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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
\(a,x\cdot\frac{1}{2}\cdot\frac{2}{3}=4\)
\(\Rightarrow x\cdot\frac{1}{3}=4\)
\(\Rightarrow x=12\)
\(b,-\frac{2}{7}\cdot\frac{5}{7}\cdot x=\frac{7}{21}\)
\(\Rightarrow-\frac{10}{49}x=\frac{7}{21}\)
\(\Rightarrow x=-\frac{49}{30}\)
k đi làm tiếp cho
Ta có: 1/2 - (1/3 + 1/4) = 1/2 - 7/12 = -1/12 ;
1/48 - (1/16 - 1/6) = 1/48 + 5/48 = 1/8
Vì \(-\frac{1}{12}< x< \frac{1}{8}\) nên x = 0
b) \(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)\)Ta có :
\(4\frac{5}{9}:2\frac{5}{18}-7=2-7=-5\)
\(\left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(-21\frac{2}{3}\right)=\left(1+\frac{38}{5}\right):\left(-21\frac{2}{3}\right)=\frac{43}{5}:\frac{-65}{3}=-\frac{129}{325}\)
Vì \(-5< x< -\frac{129}{325}\) nên \(x\in\left\{-4;-3;-2;-1\right\}\)
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
mik ko bít
I don't now
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b, ta có : x/3 = y/5 -> x/6 = y/10 ; y/2 = z/4 -> y/10 = z/20 . suy ra : x/6 = y/10 = z/20
áp dụng dãy tỉ số bằng nhau ta có : x/6 = y/10 = z/20 = 2x + y - z / 12 + 10 - 20 = 16 / 2 = 8
suy ra : x/6 = 8 -> x = 48
y = 80
z = 160
a, \(\frac{3}{x-1}\) = \(\frac{21}{16}\)
=> 3 . 16 = 21 . ( x - 1 ) = 48
x - 1 = 48 : 21 = \(\frac{16}{7}\)
x = \(\frac{16}{7}\) + 1 = \(\frac{23}{7}\)
b, \(\frac{2x-3}{x-2}\) = \(\frac{3}{4}\)
=> 4 . ( 2x - 3 ) = 3 . ( x - 2 )
=> 8x - 12 = 3x - 6
=> 8x -12 - 3x + 6 = 0
= 5x - 6 = 0; 5x =6
=> x = \(\frac{6}{5}\)