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a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)
\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3^x-9\right)=0\)
=>x=2 hoặc x=-3/2
b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)
=>2x+5=1/44 hoặc 2x+1=-1/44
=>x=-219/88 hoặc x=-221/88
mình làm lại câu b) nha
b) |x-3|=-4
th1: x-3=-4
x=3+(-4)
x=-1
th2: x-3=4
x=3+4
x=7
b) \(\left|x-3\right|=-4\)
t/h1:\(x-3=-4\)
\(x=3-\left(-4\right)\)
\(x=7\)
t/h2:\(x-3=4\)
\(x=3-4\)
\(x=-1\)
\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)
\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)
\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)
\(a,\left(x-1,2\right)^2=4\)
\(\Rightarrow x-1,2=2\)
\(\Rightarrow x=3,2\)
\(b,\left(x+1\right)^3=-125\)
\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Rightarrow x+1=-5\Rightarrow x=-6\)
\(c,\left(x-5\right)^3=2^6\)
\(\Rightarrow\left(x-5\right)^3=4^3\)
\(\Rightarrow x-5=4\Rightarrow x=9\)
\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
|2x-1|=1,5
TH(1)2x-1=1,5
2x =1,5+1
2x =2,5
x =2,5 :2
x =1,25
TH(2) 2x-1=-1,5
2x =-1,5+1
2x =-0,5
x =-0,5:2
x =-0,25
các câu khác cứ tương tự bạn nhé
b) \(7,5-\left|5-2x\right|=-4,5\)
\(\left|5-2x\right|=7,5+4,7\)
\(\left|5-2x\right|=12\)
th1 :\(5-2x=12\)
\(2x=5-12\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
th2: \(5-2x=-12\)
\(2x=5+12\)
\(2x=17\)
\(x=17:2\)
\(x=8,5\)
c) \(-3+\left|x\right|=-1\)
\(\left|x\right|=-1+3\)
\(\left|x\right|=2\)
th1: \(x=-2\)
th2 : \(x=2\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)
th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)
\(x=\dfrac{7}{3}-\dfrac{1}{2}\)
\(x=\dfrac{11}{6}\)
th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)
\(x=\dfrac{7}{3}+\dfrac{1}{6}\)
\(x=\dfrac{-5}{2}\)
e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{9}{14}\)
th1 :\(x+1=\dfrac{9}{14}\)
\(x=\dfrac{9}{14}-1\)
\(x=\dfrac{-5}{14}\)
th2 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
Bài 1:
a) \(x^2-3=1\)
\(\Rightarrow x^2=1+3=4\)
\(\Rightarrow x=\pm2\)
b)\(2x^3+12=-4\)
\(\Rightarrow2x^3=-4-12=-16\)
\(\Rightarrow x^3=-8\)
\(\Rightarrow x=-2\)
c)\(\left(2x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)
a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)
b) \(2x^3+12=-4\Rightarrow2x^3=-16\)
\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)
\(\Rightarrow x=-2\)
c) \(\left(2x-3\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d,h,i,k cững tương tự....
a) \(\frac{4}{x+5}=\frac{3}{2x-1}\)
=> 4(2x - 1) = 3(x + 5)
=> 8x - 4 = 3x + 15
=> 8x - 3x = 15 + 4
=> 5x = 19
=> x = 19/5
b) \(\frac{x+11}{19}+\frac{x+12}{20}+\frac{x+13}{21}=3\)
=> \(\left(\frac{x+11}{19}-1\right)+\left(\frac{x+12}{20}-1\right)+\left(\frac{x+13}{21}-1\right)=0\)
=> \(\frac{x-8}{19}+\frac{x-8}{20}+\frac{x-8}{21}=0\)
=> \(\left(x-8\right)\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)
=> x - 8 = 0
=> x = 8
c) \(\left(2x-1\right)^2=\left(2x-1\right)^3\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^3=0\)
=> \(\left(2x-1\right)^2.\left[1-\left(2x-1\right)\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\1-2x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
a) 4/x + 3 = 3/2x - 1
<=> 4.(2x - 1) = (x + 3).3
<=> 8x - 4 = 3x + 9
<=> 8x = 3x + 9 + 4
<=> 8x = 3x + 13
<=> 8x - 3x = 13
<=> 5x = 13
<=> x = 13/5
=> x = 13/5
c) (2x - 1)2 = (2x - 1)3
<=> 4x2 - 4x + 1 = 8x3 - 12x2 + 6x - 1
<=> 8x3 - 12x2 + 6x - 1 = 4x2 - 4x + 1
<=> 8x3 - 12x2 + 6x - 1 - 1 = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x - 4x = 4x2
<=> 8x3 - 12x2 + 10x - 2 = 4x2
<=> 8x3 - 12x2 + 10x - 2 - 4x2 = 0
<=> 8x2 - 16x2 + 10x - 2 = 0
<=> 2(x - 1)(2x - 1)2 = 0
<=> x - 1 = 0 hoặc 2x - 1 = 0
x = 0 + 1 2x = 0 + 1
x = 1 2x = 1
x = 1/2
=> x = 1 hoặc x = 1/2
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
\(A\left(x\right)-B\left(x\right)=2x^3+4x^2-x-1-\left(2x^3-2x^2-x-3\right)\\ =2x^3+4x^2-x-1-2x^3+2x^2+x+3\\ =\left(2x^3-2x^3\right)+\left(4x^2+2x^2\right)+\left(x-x\right)+\left(3-1\right)=6x^2+2\)
A(x)+B(x)
=2x^3+4x^2-x-1+2x^3-2x^2-x-3
=4x^3+2x^2-2x-4