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

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\(\left(x+1\right)\left(x+7\right)< 0\)

thì \(x+1;x+7\)khác dấu

 th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)

th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)

vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)

16 tháng 7 2019

a) (2x - 3) = 5

<=> 2x - 3 = 5

<=> 2x = 5 + 3

<=> 2x = 8

<=> x = 4

=> x = 4

b) (5x - 3) = 1/2

<=> 5x - 3 = 1/2

<=> 5x = 1/2 + 3

<=> 5x = 7/2

<=> x = 7/10

=> x = 7/10

c) (x + 1)(x + 7) < 0

<=> x = -1; -7

<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)

<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)

<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)

Vậy: -7 < x < -1

16 tháng 7 2017

a, \(x^2-9=0\Rightarrow x^2=9\Rightarrow x\pm3\)

b, \(\left(x-3\right)^2-25=0\Rightarrow\left(x-3\right)^2=25\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

c, \(\left(x-3\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)

d, \(\left(x-3\right)x-2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

e, \(3x\left(x-1\right)-5\left(1-x\right)=0\)

\(\Rightarrow3x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

g, \(x^2+6x-7=0\)

\(\Rightarrow x^2-x+7x-7=0\)

\(\Rightarrow x.\left(x-1\right)+7.\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

h,\(2x^2+5x-7=0\)

\(\Rightarrow2x^2-2x+7x-7=0\)

\(\Rightarrow2x.\left(x-1\right)+7.\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Chúc bạn học tốt!!!

16 tháng 7 2017

a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) vậy \(x=3;x=-3\)

b) \(\left(x-3\right)^2-25=0\Leftrightarrow\left(x-3\right)^2=25\Leftrightarrow\left\{{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

vậy \(x=8;x=-2\)

c) \(\left(x-3\right)\left(2x-5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{5}{2}\end{matrix}\right.\)

vậy \(x=3;x=\dfrac{5}{2}\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\) vậy \(x=2;x=3\)

e) \(3x\left(x-1\right)-5\left(1-x\right)=0\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-5}{3};x=1\)

câu e t thấy sai sai nhưng vẫn làm ; bn coi lại đề nha

g) \(x^2+6x-7=0\Leftrightarrow x^2-x+7x-7=0\)

\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(x+7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\x=1\end{matrix}\right.\) vậy \(x=-7;x=1\)

h) \(2x^2+5x-7=0\Leftrightarrow2x^2-2x+7x-7=0\)

\(\Leftrightarrow2x\left(x-1\right)+7\left(x-1\right)=0\Leftrightarrow\left(2x+7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+7=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-7}{2};x=1\)

17 tháng 1 2017

b) \(x^2-7x=0\)

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

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

Vậy \(x\in\left\{0;7\right\}\)

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

\(\Rightarrow x^2+5x=0\)

\(\Rightarrow x\left(x+5\right)=0\)

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

Vậy \(x\in\left\{0;-5\right\}\)

17 tháng 1 2017

a) \(\left(x-5\right)\left(x+4\right)=0\)

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

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

Vậy...

b) \(x^2-7x=0\)

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

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

Vậy...

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

\(\Leftrightarrow x=-5\)

d) \(x^3=x\)

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

21 tháng 7 2019

Do \(x\ge6\) nên:

\(A=\left\{6\right\}\)

________________

\(6x-3< 5x+1\\ \Leftrightarrow6x-5x< 1+3\\ \Leftrightarrow x< 4\)

Vậy \(B=\left\{0;1;2;3\right\}\)

________________

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

\(\Leftrightarrow2x^2-5x+3=0\\ \Leftrightarrow2x^2-2x-3x+3=0\\ \Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)

\(x\in N\) nên \(C=\left\{1\right\}\)

6 tháng 4 2018

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

\(x^2\) . \(x\). (1-5) = 0

\(x^3\) . (-4) = 0

\(x^3\) = 0 : (-4)

\(x^3\) = 0

\(x\) = 0

Nhớ tick cho mik nha!!!hihi

6 tháng 4 2018

ủa bạn ơi,x2 cơ mà

17 tháng 1 2018

Câu a là giá trị tuyệt đối nhé

22 tháng 4 2020

\(-5|x-5|=-10\)
\(\Rightarrow|x-5|=(-10):(-5)\)
\(\Rightarrow|x-5|=2\)
\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=5+2\\x=-2+5\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}\)
Vậy \(x\in\left\{3;7\right\}\)

22 tháng 4 2020

\(|5x-7|< -18\)
Vì \(|5x-7|\ge0\forall x\)\(\Rightarrow|5x-7|\)không thể <-18