(x^2 + x -6) . (x^2 + x -4) =0
e cảm ơn trước ạ
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`4(x-6)-x^2 (2+3x)+x(5x-4)+3x^2 (x-1)`
`=4x-24-2x^2 -3x^3 +5x^2-4x+3x^3-3x^2`
`=-24`
\(4\left(x-6\right)-2x\left(2+3x\right)+x\left(5x-4\right)+3x2\left(x-1\right)\\ =4x-24-4x-6x^2+5x^2-4x+6x^2+6x\\ =2x+5x^2-24\)
a: =>2x>-6
hay x>-3
e: =>(5-x)/x<0
=>0<x<5
h: \(\Leftrightarrow\dfrac{x+5-x-3}{x+3}< 0\)
\(\Leftrightarrow x+3< 0\)
hay x<-3
g: \(\Leftrightarrow\dfrac{2x+7}{x+4}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{7}{2}\\x< -4\end{matrix}\right.\)
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(5^n.25=125^2\)
\(\Rightarrow5^n.5^2=\left(5^3\right)^2\)
\(\Rightarrow5^n.5^2=5^6\)
\(\Rightarrow5^n=5^6:5^2\)
\(\Rightarrow5^n=5^4\)
\(\Rightarrow n=4\)
Vậy \(n=4.\)
b) \(3^n.9^2=27^3\)
\(\Rightarrow3^n.\left(3^2\right)^2=\left(3^3\right)^3\)
\(\Rightarrow3^n.3^4=3^9\)
\(\Rightarrow3^n=3^9:3^4\)
\(\Rightarrow3^n=3^5\)
\(\Rightarrow n=5\)
Vậy \(n=5.\)
c) \(2^4.4^n=8^6\)
\(\Rightarrow\left(2^2\right)^2.4^n=2^{18}\)
\(\Rightarrow4^2.4^n=\left(2^2\right)^9\)
\(\Rightarrow4^2.4^n=4^9\)
\(\Rightarrow4^n=4^9:4^2\)
\(\Rightarrow4^n=4^7\)
\(\Rightarrow n=7\)
Vậy \(n=7.\)
Chúc bạn học tốt!
1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
\(a,\Rightarrow3x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ c,Đề.sai\\ d,Sửa:\left(x-2\right)^2-16\left(5-2x\right)^2=0\\ \Rightarrow\left[x-2-4\left(5-2x\right)\right]\left[x-2+4\left(5-2x\right)\right]=0\\ \Rightarrow\left(x-2-20+8x\right)\left(x-2+20-8x\right)=0\\ \Rightarrow\left(9x-22\right)\left(18-7x\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{9}\\x=\dfrac{18}{7}\end{matrix}\right.\)
\(\dfrac{x^2+4}{4}\ge x\)
\(\Leftrightarrow\dfrac{4\left(x^2+4\right)}{4}\ge4x\)
\(\Leftrightarrow x^2+4\ge4x\)
\(\Leftrightarrow x^2-4x+4\ge0\)
\(\Leftrightarrow\left(x-2\right)^2\ge0\) (Luôn đúng)
Vậy đẳng thức ban đầu được chứng minh.
\(\dfrac{x^2+4}{4}\ge x\)
\(\Leftrightarrow\dfrac{x^2+4}{4}\ge\dfrac{4x}{4}\)
\(\Leftrightarrow x^2+4+4x\ge0\)
\(\Leftrightarrow\left(x+2\right)^2\ge0\) (luôn đúng)
4-2(x+1)=-x
<=>4-2x-2=-x
<=>-2x+x=-4+2
<=>-x =-2
<=>x=2
4-2(x+1)=-x
4 - 2x + 2 = -x
-2x + x = -2 -4
-x = -6
x = 6
hok tốt!
Ko hiểu cách này, ib chỉ cho cách khác nhé ! ( ko thể hiện )
\(\left(x^2+x-6\right)\left(x^2+x-4\right)=0\)
TH1 : \(x^2+x-6=0\)
\(\Delta=1^2-4.\left(-6\right)=1+24=25>0\)
Nên phương trình có 2 nghiệm phân biệt
\(x_1=\frac{-1-\sqrt{25}}{2}=\frac{-1-5}{2}=-\frac{6}{2}=-3\)
\(x_2=\frac{-1+\sqrt{25}}{2}=\frac{-1+5}{2}=\frac{4}{2}=2\)
TH2 : \(x^2+x-4=0\)
\(\Delta=1^2-4.\left(-4\right)=1+16=17>0\)
Nên phương trình có 2 nghiệm phân biệt
\(x_1=\frac{-1-\sqrt{17}}{2};x_2=\frac{-1+\sqrt{17}}{2}\)