\frac{2^{35}.6.7...10}{1.3.5.7.9}=\frac{2^{38}.3.4.5}{1.3.5}=240^{ }\)
\(\frac{2^{38}.3.4.5}{1.3.5}=2^{40}=\left(2^{20}\right)^2\)là số cp
suy ra C là số cp
Ta có:
C=41.42.43...79.80.811.3.5.7...77.79C=41.42.43...79.80.811.3.5.7...77.79
→C=1.2.3...80.811.2.3...401.3.5.7...77.79→C=1.2.3...80.811.2.3...401.3.5.7...77.79
→C=1.2.3...80.81.(2.4.6..80)1.2.3...40(2.4.6...80).1.3.5.7...77.79→C=1.2.3...80.81.(2.4.6..80)1.2.3...40(2.4.6...80).1.3.5.7...77.79
→C=1.2.3...80.81.240.(1.23...40)1.2.3...401.2.3...79.80→C=1.2.3...80.81.240.(1.23...40)1.2.3...401.2.3...79.80
→C=81.240→C=81.240
→C=92.(220)2→C=92.(220)2
→C=(9.220)2→C=(9.220)2
→C→C là số chính phương
\(f=\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right):\left(\frac{30}{26}+\frac{31}{27}+...+\frac{54}{50}-25\right)\)
\(f=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right):\left(\frac{30}{26}-1+\frac{31}{27}-1+...+\frac{54}{50}-1\right)\)
\(f=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):\left(\frac{4}{26}+\frac{4}{27}+...+\frac{4}{50}\right)\)
\(f=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):4\cdot\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)\)
\(f=4\)
tại sao cái ngoặc thứ nhất như thế thì xem b4
1/1.2+1/3.4+...+1/49.50=1-1/2+1/3-1/4+...+1/49-1/50=1+1/2+1/3+...+1/50-2(1/2+1/4+...+1/50)=1+1/2+1/3+...+1/50-(1+1/2+...+1/25)=1/26+1/27+...+1/50
30/26+31/27+...+54/50-25=(30/26-1)+(31/27-1)+...+(54/50-1)=4/26+4/27+...+4/50
với 30/26+31/27+...+54/50-25=4/26+4/27+...+4/50 và 1/1.2+1/3.4+...+1/49.50=1/26+1/27+...+1/50 ta có:
F=(1/26+1/27+...+1/50 ):(4/26+4/27+...+4/50)=1/26+1/27+...+1/50/4(1/26+1/27+...+1/50)=4
vậy F=4
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1008}\)
\(S=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}\)
Vậy \(S=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}\)
b4 :
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1008}\)
\(S=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}\)
có \(\left(S-P\right)^{2017}=\left(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2017}-\frac{1}{1009}-\frac{1}{1010}-...-\frac{1}{2017}\right)^{2017}=0\)
