12 x + 1 /12 <= 9 x + 1/3 -8x+1/4
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Phân số tối giản là `5/9`
\(x-\dfrac{1}{4}=\dfrac{5}{3}:\dfrac{2}{3}\)
\(x-\dfrac{1}{4}=\dfrac{5}{3}\times\dfrac{3}{2}\)
\(x-\dfrac{1}{4}=\dfrac{5}{2}\)
\(x=\dfrac{5}{2}+\dfrac{1}{4}\)
\(x=\dfrac{11}{4}\)
Gọi a,b,c,... cho dễ nhé!
a,\(7+2x=22-3x\)
\(\Leftrightarrow2x+3x=22-7\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=3\)
Vậy...
b,\(x-12+4x=25+2x-1\)
\(\Leftrightarrow x+4x-2x=25-1+12\)
\(\Leftrightarrow3x=36\)
\(\Leftrightarrow x=12\)
Vậy...
c,\(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow-2x+x=-4+4-7\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy...
d,\(8x-3=5x+12\)
\(\Leftrightarrow8x-5x=12+3\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy...
e,\(x+2x+3x-19=3x+5\)
\(\Leftrightarrow x+2x+3x-3x=5+19\)
\(\Leftrightarrow3x=24\)
\(\Leftrightarrow x=8\)
Vậy...
f,\(\left(x-1\right)-\left(2x-1\right)=9-x\)
\(\Leftrightarrow x-1-2x+1=9-x\)
\(\Leftrightarrow x-2x+x=9-1+1\)
\(\Leftrightarrow0x=9\) (Vô lý)
Vậy...
a, \(7+2x=22-3x\)
\(\Rightarrow7+2x-22+3x=0\)
\(\Rightarrow5x-15=0\)
\(\Rightarrow5x=15\Rightarrow x=3\)
b, \(x-12+4x=25+2x-1\)
\(\Rightarrow3x-12-24-2x=0\)
\(\Rightarrow x-36=0\Rightarrow x=36\)
c, \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Rightarrow7-2x-4=-x-4\)
\(\Rightarrow3-2x+x+4=0\)
\(\Rightarrow-x=-7\Rightarrow x=7\)
d, \(8x-3=5x+12\)
\(\Rightarrow8x-3-5x-12=0\)
\(\Rightarrow3x-15=0\)
\(\Rightarrow3x=15\Rightarrow x=5\)
e, \(x+2x+3x-19=3x+5\)
\(\Rightarrow6x-19-3x-5=0\)
\(\Rightarrow3x-24=0\)
\(\Rightarrow3x=24\Rightarrow x=8\)
f, \(\left(x-1\right)-\left(2x-1\right)=9-x\)
\(\Rightarrow x-1-2x+1-9+x=0\)
(hình như câu này bị sai đề rồi, bạn xem lại đề nhé)
Chúc bạn học tốt!
k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)
giúp mk câu k nhé đề bài như trên
b: \(\Leftrightarrow4x+8-9=4x-4\)
=>-1=-4(loại)
d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)
=>8x=3x-6+2x+2=5x-4
=>3x=-4
=>x=-4/3
f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)
=>3x+6+8x-12=2x-24
=>11x-6=2x-24
=>9x=-18
=>x=-2
\(a,4x-6< 7x-12\)
\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)
\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)
\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)
\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)
\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)
\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)
\(\Leftrightarrow-16x\ge11\)
\(\Leftrightarrow x\le-\frac{11}{16}\)
\(d,-12-8x>3+2x-\left(5-7x\right)\)
\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)
\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)
\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)
\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)
2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow-24x+144=-5x+30\)
\(\Leftrightarrow-24x+144+5x-30=0\)
\(\Leftrightarrow-19x+114=0\)
\(\Leftrightarrow-19x=-114\)
hay x=6
Vậy: x=6
3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9-60-32x=0\)
\(\Leftrightarrow-2x-51=0\)
\(\Leftrightarrow-2x=51\)
hay \(x=-\dfrac{51}{2}\)
Vậy: \(x=-\dfrac{51}{2}\)
4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)
\(\Leftrightarrow2x+2-x+2=6x-3\)
\(\Leftrightarrow x+4-6x+3=0\)
\(\Leftrightarrow-5x+7=0\)
\(\Leftrightarrow-5x=-7\)
hay \(x=\dfrac{7}{5}\)
Vậy: \(x=\dfrac{7}{5}\)
1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(2\left(5x-2\right)=3\left(5-3x\right)\)
\(10x-4=15-9x\)
\(10x+9x=15+4\)
\(19x=19\)
\(x=1\)
Vậy \(x=1\)
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.