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

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25 tháng 6 2023

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

\(\Leftrightarrow\left(x-5\right)^2+7\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)+7\right]=0\)

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

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

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

Vậy \(S=\left\{5;-2\right\}\)

25 tháng 6 2023

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

19 tháng 7 2018

a) \(25x^2-9=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)

Vậy \(S=\left\{\frac{3}{5};\frac{-3}{5}\right\}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\)

\(\Leftrightarrow8x=-1\)

\(\Leftrightarrow x=-\frac{1}{8}\)

Vậy.........

c)\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x=-255\)

\(\Leftrightarrow x=-127,5\)

Vậy.............

có j sai xót mong m.n bỏ qua☺

19 tháng 7 2018

a) \(25x^2-9=0\)                      

<=> \(\left(5x\right)^2=9\)

<=> \(\left(5x\right)^2=3^2\)

<=> \(5x=3\)

<=> \(x=\frac{3}{5}\)

b) \(\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\)

<=> \(x^2+2.x.4+4^2-\left(x^2-1^2\right)=16\)

<=> \(x^2+8x+16-x^2+1=16\)

<=> \(\left(x^2-x^2\right)+8x+\left(16+1\right)=16\)

<=> \(8x+17=16\)

<=> \(8x=-1\)

<=> \(x=\frac{-1}{8}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

<=> \(\left(2x\right)^2-2.2x.1+1^2+x^2+2.x.3+3^2-5\left(x^2-7^2\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5x^2+5.7^2=0\)

<=> \(\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+5.7^2\right)=0\)

<=> \(2x+245=0\)

<=> \(2x=-245\)

<=> \(x=\frac{-245}{2}\)

3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\) \(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\) \(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\) Mc: \(x.\left(x-5\right)\) \(\Leftrightarrow\) \(x^2\) - 2\(x\) - 5 = \(x\) - 5 \(\Leftrightarrow\) \(x^2\) - 2\(x\) - \(x\) - 5 + 5 = 0 \(\Leftrightarrow\) \(x^2\) - 3\(x\) = 0 \(\Leftrightarrow\) \(x\) . (\(x\) - 3) =...
Đọc tiếp

3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\)

Mc: \(x.\left(x-5\right)\)

\(\Leftrightarrow\) \(x^2\) - 2\(x\) - 5 = \(x\) - 5

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

\(\Leftrightarrow\) \(x^2\) - 3\(x\) = 0

\(\Leftrightarrow\) \(x\) . (\(x\) - 3) = 0

\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) - 3 = 0

\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) = 3

Vậy \(x\) = 0 hoặc \(x\) = 3

\(x-5\ne0\Rightarrow x\ne5\)

\(x^2-5\ne0\Rightarrow x\ne5\)\(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)

\(x\ne0\)

Vậy S = {3}

4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\frac{x.\left(x-4\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

Mc: \(x.\left(x+7\right)\)

\(\Leftrightarrow x^2-4x-x-7=-7\)

\(\Leftrightarrow x^2-4x-x=-7+7\)

\(\Leftrightarrow\) \(x^2-5x=0\)

\(\Leftrightarrow x.\left(x-5\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(x-5=0\)

\(\Leftrightarrow x=0\) hoặc \(x=5\)

Vậy \(x=0\) hoặc \(x=5\)

\(x+7\ne0\Rightarrow x\ne-7\)

\(x^2+7\ne0\Rightarrow x\ne-7\)\(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-7\end{matrix}\right.\)

\(x\ne0\)

Vậy S = {5}

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\Rightarrow TXĐ\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

Mc : \(\left(x-2\right).\left(x+2\right)\)

\(\Leftrightarrow\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

\(\Leftrightarrow x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)

\(\Leftrightarrow2x^2-8x+8=0\)

\(\Leftrightarrow\) \(2x^2-4x-4x+8=0\)

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

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

\(\Leftrightarrow2x-4=0\) hoặc \(x-2=0\)

\(\Leftrightarrow x=2\) hoặc \(x=2\)

\(\Leftrightarrow x=2\) (Loại) hoặc x = 2 (Loại)

Vậy S = \(\left\{\varnothing\right\}\)

6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)

\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

MC: \(\left(x-1\right).\left(x+1\right)\)

\(\Leftrightarrow x^2+x+x+1-x^2+x+x-1=4\)

\(\Leftrightarrow x^2-x^2+x+x+x+x+1-1-4=0\)

\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow4.\left(x-1\right)=0\)

