Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
a); b) Do tích = 0
=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)
=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)
a; *x-1=0 <=>x=1
*2x+5=0 <=>x=-2,5
*x2+2=0 <=> ko có x
b; tương tự a
T ko biết làm, chỉ hỏi liên thiên thôi :)))
Hủ phải không???? OvO Dưa Trong Cúc
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
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