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a) Ta có:
\(90.10^k-10^{k+2}+10^{k+1}\)
\(=90.10^k-10^k.10^2+10^k.10\)
\(=10^k\left(90-10^2+10\right)\)
\(=10^k.0=0\)
b) Ta có:
\(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
\(=2,5.10.5^{n-3}+5^n-6.5^{n-1}\)
\(=5.5.5^{n-3}+5^n-6.5^{n-1}\)
\(=5^2.5^{n-3}+5^n-6.5^{n-1}\)
\(=5^{n-3+2}+5^n-6.5^{n-1}\)
\(=5^{n-1}\left(1+5-6\right)\)
\(=5^{n-1}.0=0\)
a) Rút gọn biểu thức:
\(90\times10^k-10^{k+2}+10^{k+1}=90\times10^k-10^k\times10^2+10^k\times10\) \(=10^k\times\left(90-10^2+10\right)\) \(=10^k\times\left(90-100+10\right)\) \(=10^k\times0=0\)
b) Rút gọn biểu thức:
\(2,5\times5^{n-3}\times10+5^n-6\times5^{n-1}=2,5\times\dfrac{5^n}{5^3}\times10+5^n-6\times\dfrac{5^n}{5}\) \(=2,5\times\dfrac{5^n}{125}\times10+5^n-\dfrac{6}{5}\times5^n\) \(=0,2\times5^n+5^n-1,2\times5^n\) \(=5^n\times\left(0,2+1-1,2\right)=5^n\times0=0\)
\(a, 10^{n+1} -6.10 ^n\)
= \(10^n (10-6)=4.10^n\)
\(B/ 2^{n+3} + 2^{n+2} - 2^{n+1} +2^n\)
= \(2^n (2^3+2^2-2+1)\)
= \(2^n (8+4-2+1)\)
\(= 11.2^n\)
\(C/ 90.10^k - 10^{k +2} + 10^{k +1} \)
\(= 10^k(90-2+1)\)
= \(89.10^k\)
\(D/ 2,5 . 5^{n-3} . 10+5^n -6 .5^{n-1}\)
\(= 5.5.5^{n-3} +5^n-6.5^{n-1}\)
= \(5^2 .5^{n-3}+5^n-6.5^{n-1} \)
= \(5^{n-3+2}+5^n -6.5^{n-1}\)
\(= 5^{n-1}(1+5-6)\)
= \(5^{n-1}.0\)
= 0
a) \(10^{n+1}-6.10^n\)
\(=10^n.10-6.19^n\)
\(=10^n.\left(10-6\right)\)
\(=10^n.4\)
b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)
\(=2^n.2^3+2^n.2^2-2^n.2+2^n.1\)
\(=2^n.\left(2^3+2^2-2+1\right)\)
\(=2^n.11\)
c) \(90.10^k-10^{k+2}+10^{k+1}\)
\(=90.10^k-10^k.10^2+10^k.10\)
\(=10^k.\left(90-10^2+10\right)\)
\(=0\)
d) \(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
\(=\dfrac{2,5.5^n.10}{5^3}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n}{5}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n+5^{n+1}-6.5^n}{5}=\dfrac{5^n+5^n.5-6.5^n}{5}=\dfrac{5^n\left(1+5-6\right)}{5}=\dfrac{0}{5}=0\)
\(d,2,5.5^{n-3}.2.5+5^n-6.5^{n-1}=5.5.5^{n-3}+5^n-6.5^{n-1}=5^2.5^{n-3}+5^n-6.5^{n-1}\)
\(=5^{n-3+2}+5^n-6.5^{n-1}=5^{n-1}\left(1+5-6\right)=5^{n-1}.0=0\)
a, \(10^{n+1}-6.10^n=10^n\left(10-6\right)=4.10^n\)
b. \(2^{n+3}+2^{n+2}-2^{n+1}+2^n=2^n\left(2^3+2^2-2+1\right)=2^n\left(8+4-2+1\right)=11.2^n\)
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn
a: \(10^{n+1}=10^n\cdot10\)
b: \(2^{n+3}+2^{n+1}-2^{n+1}+2^n\)
\(=2^n\cdot8+2^n=9\cdot2^n\)
c: \(90\cdot10^k-10^{k+2}+10^{k+1}\)
\(=90\cdot10^k+10^k\cdot10-10^k\cdot100=0\)
a: \(=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2}\cdot\dfrac{1}{4}=\dfrac{1}{100}\)
b: \(=\dfrac{\left[5^3\left(5-1\right)\right]^3}{5^{12}}=\dfrac{5^9}{5^{12}}\cdot\dfrac{4^3}{1}=\dfrac{4^3}{5^3}\)
c: \(=\sqrt{1.8^2}=1.8\)
1.