\(\Leftrightarrow\) 4 = 0 hoặc \(x-1=0\)

\(\Leftrightarrow\) 4 = 0 hoặc \(x=1\)

\(\Leftrightarrow\) 4 = 0 (Loại) hoặc \(x=1\) (Loại)

Vậy S = \(\left\{\varnothing\right\}\)

7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)

\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\)

\(Mc:\left(x-1\right).\left(x+1\right)\)

\(\Leftrightarrow\) \(x^2+x+x+1-4x=x^2-x-x+1\)

\(\Leftrightarrow x^2-x^2+x+x-4x+x+x=-1+1\)

\(\Leftrightarrow0=0\) (Nhận)

Vậy S = \(\left\{x\in R;x\ne\pm1\right\}\)

0
16 tháng 8 2017

ANH HAY CHỊ ƠI LÀM GIÚP EM BAI LỚP 7 ĐI O DUOI DAY A

16 tháng 8 2017

a) \(\left(x-3\right)^2-4=0\)

\(\Rightarrow\left(x-3\right)^2=4\)

\(\Rightarrow\left(x-3\right)^2=2^2=\left(-2\right)^2\)

\(\Rightarrow x-3=2\)hoặc \(\left(x-3\right)=-2\)

\(\Rightarrow\hept{\begin{cases}x-3=2\\x-3=-2\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=-1\end{cases}}}\)

Vậy \(x\in\left\{5;-1\right\}\)

b) \(x^2-2x=24\)

\(\Rightarrow x.\left(x+2\right)=24\)

\(\Rightarrow x.\left(x+2\right)=4.6\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\) \(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\) \(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\) \(\Leftrightarrow x^2+5x-2x-10-5=1x+5\) \(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0 \(\Leftrightarrow\) \(x^2+2x-20=0\) \(\Leftrightarrow x^2+2x-10x-20=0\) \(\Leftrightarrow\) (x\(^2\) + 2x) - (10x +...
Đọc tiếp

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)

\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)

\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0

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

\(\Leftrightarrow x^2+2x-10x-20=0\)

\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0

\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0

\(\Leftrightarrow\)

4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)

\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)

\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)

\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0

\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0

\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0

\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)

\(\Leftrightarrow2x^2-8x+8=0\)

\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0

\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0

\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0

\(\Leftrightarrow\) 2x = 8 hoặc x = 1

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)

Vậy S = {4; 1}

6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4

\(\Leftrightarrow\) 4x - 4 = 0

\(\Leftrightarrow\) 4 (x - 1) =0

\(\Leftrightarrow\) x - 1 = 0 / 4 = 0

\(\Leftrightarrow\) x = 1 (Nhận)

Vậy S = {1}

7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)

\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)

\(\Leftrightarrow\) 0

Vậy S ={\(\varnothing\)}

0

1: =>2x-5=4 hoặc 2x-5=-4

=>2x=9 hoặc 2x=1

=>x=9/2hoặc x=1/2

2: \(\Leftrightarrow\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}=\dfrac{-1}{8}\)(vô lý)

3: \(\Leftrightarrow\left|5x-3\right|=x+5\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(5x-3-x-5\right)\left(5x-3+x+5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(4x-8\right)\left(6x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;-\dfrac{1}{3}\right\}\)

19 tháng 6 2018

Bạn coi lại đề giùm đi.

19 tháng 6 2018

\(18x^2-6x-9x+3-18x^2+2x-27x+3=-6.\)

\(-15x+12+2x=0\)

\(-13x=-12\Leftrightarrow x=\frac{13}{12}\)

21 tháng 11 2019

1) (x - 2)2 - (x - 3)(x + 3) = 17

=> x2 - 4x + 4 - x2 + 9 = 17

=> -4x = 17 - 13

=> -4x = 4

=> x = -1

2) TTT

3) x2 + 6x - 147 = 0

=> x2 + 19x - 13x - 147 = 0

=> x(x + 19) - 13(x + 19) = 0

=> (x - 13)(x + 19) = 0

=> \(\orbr{\begin{cases}x-13=0\\x+19=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=13\\x=-19\end{cases}}\)

4) (3x - 5)(2x + 3) - 6x2 = 7

=> 6x2 + 9x - 10x - 15 - 6x2 = 7

=> -x - 15 = 7

=> -x = 7 + 15

=> -x = 22

=> x = -22

5) TL

22 tháng 7 2017

a, x= -3

b, x= -3, x= 3/2

22 tháng 7 2017

Sao khó vậy mày