Tìm x:
a,\(15-\left(5-2x\right)=-4\)
b,\(\left|x-3\right|+1=4\)
c,\(5^x\cdot5^{x+1}\cdot5^{x+2}=10000000\div2^{18}\)
1. Tìm x
a) \(2^x+5=21\)
b) \(2^x-1+3^2=5^2+2\cdot5\)
c) \(\left(2x-1\right)^3+5=130\)
d) \(5^{2x-3}-2\cdot5^2=5^2\)
e) \(3^{2x+1}-2=3^2+\left[5^2-3\left(2^2-1\right)\right]\)
f) \(\left(7^x-11\right)^3=2^5\cdot5^2+200\)
g) \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
a) \(2^x+5=21\)
\(\Rightarrow2^x=21-5=16\Rightarrow2^x=2^4\)
Vậy x = 4
b) \(2^x-1+3^2=5^2+2.5\)
\(\Rightarrow2^x-1+9=35\)
\(\Rightarrow2^x=35-9+1=27\)
Vậy x không có giá trị
c;d;e;f làm tương tự
Tìm x:
\(\frac{\left(13\frac{2}{9}-15\frac{2}{3}\right)\cdot\left(30^2-5^4\right)}{\left(18\frac{3}{7}-17\frac{1}{4}\right)\cdot\left(25-12\cdot5^2\right)}\cdot x=\frac{\frac{2}{11}+\frac{3}{13}+\frac{4}{15}+\frac{5}{17}}{4\frac{1}{11}+\frac{5}{13}+\frac{9}{15}+\frac{13}{17}}\)
A=\(\dfrac{2^{30}\cdot5^7+3^{13}\cdot5^{27}}{2^{27}\cdot5^7+2^{10}\cdot5^{27}}\)
M=\(\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+6\right)^{\left(x+5\right)}}}}\) tại x=7
a: \(=\dfrac{2^{13}\cdot5^7\left(2^{17}+5^{20}\right)}{2^{10}\cdot5^7\left(2^{17}+5^{20}\right)}=2^3\)
b: \(M=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)
\(=3^{2\cdot1\cdot13\cdot12}=3^{312}\)
Bài 1: Tìm n
1/4*2/6*3/8*4/10*5/12*.....* 30/62*31/64=2^n
Bài 2: Tính
a) \(\frac{\left(-5\right)^{60}\cdot30^5}{15^5\cdot5^{61}}\)
b)\(2^3+3\cdot\left(\frac{1}{3}\right)^0-2^{-2}\cdot4+\left[\left(-2\right)^3:\frac{1}{2}\right]\cdot8\)
(GỢI Ý CÂU B:\(^{2^{-2}=\frac{1}{2^2}}\)
Bài 3: Tìm x
a)\(^{2^x=16}\)
b)\(^{\left(\frac{x}{13}\right)^2=\frac{49}{169}}\)
c)\(\left(\frac{-1}{5}\right)^x=\left(\frac{-1}{125}\right)^3\cdot x^4=\frac{-16}{625}\)
d)\(^{6^{4-x}=216}\)
Bài 4:Tìm n
a)\(3^n\cdot3^{-2}=3^5\)
b)Tìm x:
1)\(\frac{2^{4-x}}{16^5}=32^6\)
2)\(^{9\cdot5^x=6\cdot5^6+3\cdot5^6}\)
3)\(\frac{2^3}{2^x}=4^5\)
Các bạn giúp mik!!!Mik sắp kiểm tra 45p r
Tìm x:
1) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
2) \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
3) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
1) ( 2x -15 )5 = ( 2x - 15 )3
( 2x -15 )5 - ( 2x - 15 )3 = 0
( 2x - 15 )3 . [ ( 2x - 15 )2 - 1 ] = 0
\(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}}\)
\(\orbr{\begin{cases}2x=15\\2x=16\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}\)
1: \(15-\left\{2\cdot\left[x-\left(2x-4\right)\cdot5\right]\cdot3\cdot\left(x+1\right)\right\}=12-x\)
\(15-\left\{2.\left[\left(2x-4\right).5\right].3.\left(x+1\right)\right\}=12-x\)
\(15-\left\{\left[10x-20\right].6.\left(x+1\right)\right\}=12-x\)
\(15-\left\{10x-20.6x+1\right\}=12-x\)
\(15-\left\{10x-120x+1\right\}=12-x\)
\(15-\left(-110x\right)-1=12-x\)
\(15+110x-1=12-x\)
\(110x+x=12-15+1\)
\(111x=-2\)
\(x=\dfrac{-2}{111}\)
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
Tìm số đo góc nhọn x:
a) \(4\sin x-1=1\)
b) \(2\sqrt{3}-3\tan x=\sqrt{3}\)
c) \(7\sin-3\cos\left(90^o-x\right)=2,5\)
d) \(\left(2\sin-\sqrt{2}\right)\left(4\cos-5\right)=0\)
e) \(\dfrac{1}{\cos^2x}-\tan x=1\)
f) \(\cos^2x-3\sin^2x=0,19\)
a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow x=30^o\)
b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)
\(\Leftrightarrow x=30^o\)
c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)
d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)
Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(
e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)
f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)
Tìm x: a,\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
b,\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
a, Ta có \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
<=> \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)
<=> \(\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)-\left(\frac{x-3}{2009}-1\right)-\left(\frac{x-4}{2008}-1\right)=0\)
<=>\(\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> \(x-2012=0=>x=2012\)
b, \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
=>\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2x-1\right)\left(2x+1\right)}=2\cdot\frac{49}{99}\)
=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(1-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(\frac{2x}{2x+1}=\frac{98}{99}\)
=>2x = 98
=>x = 49