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a, \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(\Rightarrow x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(\Rightarrow x=\dfrac{-1}{6}\)
b, \(25-\left(5-x\right)=-7\)
\(\Rightarrow\) \(5-x=25-\left(-7\right)\)
\(\Rightarrow5-x=32\)
\(\Rightarrow x=5-32\)
\(\Rightarrow x=-27\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{-5}{7}\)
d, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e, \(\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|-7=-3\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=-3+7\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4+\dfrac{3}{4}\\\dfrac{1}{2}x=-4+\dfrac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{19}{4}\\\dfrac{1}{2}x=\dfrac{-13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{4}:\dfrac{1}{2}\\x=\dfrac{-13}{4}:\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=\dfrac{-13}{2}\end{matrix}\right.\)
a)\(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(x=\dfrac{-1}{6}\)
b)\(25-\left(5-x\right)=-7\)
\(5-x=25-\left(-7\right)\)
\(5-x=32\)
x= -27
c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
d)\(2x\left(x-\dfrac{1}{7}\right)=0\)
⇒\(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e)\(|\dfrac{1}{2}x-\dfrac{3}{7}|-7=-3\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=-3+7\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=4\)
⇒\(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4\dfrac{3}{4}\Rightarrow x=9\dfrac{1}{2}=\dfrac{19}{2}\\\dfrac{1}{2}x=-3\dfrac{1}{4}\Rightarrow x=\dfrac{-13}{2}\end{matrix}\right.\)
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
Nhiều quá, từng bài 1 nhé, bài nào làm được, tớ sẽ cố gắng.
bài 2:
a) \(x>2x\Leftrightarrow x-2x>0\Leftrightarrow-x>0\Leftrightarrow x< 0\)
Kl: x<0
b) \(a+x< a\Leftrightarrow x< 0\)
Kl: x<0
c) \(x^3>x^2\Leftrightarrow x^3-x^2>0\Leftrightarrow x^2\left(x-1\right)>0\) (*)
Mà x^2 > 0 \(\Rightarrow\) (*) \(\Leftrightarrow x-1>0\Leftrightarrow x>1\)
Kl: x>1
Câu 4:
a) \(1-2x< 7\Leftrightarrow2x>-6\Leftrightarrow x>3\)
Kl: x>3
b) \(\left(x-1\right)\left(x-2\right)>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< 2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)
Kl: x>2 hoặc x<1
c) \(\left(x-2\right)^2\left(x+1\right)\left(x+4\right)< 0\Leftrightarrow\left(x+1\right)\left(x+4\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x+4< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x+4>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x< -4\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x>-4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-1< x< -4\left(vô-lý\right)\\-4< x< -1\end{matrix}\right.\) \(\Leftrightarrow-4< x< -1\)
Kl: -4<x<-1
d) ĐK: x khác 9\(\dfrac{x^2\left(x+3\right)}{x-9}< 0\Leftrightarrow x^2\left(x+3\right)\left(x-9\right)< 0\Leftrightarrow\left(x+3\right)\left(x-9\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x-9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x-9>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x>9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3< x< 9\left(N\right)\\9< x< -3\left(vô-lý\right)\end{matrix}\right.\) \(\Leftrightarrow-3< x< 9\)
Kl: -3<x<9
e) Đk: x khác 0
\(\dfrac{5}{x}< 1\Leftrightarrow\dfrac{5}{x}< \dfrac{5}{5}\Leftrightarrow x>5\left(N\right)\)
KL: x >5
f) ĐK: x khác 1
\(\dfrac{2x-5}{x-1}< 0\Leftrightarrow\left(2x-5\right)\left(x-1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5>0\\x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5< 0\\x-1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x>1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{2}< x< 1\left(vô-lý\right)\\1< x< \dfrac{5}{2}\left(N\right)\end{matrix}\right.\)
Kl: 1< x< 5/2
a ) \(\frac{-3}{7}+x=\frac{1}{3}\Rightarrow x=\frac{1}{3}-\frac{-3}{7}\Rightarrow x=\frac{16}{21}\)
b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\Rightarrow\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\Rightarrow\frac{1}{4}:x=-\frac{7}{20}\Rightarrow x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)
c) \(1\frac{1}{3}:0,8=\frac{2}{3}:0,1x\Rightarrow\frac{2}{3}:0,1x=\frac{5}{3}\Rightarrow0,1x=\frac{2}{5}\Rightarrow x=4\)
d) \(\left|x-3\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}x-3=\frac{1}{2}\\x-3=-\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{5}{2}\end{cases}}}\)
e) \(1\frac{1}{2}.x-4=0,5\Rightarrow\frac{3}{2}x=4,5\Rightarrow x=3\)
g) \(2^{x-1}=16\Rightarrow2^{x-1}=2^4\Rightarrow x-1=4\Rightarrow x=5\)
h) \(\left(x-1\right)^2=25\Rightarrow\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=-4\end{cases}}}\)
i) \(\left|2x-1\right|=5\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
k) \(0,2-\left|4,2-2x\right|=0\Rightarrow\left|4,2-2x\right|=0,2\Rightarrow\orbr{\begin{cases}4,2-2x=0,2\\4,2-2x=-0,2\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=2,2\end{cases}}}\)
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
a: =>1/6x=-49/60
=>x=-49/60:1/6=-49/60*6=-49/10
b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2
=>x=17/15 hoặc x=-13/15
c: =>1,25-4/5x=-5
=>4/5x=1,25+5=6,25
=>x=125/16
d: =>2^x*17=544
=>2^x=32
=>x=5
i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5
=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2
=>x=14,4 hoặc x=9,6
j: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
b: \(\dfrac{2x+3}{3-x}\le0\)
\(\Leftrightarrow\dfrac{2x+3}{x-3}\ge0\)
=>x>3 hoặc x<=-3/2
c: \(\dfrac{x+5}{x+3}>1\)
\(\Leftrightarrow\dfrac{x+5-x-3}{x+3}>0\)
=>2/(x+3)>0
=>x+3>0
hay x>-3
a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)
\(5^3.21=5^3.3.7⋮7\) (đpcm)
b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)
\(=7^4.55=7^4.5.11⋮11\) (đpcm)
c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)
\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)
\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)
d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)
\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)
a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)
\(\)\(\RightarrowĐPCM\)
b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)
\(\Rightarrowđpcm\)
a: \(\Leftrightarrow4^{x-5}\cdot17=68\)
=>4^x-5=4
=>x-5=1
=>x=6
b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)
=>|2x-1|=1/3
=>2x-1=1/3 hoặc 2x-1=-1/3
=>x=2/3 hoặc x=1/3
c: =>|2x-2|=|3x+15|
=>3x+15=2x-2 hoặc 3x+15=-2x+2
=>x=-17 hoặc x=-13/5