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1) đặt 2x+1 = a => \(a^4-3a^2+2=\left(a^2-1\right)\left(a^2-2\right)=\)\(\left(a-1\right)\left(a+1\right)\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\)
=(2x+1-1)(2x+1+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\)) = 4x(x+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\))
2) =(x2-x)(x2-x-2)-3
đặt x2-x = b => b(b-2)-3 = b2-2b-3 = (b+1)(b-3) = (x2-x+1)(x2-x-3)
3) đặt x2+2x-1 = c => c2-3xc+2x2 = (c-x)(c-2x) = (x2+2x-1-x)(x2+2x-1-2x) = (x2+x-1)(x2-1) = (x2+x-1)(x-1)(x+1)
tìm x
x3-8 +(x-2)(x+1)=0 <=> (x-2)(x2+2x+4)+(x-2)(x+1)=0 <=>(x-2)(x2+2x+4+x+1)=0 <=> x=2 (vì x2+3x+5= (x+\(\frac{3}{2}\))2 +\(\frac{11}{4}\)>0)
vậy x=2
2) \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)-3\)
\(=\left(x^2-x\right)\left(x^2-x-2\right)-3\)(1)
Đặt \(x^2-x=t\)
\(\Rightarrow\left(1\right)=t\left(t-2\right)-3=t^2-2t+1-4\)
\(=\left(t-1\right)^2-4\)
\(=\left(t+3\right)\left(t-5\right)\)
Thay \(x^2-x=t\), ta được:
\(BTDNT=\left(x^2-x+3\right)\left(x^2-x-5\right)\)
Câu a đơn giản
b)
\(A=\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+\frac{x^2}{4}\right)+\left(\frac{11}{4}x^2+2x+\frac{4}{11}\right)+1-\frac{4}{11}}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x^2+x+0,25\right)+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
Vì \(\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]>0\)và \(\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}>0\)
nên \(A>0\)hay A ko âm
Nhớ k nha !
a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)
=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)
=> \(6x+6+3x-6=12-8x+8\)
=> \(17x=20\)
=> \(x=\frac{20}{17}\)
b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)
=> \(4\left(11x-1\right)=6\left(6-x\right)\)
=> \(44x-4-36+6x=0\)
=> \(\)\(50x=40\)
=> \(x=\frac{4}{5}\)
c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
Bài 3:
a: =>6x(x^2-4)=0
=>x(x-2)(x+2)=0
hay \(x\in\left\{0;2;-2\right\}\)
b: \(\Leftrightarrow9\left(x^2-1\right)-9x^2+6x-1=2\)
=>9x^2-9-9x^2+6x-1=2
=>6x-10=2
=>6x=12
=>x=2
a) Ta có :
\(3x=3\left(x+2\right)\)
\(\Leftrightarrow3x=3x+2\)
\(\Leftrightarrow0=2\) ( vô lí )
Do đó pt đã cho vô nghiệm
b) Ta có \(\left|x\right|=-x^2-2\) (1)
Nhân xét : VT (1) : \(\left|x\right|\ge0\forall x\)
VP (1) : \(-x^2\le0\Leftrightarrow-x^2-2\le-2\forall x\)
Do đó : \(VT\ne VP\)
Vì vậy pt đã cho vô nghiệm
\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)