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a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)
=>-8x+3=11
=>-8x=8
hay x=-1
b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)
=>-11x=10
hay x=-10/11
a) \(\left|2x-5\right|=x+1\)
<=> \(\orbr{\begin{cases}2x-5=x+1\left(x\ge\frac{5}{2}\right)\\5-2x=x+1\left(x< \frac{5}{2}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\left(ktm\right)\\3x=4\end{cases}}\)
<=> \(x=\frac{4}{3}\left(tm\right)\)
b) \(\left|3x-2\right|-1=2x\) <=> \(\left|3x-2\right|=2x+1\)
<=> \(\orbr{\begin{cases}3x-2=2x+1\left(x\ge\frac{2}{3}\right)\\2-3x=2x+1\left(x< \frac{2}{3}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-3\left(ktm\right)\\5x=1\end{cases}}\) <=> \(x=\frac{1}{5}\left(tm\right)\)
c) \(\left|x-5\right|+5=x\) <=> \(\left|x-5\right|=x-5\)
<=> \(\orbr{\begin{cases}x-5=x-5\left(x\ge5\right)\\5-x=x-5\left(x< 5\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}0x=0\\2x=10\end{cases}}\) <=> 0x = 0 (luôn đúng) hoặc x = 5 (ktm)
Vậy x \(\ge\)5
d) \(\left|3x-5\right|=3x-5\) <=> \(\orbr{\begin{cases}3x-5=3x-5\left(x\ge\frac{5}{3}\right)\\5-3x=3x-5\left(x< \frac{5}{3}\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}0x=0\left(luônđúng\right)\\6x=10\end{cases}}\)
<=> \(\orbr{\begin{cases}x\ge\frac{5}{3}\\x=\frac{5}{3}\left(ktm\right)\end{cases}}\)Vậy x \(\ge\)5/3
c) \(\left(3x-1\right).\left(2x+7\right)-\left(x+1\right).\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)=x+2-x+5\)
\(\Leftrightarrow18x-2-7=0\)
\(\Rightarrow x=\dfrac{9}{18}=\dfrac{1}{2}\)
b) \(2.\left(3x-1\right).\left(2x+5\right)-6.\left(2x-1\right).\left(x+2\right)=1\)
\(\Leftrightarrow\left(6x-2\right).\left(2x+5\right)-\left(12x-6\right).\left(x+2\right)=1\)
\(\Leftrightarrow12x^2+30x-4x-10-\left(12x^2+24x-6x-12\right)=1\)
\(\Leftrightarrow12x^2+26x-10-12x^2-18x +12=1\)
\(\Leftrightarrow8x+2=1\)
\(\Rightarrow x=\dfrac{-1}{8}\)
Lời giải:
a.
PT $\Leftrightarrow -5x^2+15x-5+x+5x^2=x-2$
$\Leftrightarrow 16x-5=x-2$
$\Leftrightarrow 15x=3$
$\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}$
b.
PT $\Leftrightarrow -4x^2+20x+7x^2-28x-3x^2=12$
$\Leftrightarrow -8x=12$
$\Leftrightarrow x=\frac{-3}{2}$
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1
<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5
<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3
<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5
<=> |x - 5/3|/6 = 9/5
<=> |x - 5/3| = 9/5.6
<=> |x - 5/3| = 54/5
<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5
x = 54/5 + 5/3 x = -54/5 - 5/3
x = 187/15 x = -137/15
b) 2.|3x + 1| = 1/3.|3x + 1| + 5
<=> 2.|3x + 1| - 1/3.|3x + 1| = 5
<=> 5/3.|3x + 1| = 5
<=> 5.|3x + 1| = 5.3
<=> 5.|3x + 1| = 15
<=> |3x + 1| = 15 : 5
<=> |3x + 1| = 3
3x + 1 = 3 hoặc 3x + 1 = -3
3x = 3 - 1 3x = -3 - 1
3x = 2 3x = -4
x = 2/3 x = -4/3
=> x = 2/3 hoặc x = -4/3
c) làm tương tự câu a) mình hơi lời
Làm câu c) cho
\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)
\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)
\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)
Giải tiếp nha
a: =3x^3-15x^2+21x
b: =-x^3+6x^2+5x-4x^2-24x-20
=-x^3+2x^2-19x-20
c: =9x^2+15x-3x-5-7x^2-14
=2x^2+12x-19
d: =10x^2-4x+2/3
3x - ( 1 - x ) = - x + 5
3x - 1 + x = - x + 5
3x + x + x = 5 + 1
5x = 6
x = 6/5
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