Tìm x biết
a, \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
b, 7x - 3x = 3,2
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a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)
<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)
<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)
<=> \(14\left(32x+19\right)+560\)
<=> \(36x+72-35x-140=448x+226+560\)
<=> \(-447x=894\)
<=> x = -2
Bài 1 :
a, Ta có : \(3x-1=2x+4\)
=> \(3x-2x=4+1\)
=> \(x=5\)
Vậy phương trình có tập nghiệm \(S=\left\{5\right\}\)
b, Ta có : \(5x-2=0\)
=> \(5x=2\)
=> \(x=\frac{2}{5}\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{2}{5}\right\}\)
c, Ta có : \(7x-4=3x+12\)
=> \(7x-3x=12+4\)
=> \(4x=16\)
=> \(x=4\)
Vậy phương trình có tập nghiệm \(S=\left\{4\right\}\)
d, Ta có : \(\frac{x-1}{2}+\frac{3x+2}{4}=\frac{x-7}{12}\)
=> \(\frac{6\left(x-1\right)}{12}+\frac{3\left(3x+2\right)}{12}=\frac{x-7}{12}\)
=> \(6\left(x-1\right)+3\left(3x+2\right)=x-7\)
=> \(6x-6+9x+6=x-7\)
=> \(6x+9x-x=6-7-6\)
=> \(14x=-7\)
=> \(x=-\frac{1}{2}\)
Vậy phương trình có tập nghiệm \(S=\left\{-\frac{1}{2}\right\}\)
Bài 2 :
a, ĐKXĐ : \(\left\{{}\begin{matrix}x^2-2x+1\ne0\\x-1\ne0\end{matrix}\right.\)
=> \(x-1\ne0\)
=> \(x\ne1\)
- Ta có : \(\left(\frac{x+1}{x^2-2x+1}+\frac{1}{x-1}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{x+1}{\left(x-1\right)^2}+\frac{x-1}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{2x}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\left(\frac{2x}{\left(x-1\right)^2}\right)\left(\frac{x-1}{x}\right)-\frac{2}{x-1}\)
= \(\frac{x}{x-1}-\frac{2}{x-1}\)
= \(\frac{x-2}{x-1}\)
\(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{840}-\frac{17}{210}=\frac{8x}{15}+\frac{19}{60}+\frac{2}{3}\)
\(\Leftrightarrow\frac{x}{840}.840-\frac{17}{210}.840=\frac{8x}{15}.840+\frac{19}{60}.840+\frac{2}{3}.840\)
\(\Leftrightarrow x-68=448x+226+560\)
\(\Leftrightarrow x-68=448x+826\)
\(\Leftrightarrow x=448x+826+68\)
\(\Leftrightarrow x=448x+894\)
\(\Leftrightarrow-447x=894\)
=> x = -2
\(a,4x-6< 7x-12\)
\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)
\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)
\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)
\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)
\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)
\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)
\(\Leftrightarrow-16x\ge11\)
\(\Leftrightarrow x\le-\frac{11}{16}\)
\(d,-12-8x>3+2x-\left(5-7x\right)\)
\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)
\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)
\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)
\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)
a) Qui đồng rồi khử mẫu ta được:
3(3x+2)-(3x+1)=2x.6+5.2
<=> 9x+6-3x-1 = 12x+10
<=> 9x-3x-12x = 10-6+1
<=> -6x = 5
<=> x = -5/6
Vậy ....
b) ĐKXĐ: \(x\ne\pm2\)
Qui đồng rồi khử mẫu ta được:
(x+1)(x+2)+(x-1)(x-2) = 2(x2+2)
<=> x2+3x+2+x2-3x+2 = 2x2+4
<=> x2+x2-2x2+3x-3x = 4-2-2
<=> 0x = 0
<=> x vô số nghiệm
Vậy x vô số nghiệm với x khác 2 và x khác -2
c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)
Vậy ......
d) (x+1)2-4(x2-2x+1) = 0
<=> x2+2x+1-4x2+8x-4 = 0
<=> -3x2+10x-3 = 0
giải phương trình
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
a.\(\frac{3x-1}{3x+1}+\frac{x-3}{x+3}=2\)
\(\frac{\left(3x-1\right)\left(x+3\right)+\left(3x+1\right)\left(x-3\right)}{\left(3x+1\right)\left(x+3\right)}=\frac{3x^2+8x-3+3x^2-8x-3}{\left(3x+1\right)\left(x+3\right)}=\frac{6x^2-6}{\left(3x+1\right)\left(x+3\right)}=2\)
\(6x^2-6=2\left(3x^2+10x+3\right)\)
\(6x^2-6=6x^2+20x+6\)
-20x-12=0
x=\(\frac{-3}{5}\)
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) 2/3x-1/2x=15/12
=> x.(2/3-1/2)=15/12
=> x.1/6 =15/12
=> x =15/12:1/6
=> x =5/24
b) 7x-3x=3,2
=> x.(7-3) =3,2
=> x.4 =3,2
=> x =3,2:4
=> x =0,8
a,2/3.x - 1/2 . x = 5/12
x . ( 2/3 - 1/2 ) = 5/12
x . 1/6 = 5/12
x = 5/12 : 1/6
x = 15/1