a) 4x2+8x=√2x+6 b) x2-10x-12=4√2+3
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a: (x^2+x)^2+4x^2+4x-12
=(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
b: =(x^2+8x)^2+22(x^2+8x)+105+15
=(x^2+8x)^2+22(x^2+8x)+120
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)
c: =8x^2+12x-2x-3
=(2x+3)(4x-1)
a: =(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
b: =(x^2+8x)^2+22(x^2+8x)+120
=(x^2+8x+12)(x^2+8x+10)
=(x+2)(x+6)(x^2+8x+10)
c: =8x^2+12x-2x-3
=(2x+3)(4x-1)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)
a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)
\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)
\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)
b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)
\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)
\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)
\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
c, \(4x^2-2x-1=0\)
\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)
\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)
\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)
\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
d,\(x^4-4x^2-32=0\)
đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)
\(< =>t^2-2.2t+4-6^2=0\)
\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)
\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)