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Trường hợp 1: x<-2
Phương trình sẽ là:
-2x-4-2x+3=7
=>-4x-1=7
=>-4x=8
hay x=-2(loại)
Trường hợp 2: -2<=x<3/2
Pt sẽ là:
2x+4-2x+3=7
=>7=7(luôn đúng)
Trường hợp 3: x>=3/2
Pt sẽ là 2x-3+2x+4=7
=>4x+1=7
hay x=3/2(nhận)
Vậy: -2<=x<=3/2
\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)
\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)
\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)
\(\Rightarrow7x=25x\)
\(\Rightarrow x=0\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\)
\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)
\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)
a. 0,34-2x=1,2
=> 2x=0,34-1,2
=> 2x=-0,86
=> x=-0,86:2
=> x=-0,43
b. \(\frac{3}{2x+1}=\frac{4}{7}\)
=> \(2x+1=\frac{3.7}{4}\)
=> 2x+1=5,25
=> 2x=4,25
=> x=4,25:2
=> x=2,125
c. \(\frac{x-3}{2}=\frac{5-2x}{3}\)
=> (x-3).3=(5-2x).2
=> 3x-9=10-4x
=> 3x+4x=10+9
=> 7x=19
=> x=19:7
=> x=19/7
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
\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{7^x.\left(7^2+7+1\right)}{57}=7^x\)
\(\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}=\frac{5^{2x}\left(1+5+5^3\right)}{131}=\frac{25^x.131}{131}=25^x\)
\(\Rightarrow7^x=25^x\Rightarrow x=0\)
\(\frac{x+1}{x-2}=\frac{3}{4}\) ( \(ĐKXĐ\) : \(x\ne2\) )
\(\Leftrightarrow\left(x+1\right).4=\left(x-2\right).3\)
\(\Leftrightarrow4x+4=3x-6\)
\(\Leftrightarrow4x-3x=-6-4\)
\(\Leftrightarrow x=-10\)
b ) \(\frac{2x-3}{x+1}=\frac{4}{7}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow7\left(2x-3\right)=4\left(x+1\right)\)
\(\Leftrightarrow14x-21=4x+4\)
\(\Leftrightarrow10x=25\)
\(\Leftrightarrow x=\frac{5}{2}\)
|2x - 3| + |2x + 4| = 7
+ Với \(x< -2\) thì |2x - 3| = -(2x - 3) = -2x + 3; |2x + 4| = -(2x + 4) = -2x - 4
Ta có: -2x + 3 - 2x - 4 = 7
=> -4x - 1 = 7
=> -4x = 7 + 1
=> -4x = 8
=> x = 8 : (-4)
=> x = -2, không thỏa mãn x < -2
+ Với \(-2\le x< \frac{3}{2}\) thì |2x - 3| = 3 - 2x; |2x + 4| = 2x + 4
Ta có:
3 - 2x + 2x + 4 = 7
=> 7 = 7, đúng
+ Với \(x\ge\frac{3}{2}\) thì |2x - 3| = 2x - 3; |2x + 4| = 2x + 4
Ta có: 2x - 3 + 2x + 4 = 7
=> 4x + 1 = 7
=> 4x = 7 - 1
=> 4x = 6
=> \(x=\frac{6}{4}=\frac{3}{2}\), thỏa mãn \(x\ge\frac{3}{2}\)
Vậy \(-2\le x\le\frac{3}{2}\)
sai