Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
Bài 1:
b: \(x^3-4x^2+7x-6=0\)
\(\Leftrightarrow x^3-2x^2-2x^2+4x+3x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-2x+3\right)=0\)
=>x-2=0
hay x=2
c: \(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2\left(x+1\right)\left(x^2-x+1\right)+7x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2-2x+2+7x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+4x+x+2\right)=0\)
=>(x+1)(x+2)(2x+1)=0
hay \(x\in\left\{-1;-2;-\dfrac{1}{2}\right\}\)
d: \(2x^3-9x+2=0\)
\(\Leftrightarrow2x^3-4x^2+4x^2-8x-x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+1-\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1+\dfrac{\sqrt{6}}{2}\right)\left(x+1-\dfrac{\sqrt{6}}{2}\right)=0\)
hay \(x\in\left\{2;-1-\dfrac{\sqrt{6}}{2};-1+\dfrac{\sqrt{6}}{2}\right\}\)
c)\(4x^4-101x^2+25=0\)
\(\Leftrightarrow4x^4-100x^2-x^2+25=0\)
\(\Leftrightarrow4x^2\left(x^2-25\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\\x=5\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};\frac{-1}{2};5;-5\right\}\)
a)\(\left(2x-1\right)^2=x+5\)
\(\Leftrightarrow4x^2-4x+1=x+5\)
\(\Leftrightarrow4x^2-5x-4=0\)
\(\Leftrightarrow4\left(x^2-\frac{5}{4}x-1\right)=0\)
\(\Leftrightarrow4\left(x^2-\frac{5}{4}x+\frac{25}{64}-\frac{89}{64}\right)=0\)
\(\Leftrightarrow4\left[\left(x-\frac{5}{8}\right)^2-\frac{89}{64}\right]=0\)
\(\Leftrightarrow4\left(x-\frac{5}{8}+\frac{\sqrt{89}}{8}\right)\left(x-\frac{5}{8}-\frac{\sqrt{89}}{8}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5}{8}+\frac{\sqrt{89}}{8}=0\\x-\frac{5}{8}-\frac{\sqrt{89}}{8}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5-\sqrt{89}}{8}\\x=\frac{5+\sqrt{89}}{8}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5-\sqrt{89}}{8};\frac{5+\sqrt{89}}{8}\right\}\)
1. a)\(x^2+x-3x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
a/=> 9x2 - 6x + 1 - (9x2 + 12x + 4)=0 => 9x2 - 6x + 1 - 9x2 - 12x - 4 =0 => -18x - 3 =0 => -18x = 3 => x = -1/6 b/=>4x2 + 4x + 1 - (x2 - 2x + 1)=0 => 4x2 + 4x + 1 - x2 + 2x - 1 =0 => 3x2 + 6x =0 => 3x(x+2)=0 => trường hợp 1: 3x=0=>x=0 ; trường hợp 2: x+2=0=>x=-2 c/=> x2 - 2*2*x + 22=0 => (x - 2)2 =0 => x-2=0 => x=2 d/=> x2 - 2*5*x + 52 =0 => (x - 5)2 =0 => x-5=0 => x=5 e/=> 9x2 + 6x - 3 =0 => 9x2 - 3x + 9x - 3 =0 => 3x(3x - 1) + 3(3x - 1) =0 => (3x + 3)(3x - 1) =0 => trường hợp1: 3x+3=0 =>3x=-3=>x=-1 ; trường hợp2: 3x-1=0=>x=1/3