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1.\(45^{10}.5^{30}=45^{10}.125^{10}=\left(45.125\right)^{10}=5625^{10}\)
2.a. \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
c. \(\left(2x+3\right)^2=\frac{9}{121}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
d.\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow x=\frac{1}{9}\)
4.
a.\(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Do \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b.\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)
\(\Rightarrow3^{4000}=9^{2000}\)
c.\(2^{332}=\left(2^3\right)^{110}.2^2=8^{110}.4\)
\(3^{223}=\left(3^2\right)^{110}.3^3=\left(3^2\right)^{110}.9=9^{110}.9\)
Ta thấy \(4.8^{110}< 9.9^{110}\)
Vậy \(2^{332}< 3^{223}\)
trắc nghiệm
câu 1: c
câu 2: B
câu 3: D
câu 4: A
câu 5: C
câu 6: D
tự luận
câu 1:
a)M(x) = x4 + 2x2 + 1
b) M(x) + N(x) = -4x4 + x3 + 5x2 - 2
M(x) - N(x) = 6x4 - x3 - x2 + 4
c) \(M\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^4+2\left(-\dfrac{1}{2}\right)^2+1=\dfrac{25}{16}\)
a: \(=\dfrac{2^6\cdot3^3}{3^{-4}\cdot2^6}=\dfrac{3^3}{3^{-4}}=3^7\)
c: \(=5^4\cdot5^3\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{4}{100}\)
\(=5^7\cdot\left(\dfrac{2}{5}\right)^5\cdot\left(\dfrac{1}{5}\right)^2\)
\(=5^2\cdot\left(\dfrac{1}{5}\right)^2\cdot5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
\(a, 10^{n+1} -6.10 ^n\)
= \(10^n (10-6)=4.10^n\)
\(B/ 2^{n+3} + 2^{n+2} - 2^{n+1} +2^n\)
= \(2^n (2^3+2^2-2+1)\)
= \(2^n (8+4-2+1)\)
\(= 11.2^n\)
\(C/ 90.10^k - 10^{k +2} + 10^{k +1} \)
\(= 10^k(90-2+1)\)
= \(89.10^k\)
\(D/ 2,5 . 5^{n-3} . 10+5^n -6 .5^{n-1}\)
\(= 5.5.5^{n-3} +5^n-6.5^{n-1}\)
= \(5^2 .5^{n-3}+5^n-6.5^{n-1} \)
= \(5^{n-3+2}+5^n -6.5^{n-1}\)
\(= 5^{n-1}(1+5-6)\)
= \(5^{n-1}.0\)
= 0
1/ 106=(5x2)6=56x26=56x64=>106-57=56x(64-5)=56x59. Vậy ta có điều phải chứng minh
Ta có:\(2^n+2^{n+1}+2^{n+2}+...+2^{n+m}=2^{n+m+1}-2^n\)
Áp dụng:
\(A=1+2+2^2+...+2^{30}=2^{31}-1\)
\(\Rightarrow A+1=2^{31}\)