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e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
Bài 1 : \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
Bài 2 : a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)=> \(x^8=x^7\)
=> \(x^8-x^7=0\)
=> \(x^7\left(x-1\right)=0\)
=> \(x-1=0\Rightarrow x=1\)(vì x7 = 0 => x = 0 mà x \(\ne\)0 nên loại)
b) \(x^{10}-25x^8=0\)
=> \(x^8\left(x^2-25\right)=0\)
=> x8 = 0 hoặc x2 - 25 = 0
=> x = 0 hoặc x2 = 25
=> x = 0 hoặc x = \(\pm\)5
Bài 3 : a) \(\left(2x+3\right)^2=\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)
=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
=> 3x - 1 = -2/3
=> 3x = 1/3
=> x = 1/3 : 3 = 1/9
1) Ta có \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{30}+1\right)}=2^{10}=1024\)
2) a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)
=> x8 = x7
=> x8 - x7 = 0
=> x7(x - 1) = 0
=> \(\orbr{\begin{cases}x^7=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x \(\in\left\{0;1\right\}\)
b) x10 = 25x8
=> x10 - 25x8 = 0
=> x8(x2 - 25) = 0
=> \(\orbr{\begin{cases}x^8=0\\x^2-25=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
Vậy \(x\in\left\{0;5;-5\right\}\)
3) \(\left(2x+3\right)^2=\frac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)
=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{-30}{11}\\2x=-\frac{36}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
=> \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-1=-\frac{2}{3}\)
=> \(3x=\frac{1}{3}\)
=> \(x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
\(a,\left(x+1\right)^2=81\)
\(\left(x+1\right)^2=9^2\) Hoặc \(\left(x+1\right)^2=\left(-9\right)^2\)
\(\left(x+1\right)=9\) \(x+1=-9\)
\(x=8\) \(x=-10\)
b,\(\left(x+5\right)^{^{ }3}=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(x+5=-4\)
=> \(x=-9\)
c,\(\left(2x-3\right)^2=9\)
=>\(\left(2x-3\right)^2=3^2\)Hoặc \(\left(2x-3\right)^2=\left(-3\right)^2\)
\(2x-3=3\) \(2x-3=-3\)
\(2x=6\) \(2x=0\)
=> \(\hept{\begin{cases}x=3\\x=0\end{cases}}\)
d, \(\left(4x+1\right)^3=27\)
\(\left(4x+1\right)^{^{ }3}=3^3\)
\(4x+1=3\)
\(4x=2\)
\(x=\frac{1}{2}\)
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{8^6}{4}=\frac{\left(2^3\right)^6}{2^2}=\frac{2^{18}}{2^2}=2^{16}\)
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{15}+4^{10}}{4^6+4^{11}}=\frac{4^{10}.4^5+4^{10}}{4^6+4^6.4^5}=\frac{4^{10}.\left(4^5+1\right)}{4^6.\left(4^5+1\right)}=\frac{4^{10}}{4^6}=4^4=256\)
phần D trên mk làm sai xin lỗi nha
a)
\((3x-7)^5=0\Rightarrow 3x-7=0\Rightarrow x=\frac{7}{3}\)
b)
\(\frac{1}{4}-(2x-1)^2=0\)
\(\Leftrightarrow (2x-1)^2=\frac{1}{4}=(\frac{1}{2})^2=(-\frac{1}{2})^2\)
\(\Rightarrow \left[\begin{matrix} 2x-1=\frac{1}{2}\\ 2x-1=\frac{-1}{2}\end{matrix}\right.\Rightarrow \Rightarrow \left[\begin{matrix} x=\frac{3}{4}\\ x=\frac{1}{4}\end{matrix}\right.\)
c)
\(\frac{1}{16}-(5-x)^3=\frac{31}{64}\)
\(\Leftrightarrow (5-x)^3=\frac{1}{16}-\frac{31}{64}=\frac{-27}{64}=(\frac{-3}{4})^3\)
\(\Leftrightarrow 5-x=\frac{-3}{4}\)
\(\Leftrightarrow x=\frac{23}{4}\)
d)
\(2x=(3,8)^3:(-3,8)^2=(3,8)^3:(3,8)^2=3,8\)
\(\Rightarrow x=3,8:2=1,9\)
e)
\((\frac{27}{64})^9.x=(\frac{-3}{4})^{32}\)
\(\Leftrightarrow [(\frac{3}{4})^3]^9.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow (\frac{3}{4})^{27}.x=(\frac{3}{4})^{32}\)
\(\Leftrightarrow x=(\frac{3}{4})^{32}:(\frac{3}{4})^{27}=(\frac{3}{4})^5\)
f)
\(5^{(x+5)(x^2-4)}=1\)
\(\Leftrightarrow (x+5)(x^2-4)=0\)
\(\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x+5=0\\ x^2=4=2^2=(-2)^2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=-5\\ x=\pm 2\end{matrix}\right.\)
g)
\((x-2,5)^2=\frac{4}{9}=(\frac{2}{3})^2=(\frac{-2}{3})^2\)
\(\Rightarrow \left[\begin{matrix} x-2,5=\frac{2}{3}\\ x-2,5=\frac{-2}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{19}{6}\\ x=\frac{11}{6}\end{matrix}\right.\)
h)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
a/ \(\Rightarrow x^{10}-25x^8=0\Rightarrow x^8\left(x^2-25\right)=0\)
\(\Rightarrow x^8=0\Rightarrow x=0\)
hoặc \(x^2-25=0\Rightarrow x=5;x=-5\)
Vậy x = 0 ; x = 5; x = -5
b/ \(\Rightarrow2x+3=\frac{3}{11}\Rightarrow2x=-\frac{30}{11}\Rightarrow x=-\frac{15}{11}\)
hoặc \(2x+3=-\frac{3}{11}\Rightarrow2x=-\frac{36}{11}\Rightarrow x=-\frac{18}{11}\)
Vậy x = -15/11 ; x = -18/11
c/ \(\Rightarrow\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\Rightarrow3x-1=-\frac{2}{3}\Rightarrow3x=\frac{1}{3}\Rightarrow x=\frac{1}{9}\)
Vậy x = 1/9
a, x10 = 25.x8
=> x2 = 25 = 52 = (-5)2
=> x = + 5
b, \(\left(2x+3\right)^2=\frac{9}{121}=\left(\frac{3}{11}\right)^2=\left(-\frac{3}{11}\right)^2\)
=> 2x + 3 = + \(\frac{3}{11}\)
TH1: 2x + 3 = \(\frac{3}{11}\)
=> 2x = \(\frac{-30}{11}\)
=> x = \(\frac{-15}{11}\)
TH2: 2x + 3 = \(\frac{-3}{11}\)
=> 2x = \(\frac{-36}{11}\)
=> x = \(\frac{-18}{11}\)
(3x - 1)3 = \(\frac{-8}{27}\) = \(\left(\frac{-2}{3}\right)^3\)
=> 3x - 1 = \(\frac{-2}{3}\)
=> 3x = \(\frac{1}{3}\)
=> x = \(\frac{1}{9}\)