Tìm x biết : (\(\frac{4}{9}\) )x = ( \(\frac{8}{27}\))10
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\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^{10}\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{30}\)
\(\Leftrightarrow2x=30\Leftrightarrow x=15\)
Bài làm
Ta có: \(\left(\frac{4}{9}\right)^x=\left(\frac{2}{3}\right)^{2x}\)
\(\left(\frac{8}{27}\right)^{10}=\left(\frac{2}{3}\right)^{3.10}=\left(\frac{2}{3}\right)^{30}\)
\(\Rightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{30}\)
\(\Rightarrow2x=30\)
\(\Rightarrow x=15\)
Vậy \(x=15\)
# Học tốt #
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
a) \(\left(2x+3\right)^2=\frac{9}{144}\)
\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+3=\frac{1}{4}\\2x+3=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{-11}{4}\\2x=\frac{-13}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-11}{8}\\x=\frac{-13}{8}\end{cases}}}\)
Vậy ...
b) Ta có: \(\left(3x-1\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)
Vậy ....
c) \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25\Leftrightarrow x=\left\{5;-5\right\}\)
Vậy ...
d) \(\frac{x^7}{81}=27\Leftrightarrow x^7=27.81=2187\)
Mà 37 = 2187 => x7 = 37 => x = 3
Vậy ....
e) \(\frac{x^8}{9}=729\Leftrightarrow x^8=729.9=6561\)
Mà 38 = (-3)8 = 6561
=> x8 = 38 = (-3)8
=> x = {-3;3}
Vậy ...
Lời giải: Giải phương trình với tập xác định
Tập xác định của phương trình
\(x\in\infty-\infty\)
\(\frac{19x+67}{90}=\frac{15x+83}{56}\Rightarrow\left(19x=67\right)56=90\left(15x+83\right)\)
Kết quả : \(-13\)
kq đúng nhưng mk k biết mấy cái phương trình đó vì mk mới lớp 7
a) \(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\Leftrightarrow\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Leftrightarrow2x=18\)
\(\Leftrightarrow x=9\)
b) \(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^{22}\)
\(\Leftrightarrow\left(\frac{1}{9}\right)^x=\left(\frac{1}{3}\right)^{66}\)
\(\Leftrightarrow x=66\)
\(\Leftrightarrow\frac{x+4}{9}+\frac{x+11}{8}+\frac{x+16}{7}+\frac{x+19}{6}=10\)
\(\Leftrightarrow\left(\frac{x+4}{9}-1\right)+\left(\frac{x+11}{8}-2\right)+\left(\frac{x+16}{7}-3\right)+\left(\frac{x+19}{6}-4\right)=0\)
\(\Leftrightarrow\frac{x+4-9}{9}+\frac{x+11-16}{8}+\frac{x+16-21}{7}+\frac{x+19-24}{6}=0\)
\(\Leftrightarrow\frac{x-5}{9}+\frac{x-5}{8}+\frac{x-5}{7}+\frac{x-5}{6}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6}\right)=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
V...
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
(x+2)/17+(x+4)/15+(x+6)/13=(x+8)/11+(x+10)/9+(x+12)/7
=>(x+2+17)/17+(x+4+15)/15+(x+6+13)/13=(x+8+11)/11+(x+10+9)/9+(x+12+7)/7
=>(x+19)/17+(x+19)/15+(x+19)/13=(x+19)/11+(x+19)/9+(x+19)/7
=>(x+19)/17+(x+19)/15+(x+19)/13-(x+19)/11-(x+19)/9-(x+19)/7=0
=>(x+19)*(1/17+1/15+1/13-1/11-1/9-1/7)=0
=>x+19=0
=>x=19
áp dụng tc tỉ lệ thức ta có :
\(\Leftrightarrow\frac{671x+2804}{3315}=\frac{239x+2462}{693}\Rightarrow\left(671x+2804\right)693=3315\left(239x+2462\right)\)
=>(671x+2804)693=693(671x+2804) (VT)
<=>693(671x+2804)=3315(239x+2462)
=>465003x+1943172=792285x+8161530
=>-327282x=621835
=>x=621835:(-327282)
=>x=-19
\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^{10}\)
\(\left(\frac{2^2}{3^2}\right)^x=\left(\frac{2^3}{3^3}\right)^{10}\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{30}\)
\(\Rightarrow2x=30\)
\(\Rightarrow x=15\)
vay \(x=15\)
(4/9)x=(8/27)10
=> [(2/3)2 ]x=[(2/3)3 ]10
=> (2/3)2.x=(2/3)3.10
=> 2.x=3.10
=> 2x=30
=> x=30:2
=> x=15