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1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7
19.(2+3+4−5+6−7)2−9.(7x−2)=0
19.32−9(7x−2)=0
19.9−9(7x−2)=0
171−9(7x−2)=0
9(7x-2)=171
7x−2=171:9
7x−2=19
7x=19+2
7x=21
x=21:7
x=3x=3
Vậy x=3
Bài giải
a, \(\frac{2}{7}x+\frac{1}{2}=-\frac{3}{4}\)
\(\frac{2}{7}x=-\frac{3}{4}-\frac{1}{2}\)
\(\frac{2}{7}x=-\frac{5}{4}\)
\(x=-\frac{5}{4}\text{ : }\frac{2}{7}\)
\(x=-\frac{35}{8}\)
b, \(\left(6x+\frac{2}{5}\right)=-\frac{8}{125}\)
\(6x=-\frac{8}{125}-\frac{2}{5}\)
\(6x=-\frac{58}{125}\)
\(x=-\frac{58}{125}\text{ : }6\)
\(x=\frac{-29}{375}\)
c, \(\left|x-\frac{2}{3}\right|\cdot\left(18-6x^2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left|x-\frac{2}{3}\right|=0\\18-6x^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=0\\6x^2=18\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\sqrt{3}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{2}{3}\text{ ; }\sqrt{3}\right\}\)
\(x=-2+\sqrt{5}>0\Rightarrow x+2=\sqrt{5}\)
\(\Rightarrow\left(x+2\right)^2=5\Rightarrow x^2+4x=1\)
Ta có:
\(3x^5+12x^4-8x^3-23x^2-7x+1\)
\(=3x^3\left(x^2+4x\right)-8x^3-23x^2-7x+1\)
\(=-5x^3-23x^2-7x+1=-5x\left(x^2+4x\right)-3x^2-7x+1\)
\(=-3x^2-12x+1=-3\left(x^2+4x\right)+1=-3+1=-2\)
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
\(\dfrac{3x^5+5x^3+1}{4x^4-7x^2+2}.\dfrac{x}{2x+3}.\dfrac{4x^4-7x^2+2}{3x^5+5x^3+1}\) ( sửa đề )
\(=\left[\dfrac{3x^5+5x^3+1}{4x^4-7x^2+2}.\dfrac{4x^4-7x^2+2}{3x^5+5x^3+1}\right].\dfrac{x}{2x+3}\)
\(=\dfrac{x}{2x+3}\)
\(\left|7x+1\right|-4^2=2^2\)
\(\Leftrightarrow\left|7x+1\right|-16=4\)
\(\Leftrightarrow\left|7x+1\right|=4+16\)
\(\Leftrightarrow\left|7x+1\right|=20\)
\(\Leftrightarrow\orbr{\begin{cases}7x+1=20\\7x+1=-20\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}7x=19\\7x=-21\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{19}{7}\\x=-3\end{cases}}\)
K/L: Vậy x=\(\frac{19}{7}\)hoặc x= -3