Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)
=>(-2x+12)(4x+12)=0
=>x=-3 hoặc x=6
b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)
=>\(x\simeq0.93\)
d: =>-4x+28+11x=-x+3x+15
=>7x+28=2x+15
=>5x=-13
=>x=-13/5
e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)
=>-9x=-3x+5
=>-6x=5
=>x=-5/6
Bài 1:
a. A = x^2 - 5x - 1
\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)
\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)
\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)
Dấu = khi x=5/2
Vậy MinC=-29/4 khi x=5/2
2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )
=>4x2-12x+9+1-16x2=-14x2+13x-3
=>-12x2-12x+10=-14x2+13x-3
=>2x2-25x+13=0
\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)
\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)
\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)
\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)
c. 4.( x - 3 ) - ( x + 2 ) = 0
=>4x-12-x-2=0
=>3x-14=0
=>3x=14
=>x=14/3
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
a) \(4x^2+16x+3=0\)
\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)
b) \(7x^2+16x+2=1+3x^2\)
\(4x^2+16x+1=0\)
\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)
c) \(4x^2+20x+4=0\)
\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow x^2+5x+1=0\)
\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)