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1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0
\(\Rightarrow\)(x3+64)-x(x2+2x+x+2)+3x2=0
\(\Rightarrow\)x3+64-x3-2x2-x2-2x+3x2=0
\(\Rightarrow\)-2x+64=0
\(\Rightarrow\)-2x=-64
\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50
\(\Rightarrow\)8x-24x2+2-6x+24x2-60x-4x+10=50
\(\Rightarrow\)-62x+12=50
\(\Rightarrow\)-62x=50-12
\(\Rightarrow\)-62x=38
\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
a: \(x^3+8x=5x^2+4\)
=>\(x^3-5x^2+8x-4=0\)
=>\(x^3-x^2-4x^2+4x+4x-4=0\)
=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>\(\left(x-1\right)\left(x-2\right)^2=0\)
=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2: \(x^3+3x^2=x+6\)
=>\(x^3+3x^2-x-6=0\)
=>\(x^3+2x^2+x^2+2x-3x-6=0\)
=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
3: ĐKXĐ: x>=0
\(2x+3\sqrt{x}=1\)
=>\(2x+3\sqrt{x}-1=0\)
=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)
=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)
=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)
=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)
4: \(x^4+4x^2+1=3x^3+3x\)
=>\(x^4-3x^3+4x^2-3x+1=0\)
=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)
=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)
=>(x-1)^2=0
=>x-1=0
=>x=1
a.
\(x^3+8x=5x^2+4\)
\(\Leftrightarrow x^3-5x^2+8x-4=0\)
\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
b.
\(x^3+3x^2-x-6=0\)
\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)
\(\Leftrightarrow x\left(x^2+x-3\right)+2\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)
\(3x^2-6x+3-3x^2+15x-2=0\)
\(9x+1=0\)
\(x=-\frac{1}{9}\)
\(b.4x^2-12x+9=0\)
\(4x^2-6x-6x+9=0\)
\(2x\left(x-3\right)-3\left(x-3\right)=0\)
\(\left(2x-3\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
\(c.\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\left(x-8\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
Câu a : \(\left(3x+1\right)\left(4x-2\right)-12x\left(x-3\right)=5\)
\(\Leftrightarrow12x^2-2x-2-12x^2+36x-5=0\)
\(\Leftrightarrow34x-7=0\)
\(\Leftrightarrow x=\dfrac{7}{34}\)
Vậy \(x=\dfrac{7}{34}\)
Câu b : \(\left(2x+1\right)\left(4x-2\right)-\left(8x+2\right)\left(x+5\right)=4\)
\(\Leftrightarrow8x^2-2-8x^2-42x-10-4=0\)
\(\Leftrightarrow-42x-16=0\)
\(\Leftrightarrow x=-\dfrac{16}{42}=-\dfrac{8}{21}\)
Vậy \(x=-\dfrac{8}{21}\)
Chúc bạn học tốt !
\(a,2\left(x^3-1\right)-2x^2\left(x+2x^4\right)+x\left(4x^5+4\right)=6\\ \Leftrightarrow2x^3-2-2x^3-4x^6+4x^6+4x-6=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow x=2\\ b,\left(2x\right)^2\left(4x-2\right)-\left(x^3-8x^3\right)=15\\ \Leftrightarrow4x^2\left(4x-2\right)+7x^3-15=0\\ \Leftrightarrow16x^3-8x^2+7x^3-15=0\\ \Leftrightarrow23x^3-8x^2-15=0\\ \Leftrightarrow23x^3-23x^2+15x^2-15x+15x-15=0\\ \Leftrightarrow\left(x-1\right)\left(23x^2+15x-15\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\left(23x^2+15x-15>0\right)\end{matrix}\right.\)
Bài 1:
a: Ta có: \(2\left(x^3-1\right)-2x^2\left(2x^4+x\right)+x\left(4x^5+4\right)=6\)
\(\Leftrightarrow2x^3-2-4x^6-2x^3+4x^6+4x=6\)
\(\Leftrightarrow4x=8\)
hay x=2
b: Ta có: \(\left(2x\right)^2\cdot\left(4x-2\right)-\left(x^3-8x^3\right)=15\)
\(\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^3=15\)
\(\Leftrightarrow16x^3-8x^2+7x^3=15\)
\(\Leftrightarrow23x^3-8x^2-15=0\)
\(\Leftrightarrow23x^3-23x^2+15x^2-15=0\)
\(\Leftrightarrow23x^2\left(x-1\right)+15\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(23X^2+15x+15\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
1: =>x^2-x=3-x
=>x^2=3
=>x=căn 3 hoặc x=-căn 3
2: =>x^2-4x+3=x^2-4x+4 và x>=2
=>3=4(vô lý)
3: =>2|x-1|=6
=>|x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2 hoặc x=4
4: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc x=5/3
5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
=>x+2=0
=>x=-2
1: \(\Leftrightarrow x^2-25-x^2-8x-16+\left(4x+1\right)^3=64x^3+8+48x^2-12x\)
\(\Leftrightarrow-8x-41+64x^3+48x^2+12x+1=64x^3+48x^2-12x+8\)
=>4x-40=-12x+8
=>16x=48
hay x=3
2: \(\Leftrightarrow12x^2-48x-x^3+1+x^3-12x^2+48x-64=x^2-2x-3-x^2-10x-25\)
\(\Leftrightarrow-63=-12x-28\)
=>12x+28=63
=>12x=35
hay x=35/12