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28 tháng 5 2021

ĐK: ` x \ne \pm 3`

`(x^2-x)/(x+3)-(x^2)/(x-3)=(7x^2-3x)/(9-x^2)`

`<=> (x^2-x)(x-3)-x^2 (x+3) = -(7x^2-3x)`

`<=> −7x^2+3x=-7x^2+3x`

`<=> 0x=0 forall x`

Vậy `S=RR \\ {+-3}`.

28 tháng 11 2023

\(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\\ \Leftrightarrow\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=-\dfrac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ đkxđ:\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\\ \Leftrightarrow\dfrac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{0}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow0=0\left(luon.dung\right)\)

a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)

\(\Leftrightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow39x-84x=1092-73\)

=>-45x=1019

hay x=-1019/45

b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

=>21x+63-14=20x+36-49x+63

=>21x+49=-29x+99

=>50x=50

hay x=1

c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)

=>14x+7-15x-6-21x-63=0

=>-22x-64=0

hay x=-32/11

d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)

=>70x-105-30x-45=84x+63-1785

=>40x-150-84x+1722=0

=>-44x+1572=0

hay x=393/11

19 tháng 2 2022

a, msc 12.7=84 

Chuyển vế về =0 rồi làm

b,msc 28

c,làm tương tự

5 tháng 2 2022

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

a: =>10x-4=15-9x

=>19x=19

hay x=1

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x-32x=60-9

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

=>3x=6/5

hay x=2/5

d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)

=>21x-120x+1080=80x+60

=>-179x=-1020

hay x=1020/179

e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>95x+6x=96+5

=>x=1

f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+96=-5x+30

=>-19x=-66

hay x=66/19

1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(3x+9+4x-12=3x-7\)

\(\Leftrightarrow4x=-7+12-9=-4\)

hay \(x=-1\left(nhận\right)\)

2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)

\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(3x+12-4x+16=3x-4\)

\(\Leftrightarrow28-4x=-4\)

\(\Leftrightarrow4x=32\)

hay \(x=8\left(tm\right)\)

3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

Suy ra: \(5x^2-12+3x+3=5x^2-5x\)

\(\Leftrightarrow3x-9+5x=0\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(nhận\right)\)

26 tháng 2 2023

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{x^2-9}\left(x\ne3;x\ne-3\right)\\ < =>\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{\left(x-3\right)\left(x+3\right)}\)

suy ra:

`2(x+3)+3(x-3)=7x+5`

`<=>2x+6+3x-9=7x+5`

`<=>2x+3x-7x=5-6+9`

`<=> -2x=8`

`<=> x=-4(tm)`

 

NV
26 tháng 2 2023

ĐKXĐ: \(x\ne\pm3\)

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{7x+5}{x^2-9}\)

\(\Leftrightarrow\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{7x+5}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2\left(x+3\right)+3\left(x-3\right)=7x+5\)

\(\Leftrightarrow2x+6+3x-9=7x+5\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\) (thỏa)

Vậy pt có nghiệm \(x=-4\)

20 tháng 4 2023

\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)

\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)

`<=> 20(2x-1) +15(3x-2) =12(4x-3)`

`<=> 40x - 20 + 45x - 30 = 48x - 36`

`<=> 85x -50 = 48x - 36`

`<=> 85x-48x = -36+50`

`<=> 37x =14`

`<=> x= 14/37`

Vậy phương trình có nghiệm `x=14/37`

__

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)

\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

`=> 5x + 15 + 4x -12=x-6`

`<=> 9x + 3=x-6`

`<=> 9x-x=-6-3`

`<=> 8x = -9`

`<=>x=-9/8(tm)`

Vậy phương trình có nghiệm `x=-9/8`

` @ yngoc`

bài 2 giải các phương trình saub,\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)               m,\(\dfrac{3x-1}{x+1}=\dfrac{2x+1}{x-1}\)d,\(\dfrac{3x-14}{x+5}=\dfrac{2}{3}\)                   p,\(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)f,\(\dfrac{6}{x}-1=\dfrac{2x-3}{3}\)               r,\(\dfrac{1}{x+3}+\dfrac{1}{x-1}=\dfrac{10}{\left(x+3\right)\left(x-1\right)}\)h,\(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\)       ...
Đọc tiếp

bài 2 giải các phương trình sau

b,\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)               m,\(\dfrac{3x-1}{x+1}=\dfrac{2x+1}{x-1}\)

d,\(\dfrac{3x-14}{x+5}=\dfrac{2}{3}\)                   p,\(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)

f,\(\dfrac{6}{x}-1=\dfrac{2x-3}{3}\)               r,\(\dfrac{1}{x+3}+\dfrac{1}{x-1}=\dfrac{10}{\left(x+3\right)\left(x-1\right)}\)

h,\(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\)         t,\(\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)

j,\(\dfrac{5}{3x+2}=2x-1\)              u,\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)

w,\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\)         s, \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{2x}{\left(x-1\right)\left(x-3\right)}\)

ơ,\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\)          v,\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

z,\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)             ư,\(\dfrac{x+2}{x-2}-\dfrac{-2}{x^2-2x}=\dfrac{1}{x}\)

o,\(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\)          ô,\(1-\dfrac{1}{1-x}=\dfrac{x^2}{x^2-1}\)       zz,\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)

2
13 tháng 1 2023

Bạn chia nhỏ các phần ra nhé.

13 tháng 1 2023

uh mk biết lần sau mk rút kinh nghiệm