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a, \(\left|x-3\right|-3x=7\Leftrightarrow\left|x-3=7+3x\right|\)
xét \(x-3\ge0\Rightarrow\hept{\begin{cases}x\ge3\\x-3=7+3x\end{cases}}\)
x - 3 = 7 + 3x
<=> -2x = 10
<=> x = -5 (loại)
xét \(x-3< 0\Rightarrow\hept{\begin{cases}x< 3\\x-3=-7-3x\end{cases}}\)
x - 3 = -7-3x
<=> 4x = -4
<=> x = -1 ((tm)
vậy_
b, làm tương tự a
Bài giải
a, \(\left|x-3\right|-3x=7\)
\(\left|x-3\right|=7+3x\)
\(\Rightarrow\orbr{\begin{cases}x-3=-7-3x\\x-3=7+3x\end{cases}}\Rightarrow\orbr{\begin{cases}-4x=4\\2x=10\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-1\text{ ; }5\right\}\)
b, \(\left|5+x\right|+3x=17\)
\(\left|5+x\right|=17-3x\)
\(\Rightarrow\orbr{\begin{cases}5+x=-17+3x\\5+x=17-3x\end{cases}}\Rightarrow\orbr{\begin{cases}2x=22\\-4x=-12\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=3\end{cases}}\)
\(x\in\left\{11;3\right\}\)
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a) đk: \(x\ge-\frac{7}{3}\)
\(\left|x-3\right|-3x=7\)
\(\Leftrightarrow\left|x-3\right|=3x+7\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=3x+7\\x-3=-3x-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-10\\4x=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}\)
Vậy x = -1
b) đk: \(x\le\frac{17}{3}\)
\(\left|5+x\right|+3x=17\)
\(\Leftrightarrow\left|x+5\right|=17-3x\)
\(\Leftrightarrow\orbr{\begin{cases}x+5=17-3x\\x+5=3x-17\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=12\\2x=22\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\left(tm\right)\\x=11\left(ktm\right)\end{cases}}\)
Vậy x = 3
Ta có |x - 3| - 3x = 7
=> |x - 3| = 3x + 7
ĐK : \(3x+7\ge0\Rightarrow x\ge-\frac{7}{3}\)
Khi đó |x - 3| = 3x + 7
<=> \(\orbr{\begin{cases}x-3=3x+7\\x-3=-3x-7\end{cases}}\Rightarrow\orbr{\begin{cases}-2x=10\\4x=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\left(\text{loại}\right)\\x=-1\left(\text{tm}\right)\end{cases}}\)
Vậy x = -1
b) Ta có |5 + x| + 3x = 17
<=> |5 + x| = 17 - 3x
ĐK \(17-3x\ge0\Rightarrow x\le\frac{17}{3}\)
Khi đó |5 + x| = 17 - 3x
<=> \(\orbr{\begin{cases}5+x=17-3x\\5+x=17+3x\end{cases}}\Rightarrow\orbr{\begin{cases}4x=22\\-2x=12\end{cases}}\Rightarrow\orbr{\begin{cases}x=5,5\left(\text{tm}\right)\\x=-6\left(\text{tm}\right)\end{cases}}\)
Vậy \(x\in\left\{5,5;6\right\}\)
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a) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 4
<=> x3 - 9x2 + 27x - 27 - ( x3 - 27 ) + 9( x2 + 2x + 1 ) = 4
<=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9x2 + 18x + 9 = 4
<=> 45x + 9 = 4
<=> 45x = -5
<=> x = -5/45 = -1/9
b) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 17
<=> x( x2 - 25 ) - ( x3 + 8 ) = 17
<=> x3 - 25x - x3 - 8 = 17
<=> -25x - 8 = 17
<=> -25x = 25
<=> x = -1
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a)\(\left(x^2-x+1\right).\left(x+1\right)-x^3+3x=15\)
\(x^3+x^2-x^2-x+x+1-x^3+3x=15\)
\(1+3x=15\)
\(3x=15-1\)
\(3x=14\)
\(x=\frac{14}{3}\)
b) \(\left(x+3\right).\left(x-2\right)+3x=4\left(x+\frac{3}{4}\right)\)
\(x^2-2x+3x-6+3x=4x+3\)
\(x^2-2x+3x+3x-4x=6+3\)
\(x^2=9\)
\(x^2=3^2\) hoặc \(x^2=\left(-3\right)^2\)
vậy x=3 hoặc x=-3
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a,
(x2-x+1)(x+1)-x3+3x=15
x3-x2+x+x2-x+1-x3+3x=15
x3-x3-x2+x2+x-x+3x+1=15
3x+1=15
3x=15-1
3x=14
x=14/3
b,
(x+3)(x-2)+3x=\(\frac{4}{x+\frac{3}{4}}\)
x2-2x+3x-6+3x=\(\frac{4}{x+\frac{3}{4}}\)
x2-2x+3x+3x-6=\(\frac{4}{x+\frac{3}{4}}\)
Tới đây hết biết , đề có gì sai sai sao ý !
c,
(x2-5)(x+2)+5x=2x2+17
x3+2x2-5x-10+5x=2x2+17
x3+2x2-5x+5x-10=2x2+17
x3+2x2-10=2x2+17
x3-10=17
x3=17+10
x3=27
\(\Rightarrow x=3\)(Vì : 33=27)
_k_ nhé bn
Nhân ra thôi bạn, có hằng đẳng thức gì đâu !
a) \(\left(x^2-x+1\right)\left(x+1\right)-x^3+3x=15\)
\(\Leftrightarrow\left(x^2-x+1\right)\cdot x+x^2-x+1-x^3+3x=15\)
\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+3x=15\)
\(\Leftrightarrow1+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)
b) \(\left(x+3\right)\left(x-2\right)+3x=4\cdot\left(x+\frac{3}{4}\right)\)
\(\Leftrightarrow x^2+3x-2x-6+3x=4x+3\)
\(\Leftrightarrow x^2+4x-6=4x+3\)
\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=-3\\x=3\end{cases}}\)
c) \(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)
\(\Leftrightarrow x^3-5x+2x^2-10+5x=2x^2+17\)
\(\Leftrightarrow x^3=27\Leftrightarrow x=3\)
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Chúng ta sẽ sử dụng hằng đẳng thức em nhé :)
a. \(x^3+1-x^3+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)
b. \(x^2+x-6+3x=4x+3\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
c. \(x^3+2x^2-5x-10+5x=2x^2+17\Leftrightarrow x^3=27\Leftrightarrow x=3\)
x=-1
x=3
x=-1
x=3
sk ưell