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3/ ta để ý thấy ở số mũ sẽ có thừa số 1000-103=0
nên số mũ chắc chắn bằng 0
mà số nào mũ 0 cũng bằng 1 nên A=1
5/ vì |2/3x-1/6|> hoặc = 0
nên A nhỏ nhất khi |2/3x-6|=0
=>A=-1/3
6/ =>14x=10y=>x=10/14y
23x:2y=23x-y=256=28
=>3x-y=8
=>3.10/4y-y=8
=>6,5y=8
=>y=16/13
=>x=10/14y=10/14.16/13=80/91
8/106-57=56.26-56.5=56(26-5)=59.56
có chứa thừa số 59 nên chia hết 59
4/ tính x
sau đó thế vào tinh y,z
Câu 2: A = \(^{1+2+2^2+2^{ }^3+...+2^{2017}}\)
2A = \(2+2^2+2^3+...+2^{2018}\)
Suy ra 2A - A =\(2^{2018}-1\) Do đó A < B
1. Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=t\Rightarrow a=2016t,b=2017t,c=2018t\)
\(\left(a-c\right)^3=\left(2016t-2018t\right)^3=\left(-2t\right)^3=-8t^3\)
\(8\left(a-b\right)^2\left(b-c\right)=8\left(2016t-2017t\right)^2\left(2017t-2018t\right)=8.\left(-t\right)^2.\left(-t\right)=-8t^3\)
Vậy \(\left(a-c\right)^3=8\left(a-b\right)^2\left(b-c\right)\)
a) ( x - 1/5 )2 = 0
<=> x - 1/5 = 0
<=> x = 1/5
b) ( x - 2 )2 = 1
<=> ( x - 2 )2 = ( ±1 )2
<=> x - 2 = 1 hoặc x - 2 = -1
<=> x = 3 hoặc x = 1
c) ( 2x - 1 )3 = -8
<=> ( 2x - 1 )3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = -1/2
d) ( x4 )2 = x12/x5
<=> x8 = x7
<=> x8 - x7 = 0
<=> x7( x - 1 ) = 0
<=> x7 = 0 hoặc x - 1 = 0
<=> x = 0 hoặc x = 1
e) x10 = 25x8
<=> x10 - 25x8 = 0
<=> x8( x2 - 25 ) = 0
<=> x8 = 0 hoặc x2 - 25 = 0
<=> x = 0 hoặc x = ±5
f) ( 2x + 3 )2 = 9/121
<=> ( 2x + 3 )2 = ( ±3/11 )2
<=> 2x + 3 = 3/11 hoặc 2x + 3 = -3/11
<=> x = -15/11 hoặc x = -18/11
a) \(\left(x-\frac{1}{5}\right)^2=0\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3+8=0\)
\(\Leftrightarrow\left(2x-1+8\right)\left[\left(2x-1\right)^2-8\left(2x-1\right)+64\right]=0\)
\(\Leftrightarrow2x+7=0\)
\(\Leftrightarrow x=\frac{-7}{2}\)
d) ĐKXĐ : \(x\ne0\)
\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)
\(\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\)
\(\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=1\left(tm\right)\end{cases}\Leftrightarrow x=1}\)
e) ĐKXĐ : x khác 0
\(x^{10}=25x^8\)
\(\Leftrightarrow x^2=25\Leftrightarrow x=5\)
f) \(\left(2x+3\right)^2=\frac{9}{121}\)
\(\Leftrightarrow\left(2x+3+\frac{3}{11}\right)\left(2x+3-\frac{3}{11}\right)=0\)
\(\Leftrightarrow\left(2x+\frac{36}{11}\right)\left(2x+\frac{30}{11}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-18}{11}\\x=-\frac{15}{11}\end{cases}}\)
a.
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
b.
\(\left(x-2\right)^2=1\)
\(x-2=\pm1\)
TH1:
\(x-2=1\)
\(x=1+2\)
\(x=3\)
TH2:
\(x-2=-1\)
\(x=-1+2\)
\(x=1\)
Vậy x = 3 hoặc x = 1
c.
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d.
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\pm\frac{1}{4}\)
TH1:
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{1}{4}\)
TH2:
\(x+\frac{1}{2}=-\frac{1}{4}\)
\(x=-\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{3}{4}\)
Vậy \(x=-\frac{1}{4}\) hoặc \(x=-\frac{3}{4}\)
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
Câu 2a đánh thiếu đề rồi : I x+1I + I x+2I + I x+3 I = x
2c)
Ta có: \(25-y^2\le25\Rightarrow8\left(x-2012\right)^2\le25\)
\(\Rightarrow\left(x-2012\right)^2\le3\)
\(\Rightarrow\left[\begin{matrix}\left(x-2012\right)^2=0\\\left(x-2012\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x-2012=0\\\left[\begin{matrix}x-2012=1\\x-2012=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=2012\\\left[\begin{matrix}x=2013\\x=2011\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}y=5\\\left[\begin{matrix}y=\sqrt{17}\\y=\sqrt{17}\end{matrix}\right.\end{matrix}\right.\)(loại)
Vậy x=2012,y=5
a)Viết dưới dạng phân số rồi sử dụng tích chéo ý
b)\(\frac{-1}{7}.2^3-2x:1\frac{4}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-2x:\frac{7}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-\frac{6x}{7}=-2^{x-1}\)
\(\Rightarrow\frac{-8-6x}{7}=\frac{2^{x-1}}{-1}\)
\(\Rightarrow-1\left(-8-6x\right)=7.2^{x-1}\)
\(\Rightarrow6x+8=7.2^{x-1}\)
.........
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)
\((2x+\frac{1}{3})^3=\frac{8}{27}=(\frac{2}{3})^3\)
\(\Rightarrow 2x+\frac{1}{3}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
1) a) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)
b) \(x^3=x\Leftrightarrow x^3-x=0\Leftrightarrow x\left(x^2-1\right)=0\) \(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\) vậy \(x=0;x=-1;x=1\)
\(x^2=2x\Rightarrow x^2-2x=0\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)
\(x^3=x\Rightarrow x^3-x=0\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\left(\dfrac{1}{25}-1\right)...\left(\dfrac{1}{121}-1\right)\)
\(A=\dfrac{-3}{4}.\dfrac{-8}{9}.\dfrac{-15}{16}.\dfrac{-24}{25}...\dfrac{-120}{121}\)
\(A=\dfrac{3.8.15.24....120}{4.9.16.25...121}\)
\(A=\dfrac{1.3.2.4.3.5.4.6....10.12}{2.2.3.3.4.4.5.5....11.11}\)
\(A=\dfrac{1.2.4....10}{2.3.4.5...11}.\dfrac{3.4.5....12}{2.3.4.5....11}\)
\(A=\dfrac{1}{11}.6=\dfrac{6}{11}\)
3) Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{8^{2017}+1}{8^{2018}+1}< 1\)
\(B< \dfrac{8^{2017}+1+8}{8^{2018}+1+8}\)
\(B< \dfrac{8^{2017}+8}{8^{2018}+8}\)
\(B< \dfrac{8\left(8^{2016}+1\right)}{8\left(8^{2017}+1\right)}\)
\(B< \dfrac{8^{2016}+1}{8^{2017}+1}=A\)
\(B< A\)