\(10x=|x+\dfrac{1}{10}|+|x+\dfrac{2}{10}|+...+|x+\dfrac{9}{10}| \ge 0\)
\(\Rightarrow x\ge0\)
\(pt\Leftrightarrow x+\frac{1}{10}+x+\frac{2}{10}+...+x+\frac{9}{10}=10x\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{2}{10}+...+\frac{9}{10}=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}\)
4.
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{b+3c}=\frac{b}{c+3a}=\frac{c}{a+3b}=\frac{a+b+c}{4\left(a+b+c\right)}=\frac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}4a=b+3c\left(1\right)\\4b=c+3a\left(2\right)\\4c=a+3b\left(3\right)\end{matrix}\right.\)
Từ \(\left(1\right);\left(2\right)\Rightarrow4a=b+3\left(4b-3a\right)\)
\(\Rightarrow12a=12b\Rightarrow a=b\left(4\right)\)
Từ \(\left(1\right);\left(3\right)\Rightarrow4c=a+3\left(4a-3c\right)\)
\(\Rightarrow12a=12c\Rightarrow a=c\left(5\right)\)
Từ \(\left(4\right);\left(5\right)\Rightarrow a=b=c\left(đpcm\right)\)
Câu 3:
a: Theo đề ta có: P(2014)=0
\(\Leftrightarrow2014^2-2014k+2014=0\)
=>4058210-2014k=0
=>k=2015
Vậy: \(P\left(x\right)=x^2-2015x+2014\)
b: \(P\left(1\right)=1-2015+2014=0\)
nên x=1 là nghiệm của P(x)
1a) \(10^{n+1}-6\cdot10^n\)
\(=10^n\cdot10-6\cdot10^n\)
= \(10^n\left(10-6\right)\)
\(=10^n\cdot4\)
b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)
\(=2^n\cdot2^3+2^n\cdot2^2-2^n\cdot2+2^n\)
\(=2^n\left(2^3+2^2-2+1\right)\)
\(=2^n\cdot11\)
c) \(90\cdot10^k-10^{k+2}+10^{k+1}\)
\(=90\cdot10^k-10^k\cdot10^2+10^k\cdot10\)
\(=10^k\left(90-10^2+10\right)=0\)
d) \(2,5\cdot5^{n-3}\cdot10+5^n-6\cdot5^{n-1}\)
\(=\dfrac{2,5\cdot10\cdot5^n}{5^3}+5^n-\dfrac{6\cdot5^n}{5}\)
\(=\dfrac{5^n}{5}+5^n-\dfrac{6\cdot5^n}{5}\)
\(=\dfrac{5^n+5^n\cdot5-6\cdot5^n}{5}=\dfrac{5^n\left(5-6\right)+5^n}{5}=0\)
2. \(M+\left(6x^2-4xy\right)=7x^2-8xy+y^2\)
\(M=\left(7x^2-8xy+y^2\right)-\left(6x^2-4xy\right)\)
\(M=7x^2-8xy+y^2-6x^2+4xy\)
\(M=7x^2-6x^2-8xy+4xy+y^2\)
\(M=x^2-4xy+y^2\)
Mk cảm ơn bn nhiều lắm ạ Lê Mỹ Linh