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1 tháng 7 2018

\(\left(5x-3\right)^2-\left(4x-7\right)^2=0\\ \Leftrightarrow25x^2-30x+9-\left(16x^2-56x+49\right)=0\\ \Leftrightarrow25x^2-30x+9-16x^2+56x-49=0\\ \Leftrightarrow9x^2+26x-40=0\\ \Leftrightarrow x=\dfrac{-26\pm\sqrt{26^2-4.9.\left(-40\right)}}{2.9}\\ \Leftrightarrow x=\dfrac{-26\pm\sqrt{676+1440}}{18}\\ \Leftrightarrow x=\dfrac{-26\pm\sqrt{2116}}{18}\\ \Leftrightarrow x=\dfrac{-26\pm46}{18}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-26+46}{18}\\x=\dfrac{-26-46}{18}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{9}\\x=-4\end{matrix}\right.\)

Vậy ...

1 tháng 7 2018

\(\left(5x-3\right)^2-\left(4x-7\right)^2=0\)

\(\Leftrightarrow\left(5x-3-4x+7\right)\left(5x-3+4x-7\right)=0\)

\(\Leftrightarrow\)\(\left(x+4\right)\left(9x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\9x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\9x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{9}\end{matrix}\right.\)

6 tháng 2 2022

Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)

a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)

b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

Vậy: \(S=\left\{3;20\right\}\)

c) Vì \(x^2+1\ge1>0\forall x\)

\(\Rightarrow4x+2=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)

a: =>3x-2=0 hoặc 4x+5=0

=>x=2/3 hoặc x=-5/4

b: =>(x-3)(x+20)=0

=>x=3 hoặc x=-20

c: =>4x+2=0

hay x=-1/2

d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0

=>x=-7/2 hoặc x=5 hoặc x=-1/5

31 tháng 1 2019

câu a tự quy đồng cùng  mẫu rồi làm thôi :"))

b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)

\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)

Đặt \(x^2-x=k\), ta có:

\(k.\left(k-2\right)=24\)

\(\Leftrightarrow k^2-2k+1=25\)

\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)

\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)

\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)

c)\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)

\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)

p/s: bn tự kết luận nha :))

8 tháng 2 2018

\(\left(3x-2\right)\left[\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right]=0\)

\(\left(3x-2\right).\frac{10\left(x+3\right)-7\left(4x-3\right)}{35}=0\)

\(\left(3x-2\right)\left(10x+30-28x+21\right)=0\)

\(\left(3x-2\right)\left(51-18x\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-2=0\\51-18x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\-18x=-51\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}}\)

Vậy \(S=\left\{\frac{2}{3};\frac{17}{6}\right\}\)

8 tháng 2 2018

\(\left(3x-2\right)\left[\frac{2\left(x+3\right)}{7}-\frac{4x-3}{5}\right]=0\)

\(\Leftrightarrow\left(3x-2\right)\left[\frac{2.5\left(x+3\right)}{35}-\frac{7\left(4x-3\right)}{35}\right]=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\\frac{-18x+51}{35}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\-18x+51=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}}\)

Vậy \(x=\left\{\frac{2}{3};\frac{17}{6}\right\}\)

6 tháng 2 2022

d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

e) \(PT\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{7;1\right\}\)

f) \(PT\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{1;3\right\}\)

6 tháng 2 2022

\(d,x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(x-2\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

 

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

11 tháng 2 2018

khó thể xem trên mạng

11 tháng 2 2018

bài 1 câu a bỏ x= nhé !

31 tháng 7 2018

a) ta có : \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x=30\Leftrightarrow15x=30\Leftrightarrow x=2\)

b) điều kiện : \(x\ne\dfrac{1}{5};x\ne1;x\ne\dfrac{3}{5}\)

ta có : \(\dfrac{3}{5x-1}+\dfrac{2}{3-3x}=\dfrac{4}{\left(1-5x\right)\left(5x-3\right)}\)

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

\(\Leftrightarrow\dfrac{x+7}{3-3x}=\dfrac{4}{3-5x}\Leftrightarrow\left(x+7\right)\left(3-5x\right)=4\left(3-3x\right)\)

\(\Leftrightarrow-5x^2-20+9=0\)

ta có : \(\Delta'=\left(10\right)^2+5\left(9\right)=145>0\) \(\Rightarrow\) phương trình có 2 nghiệm phân biệt

\(x=\dfrac{10+\sqrt{145}}{-5};x=\dfrac{10-\sqrt{145}}{-5}\)

27 tháng 4 2018
https://i.imgur.com/gGiBWDN.jpg