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Áp dụng tính chất tỉ lệ thức và tính chất dãy tỉ số bằng nhau:
\(ay^2=bx^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)
\(\Rightarrow\left(\dfrac{x^2}{a}\right)^{1000}=\left(\dfrac{y^2}{b}\right)^{1000}=\dfrac{1}{\left(a+b\right)^{1000}}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)
\(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\Leftrightarrow\dfrac{2x^2}{18}=\dfrac{y^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x^2}{18}=\dfrac{y^2}{16}=\dfrac{2x^2+y^2}{18+16}=\dfrac{136}{34}=4\)
Suy ra: \(\left\{{}\begin{matrix}x^2=4.9=36\\y^2=4.16=64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm6\\y=\pm8\end{matrix}\right.\)
2) Ta có: \(2^{20}=\left(2^4\right)^5=16^5\)
Được biết số có tận cùng là \(6\) thì lũy thừa bao nhiêu cũng bằng \(6\)
Nên \(16^5=\overline{...6}\Leftrightarrow16^5-1=\overline{.....5}⋮5\)
Nên \(\dfrac{2^{20}-1}{5}\) là số nguyên
3)
Ta có:
\(A=100^2+200^2+...+1000^2\)
\(A=\left(1.100\right)^2+\left(2.100\right)^2+...+\left(10.100\right)^2\)
\(A=1^2.100^2+2^2.100^2+....+10^2.100^2\)
\(A=100^2\left(1^2+2^2+...+100^2\right)\)
\(A=10000.385=3850000\)
1. a) (x-2)2 =1
=> x - 2 = \(\pm\sqrt{1}\)
=> x - 2 = 1 hoặc -1
=> x = 3 hoặc 1
b) 2x - 1= -8
=> 2x = -7
=>x = \(\dfrac{-7}{2}\)
c)thiếu đề
d) (x-1)x+2 = (x-1)x+4
(x-1)x+2 = (x-1)x+2+2
(x-1)x+2 = (x-1)x+2. (x-1)2
(x-1)x+2 - (x-1)x+2. (x-1)2 = 0
=> (x-1)x+2. [1 - (x-1)2] = 0
\(\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2a) \(\dfrac{45^{10}.5^{10}}{75^{10}}\) = \(\dfrac{\left(3.3.5\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\dfrac{3^{10}.3^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(3^{10}\)
b) \(\dfrac{2^{15}.9^4}{6^6.8^3}\)=\(\dfrac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}\)=\(\dfrac{2^{15}.3^8}{2^6.3^6.2^9}\)=\(3^2\)
a)\(VT=\left(-\dfrac{1}{8}\right)^{100}=\dfrac{1}{8^{100}}=\dfrac{1}{\left(2^3\right)^{100}}=\dfrac{1}{2^{300}}\)
\(VP=\left(-\dfrac{1}{4}\right)^{200}=\dfrac{1}{\left(2^2\right)^{200}}=\dfrac{1}{2^{400}}\)
\(\Rightarrow VT>VP\)
b) \(VT=4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}=VP\)
c) \(VT=5^{2000}=\left(5^2\right)^{1000}=25^{1000}>10^{1000}=VP\)
d) \(VT=31^5< 32^5=\left(2^5\right)^5=2^{25}\)
\(VP=17^7>16^7=\left(2^4\right)^7=2^{28}\)
\(VP>VT\)
Bài 2:
\(2^{91}\) và \(5^{35}\)
Ta có:
\(2^{91}=\left(2^{13}\right)^7\) \(=8192^7\)
\(5^{35}=\left(5^5\right)^7\) =\(3125^7\)
Vì 8192\(^7\) >3125\(^7\) nên \(2^{91}>5^{35}\)
Bài 3:
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)
VT=\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(=\dfrac{a^2-2ab+b^2}{c^2-2cd+d^2}\)
Mới biết làm đến đó thôi à!
2)
\(2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
Vì \(8192>3125\)
Nên \(8192^7>3125^7\)
Vậy \(2^{91}>2^{35}\)
1. Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) \(\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) \(\left(2\right)\)
Từ \(\left(1\right)\text{và (2)}\) \(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
2. \(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0\)
\(\text{Mà }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\\dfrac{2}{7}y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\\dfrac{2}{7}x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
3. \(\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}\)
\(\text{Mà }a-b=15\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}a=60\\b=45\\c=40\end{matrix}\right.\)
\(bx^2=ay^2\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)
\(\Rightarrow\left(\dfrac{x^2}{a}\right)^{1000}=\left(\dfrac{y^2}{b}\right)^{1000}=\left(\dfrac{1}{a+b}\right)^{1000}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}=\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{1}{\left(a+b\right)^{1000}}+\dfrac{1}{\left(a+b\right)^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)