K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

22 tháng 10 2016

\(4x^2-9-x\left(2x-3\right)=0\)

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

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

\(\Leftrightarrow\left(2x-3\right)\left[\left(2x+3\right)-x\right]=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-3\end{cases}}\)

23 tháng 10 2016

-_- bài này hôm qua lm rùi

30 tháng 12 2016

b)   ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0

<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0

<=> ( 2x - 3 )( 1 + x - 1 ) = 0

<=> x( 2x - 3 ) = 0

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

Vậy .....

30 tháng 12 2016

a, 25x^2 - 1 - (5x -1)(x+2)=0

=> (5x)^2 - 1 + (5x-1)(x+2) = 0

=> (5x-1)(5x+1) + (5x-1)(x+2) = 0

=> (5x-1)(5x+1+x+2) = 0

=> (5x-1)(6x+3) = 0

=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)

14 tháng 10 2018

a) \(x^2-4x=0\)

\(x\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

b) \(4x^2-9=0\)

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

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

d) \(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-2\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

e) \(\left(2x-1\right)^2-\left(x+2\right)^2=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}}\)

14 tháng 10 2018

\(x^2-4x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-4=0\Leftrightarrow x=4\end{cases}}\)

\(4x^2-9=0\)

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

\(\left(2x\right)^2-9=0\)

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

\(\Rightarrow\orbr{\begin{cases}\left(2x\right)^2=\left(-3\right)^2\\\left(2x\right)^2=3^2\end{cases}\Rightarrow\orbr{\begin{cases}2x=-3\\2x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{3}{2}\end{cases}}}}\)

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

\(\left(x-3\right)\cdot\left(2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+3\\2x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{2}\end{cases}}}\)

\(x\left(2x+9\right)-4x-18=0\)

\(x\left(2x+9\right)-\left(4x+18\right)=0\)

\(x\left(2x+9\right)-\left(2\cdot2x+2\cdot9\right)=0\)

\(x\left(2x+9\right)-2.\left(2x+9\right)=0\)

\(\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}2x+9=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-9\\x=0+2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-9}{2}\\x=2\end{cases}}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x-1=x+2\\2x-1=-x+2\end{cases}\Rightarrow\orbr{\begin{cases}2x=3+x\\2x=-x+3\end{cases}\Rightarrow\orbr{\begin{cases}2x-x=3\\2x+x=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}}}\)

\(\)

17 tháng 9 2018

\(a)\)\(x^3-x^2-x+1=0\)

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)

Vậy \(x=1\) hoặc \(x=-1\)

Chúc bạn học tốt ~ 

17 tháng 9 2018

a) x3-x2-x+1 = 0 \(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)\(\Leftrightarrow x^2-1=0\)hoặc x-1=0 

\(\Leftrightarrow x=1\)

23 tháng 7 2015

 

4x^2-9-x.(2x-3)=0

<=>4x2-9-2x2+3x=0

<=>2x2+3x-9=0

<=>2x2-3x+6x-9=0

<=>x.(2x-3)+3.(2x-3)=0

<=>(2x-3)(x+3)=0

<=>2x-3=0 hoặc x+3=0

<=>x=3/2 hoặc x=-3

 

5 tháng 8 2016

1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

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

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

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

\(\Leftrightarrow-4x+10=0\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}.\)

Vậy \(S=\left\{\frac{5}{2}\right\}\)

2)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right).\left(x^2-3x+9+x-9\right)=0\)

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

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

\(\Leftrightarrow x+3=0\)hoặc \(x=0\)hoặc \(x-2=0\)

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

Vậy \(S=\left\{-3;0;2\right\}\)

30 tháng 10 2016

\(d,x\left(x-3\right)-7x+21=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)

30 tháng 10 2016

\(a,2x\left(x-7\right)+5x-35=0\)

 \(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)

\(c,4x^2+12x+9=0\)

\(\Leftrightarrow4x^2+6x+6x+9=0\)

\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)

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

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow x=-\frac{3}{2}\)