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|5x-3| - 3x = 7
*Nếu \(x\ge\frac{3}{5}\)
5x - 3 - 3x = 7
2x = 10
x = 5 ( tm)
*Nếu \(x< \frac{3}{5}\)
3 - 5x - 3x = 7
-8x = 4
x = \(-\frac{1}{2}\)( tm )
Làm hơi khó nhìn , thông cảm. Mệt rùi :)
|x - 3| + |x - 5| - 4x = -28
*Nếu x < 3
3 - x + 5 - x - 4x = -28
-6x = -36
x = 6 ( loại do ko tm khoảng đang xét )
* nếu 3 < x < 5
x - 3 + 5 - x - 4x = -28
-4x = -30
x= \(\frac{15}{2}\) ( loại do ko tm khaongr đang xét )
*Nếu x > 5
x - 3 + x - 5 - 4x = -28
-2x = -20
x = 10 ( tm)
Vậy x =10
\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)
\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)
\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)
\(\Rightarrow7x=25x\)
\(\Rightarrow x=0\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\)
\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)
\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)
vì \(\left|x+2\right|\ge0\forall x\)
\(\left|x+\frac{3}{5}\right|\ge0\forall x\)
\(\left|x+\frac{1}{2}\right|\ge0\forall x\)
\(\Rightarrow\left|x+2\right|+\left|x+\frac{3}{5}\right|+\left|x+\frac{1}{2}\right|\ge0\forall x\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x+2+x+\frac{3}{5}+x+\frac{1}{2}=4x\)
\(3x+3.1=4x\)
x = 3.1
vậy x = 3.1
Vì \(|x+2|\ge0;\forall x\);\(|x+\frac{3}{5}|\ge0;\forall x\); \(|x+\frac{1}{4}|\ge0;\forall x\)
suy ra \(|x+2|+\)\(|x+\frac{3}{5}|+\)\(|x+\frac{1}{4}|\ge0;\forall x\)
suy ra \(4x\ge0;\forall x\Rightarrow x\ge0,\forall x\)
Với \(x\ge0\)
suy ra \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)
suy ra \(3x+\frac{31}{10}=4x\)
suy ra \(\frac{31}{10}=x\)(thỏa mãn)
Vậy x= 31/10
Theo đề bài ta có: \(4x-3y=5\)
\(\frac{x-1}{2}=\frac{y-2}{3}\Rightarrow\frac{4x-4}{8}=\frac{3y-6}{9}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{4x-4}{8}=\frac{3y-6}{9}=\frac{4x-4-3y+6}{-1}=\frac{\left(4x-3y\right)+\left(-4+6\right)}{-1}=\frac{5+2}{-1}=-7\)
\(\Rightarrow\begin{cases}\frac{x-1}{2}=-7\rightarrow x=\left(-7\right)\cdot2+1=-13\\\frac{y-2}{3}=-7\rightarrow y=\left(-7\right)\cdot3+2=-19\end{cases}\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{4\left(x-1\right)}{2.4}=\frac{3.\left(y-2\right)}{3.3}=\frac{4x-4}{8}=\frac{3y-6}{9}=\frac{4x-4-3y+6}{8-9}=\frac{\left(4x-3y\right)-\left(4-6\right)}{-1}\)
\(=\frac{5-\left(-2\right)}{-1}=\frac{7}{-1}=-7\)
+) \(\frac{x-1}{2}=-7\Rightarrow x-1=-14\Rightarrow x=-13\)
+) \(\frac{y-2}{3}=-7\Rightarrow y-2=-21\Rightarrow y=-19\)
Vậy cặp số \(\left(x;y\right)\) là \(\left(-13;-19\right)\)
\(a)\) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}\Leftrightarrow\orbr{\begin{cases}4x-\frac{3}{2}x=\frac{1}{2}+1\\\frac{3}{2}x+4x=1-\frac{1}{2}\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}}\)
Vậy \(x=\frac{3}{5}\) hoặc \(x=\frac{1}{11}\)
Chúc bạn hojc tốt ~
\(b)\) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x-\frac{7}{2}\right|=0\)
\(\Leftrightarrow\)\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x-\frac{7}{2}\right|\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x-\frac{7}{2}\\\frac{5}{4}x-\frac{7}{2}=\frac{7}{2}-\frac{5}{8}x\end{cases}\Leftrightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{5}{8}x=-\frac{7}{2}+\frac{7}{2}\\\frac{5}{4}x+\frac{5}{8}x=\frac{7}{2}+\frac{7}{2}\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\frac{5}{8}x=0\\\frac{15}{8}x=7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{56}{15}\end{cases}}}\)
Vậy \(x=0\)\(x=\frac{56}{15}\)
Chúc bạn học tốt ~
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
a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)
=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)
=> \(\left|2-\frac{3}{2}x\right|=x+6\)
ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)
Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)
=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)
=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)
b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)
=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)
=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)
=> x = 1/4
hoặc x = 0 hoặc x = 1/2
\(\frac{x+2}{4x-1}=\frac{x-5}{4x+1}\) ( đkxđ : \(x\ne\pm\frac{1}{4}\))
<=> \(\frac{\left(x+2\right)\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{\left(x-5\right)\left(4x-1\right)}{\left(4x-1\right)\left(4x+1\right)}\)
<=> \(4x^2+9x+2=4x^2-21x+5\)
<=> \(4x^2+9x+2-4x^2+21x-5=0\)
<=> \(30x-3=0\)
<=> \(30x=3\)
<=> \(x=\frac{3}{30}=\frac{1}{10}\)( tmđk )