Bài 5 Tìm x , biết
a 4x -15 = - 75 - x
b 72 - 3x = 5x + 8
c 3 | x - 7 | = 21
d -7 | x + 3 | = - 49
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a, 4x -15=-75-x b,72-3x= 5x+8
4x+x=-75+15 -3x-5x=8-72
5x=-60 -8x=-64
x=-60:5 8x=64
x=-14 x=64:8
x=8
c,3Ix-7I=21 d,-7Ix+3I=-49
Ix-7I=21:3 Ix+3I=-49:-7
Ix-7I=7 Ix+3I=7
x-7=7 hoặc x-7=-7 x+3=7 hoặc x+3=-7
x=14 hoặc x=0 x=4 hoặc x=-10
a, 4x - 15 = ( -75 ) - x
<=> 4x +x = -75 + 15
<=> 5x = -60
<=> x = -12
b, 72 - 3x = 5x +8
<=> 72 - 8 = 5x + 3x
<=> 8x = 64
<=> x = 8
c, 3.|x-7| = 21
* Nếu x - 7 \(\ge\)0 <=> x \(\ge\)7 thì
PT <=> 3 .( x- 7 ) = 21
<=> x - 7 = 7
<=> x = 14 ( tm )
* Nếu x < 7 <=> x < 7 thì
PT <=> 3. [ - ( x- 7 ) ] = 21
<=> -x +7 = 7
<=> -x = 0
<=> x = 0 ( tm )
d, \(-7.|x+3|=-49\).
\(\Leftrightarrow|x+3|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}}\)
e, c-12.(x-5)+7.(3-x)=5
<=> -12x + 60 + 21 - 7x = 5
<=> -19x + 81 = 5
<=> -19x = -76
<=> x = 4
a, 4x -15=-75-x b,72-3x= 5x+8
4x+x=-75+15 -3x-5x=8-72
5x=-60 -8x=-64
x=-60:5 8x=64
x=-14 x=64:8
x=8
c,3Ix-7I=21 d,-7Ix+3I=-49
Ix-7I=21:3 Ix+3I=-49:-7
Ix-7I=7 Ix+3I=7
x-7=7 hoặc x-7=-7 x+3=7 hoặc x+3=-7
x=14 hoặc x=0 x=4 hoặc x=-10
k nha
a) \(\Rightarrow\left(x-2\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)
a/ \(4x-15=75-x\)
\(\Leftrightarrow4x+x=75+15\)
\(\Leftrightarrow5x=90\)
\(\Leftrightarrow x=19\)
Vậy ...
b/ \(72-3x=5x+8\)
\(\Leftrightarrow72-8=5x+3x\)
\(\Leftrightarrow64=8x\)
\(\Leftrightarrow x=8\)
Vậy ....
c/ \(3\left|x-7\right|=21\)
\(\Leftrightarrow\left|x-7\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=7\\x-7=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=0\end{matrix}\right.\)
d/ \(x-3\inƯ\left(7\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=7\\x-3=-1\\x-3=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=10\\x=2\\x=-4\end{matrix}\right.\)
Vậy ..
a) \(4x-15=-75-x\)
\(\Leftrightarrow4x+x=-75+15\)
\(\Leftrightarrow5x=-60\)
\(\Leftrightarrow x=-12\)
Vậ x=-12
b) \(72-3x=5x+8\)
\(\Leftrightarrow-3x-5x=8-72\)
\(\Leftrightarrow-8x=-64\)
\(\Leftrightarrow x=8\)
Vậy x=8
c) \(3\left|x-7\right|=21\)
\(\Leftrightarrow\left|x-7\right|=7\Leftrightarrow\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=14\\x=0\end{cases}}}\)
Vậy x=14 hoặc x=0
d) \(-7\left|x+3\right|=-49\)
\(\Leftrightarrow\left|x+3\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-10\end{cases}}}\)
Vậy x=4 hoặc x=-10
a) \(4x-15=-75-x\Leftrightarrow3x=-60\Leftrightarrow x=-20\)
b) \(72-3x=5x+8\Leftrightarrow64=8x\Leftrightarrow x=8\)
c) \(3|x-7|=21\Leftrightarrow|x-7|=7\). Xét 2TH ra được \(x=14\) và \(x=0\)
d) \(-7|x+3|=-49\Leftrightarrow|x+3|=7\). Xét 2 TH ra được \(x=4\) và \(x=-10\)