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\(f\left(x\right)=x^2-10x+27=0\Leftrightarrow x^2-10x+25+2=0\Leftrightarrow\left(x-5\right)^2+2=0\Leftrightarrow x-5=\sqrt{-2}\)=> x vô nghiệm vì không thể có cân của số âm.
\(g\left(x\right)=x^2+\frac{2}{3}x+\frac{4}{9}=0\Leftrightarrow x^2+2×\frac{1}{3}x+\frac{1}{9}+\frac{1}{3}=0\Leftrightarrow\left(x+\frac{1}{3}\right)^2+\frac{3}{9}=0\Leftrightarrow x+\frac{1}{3}=\sqrt{\frac{-3}{9}}\)=> x vô nghiệm
a) \(x^2-10x+9=x^2-x-9x+9=x\left(x-1\right)-9\left(x-1\right)=\left(x-1\right)\left(x-9\right)\)
b)\(x^2-10x+21=x^2-3x-7x+21=x\left(x-3\right)-7\left(x-3\right)=\left(x-3\right)\left(x-7\right)\) c)\(x^2-2x-3=x^2+x-3x-3=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)d)\(x^2-10x+16=x^2-2x-8x+16=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)e)\(x^2-2x-8=x^2+2x-4x-8=x\left(x+2\right)-4\left(x+2\right)=\left(x+2\right)\left(x-4\right)\)f)\(x^2-2x-48=x^2+6x-8x-48=x\left(x+6\right)-8\left(x+6\right)=\left(x+6\right)\left(x-8\right)\)g)\(x^2-10x+24=x^2-4x-6x+24=x\left(x-4\right)-6\left(x-4\right)=\left(x-4\right)\left(x-6\right)\)i)mình nghĩ câu này bị sai nên mình k giải đc. Mình nghĩ đề là \(x^4+3x^2-4\)
j)\(x^2-2x-15=x^2-3x+5x-15=x\left(x-3\right)+5\left(x-3\right)=\left(x-3\right)\left(x+5\right)\)
chúc bạn học tốt........
a,x2 - 10x + 9 = x2 - x - 9x + 9 = x(x - 1) - 9(x - 1) = (x - 9)(x - 1)
b,x2 - 10x + 21 = x2 - 3x - 7x + 21 = x(x - 3) - 7(x - 3)
c,x2 - 2x - 3 = x2 + x - 3x - 3 = x(x + 1) - 3(x + 1) = (x - 3)(x + 1)
d,x2 - 10x + 16 = x2 - 2x -8x + 16= x(x - 2) - 8(x - 2) = (x - 8)(x - 2)
e,x2 - 2x - 8 = x2 + 2x - 4x - 8 = x(x + 2) - 4(x + 2) = (x - 4)(x + 2)
f,x2 - 2x - 48 = x2 - 8x + 6x - 48 = x(x - 8) + 6(x - 8) = (x + 6)(x - 8)
g,x2 - 10x + 24 = x2 - 4x - 6x + 24 = x(x - 4) - 6(x - 4) = (x - 6)(x - 4)
j,x2 - 2x - 15 = x2 + 3x - 5x -15 = x(x + 3) - 5(x + 3) = (x - 5)(x + 3)
\(x^2-3x+4=x^2-2.x.\frac{3}{2}+\frac{9}{4}+4-\frac{9}{4}.\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)( vô nghiệm )
\(\Rightarrow x^2-3x+4\)vô nghiệm
\(x^2-3x+4\)
\(=x^2-2.x.\frac{3}{2}+\left(\frac{2}{3}\right)^2+\frac{7}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0;\frac{7}{4}>0\)
=> Đa thưc vô nghiệm
\(x^2-3x+4=x^2-2.x.\frac{3}{2}+\frac{9}{4}+4-\frac{9}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\) ( vô nghiệm )
Vậy \(x^2-3x+4\) vô nghiệm
a) \(x^2+2x+9=\left(x^2+2x+1\right)+8=\left(x+1\right)^2+8\)
Ta có :
\(\left(x+1\right)^2\ge0\)
\(\Rightarrow\left(x+1\right)^2+8\ge8>0\)
Do đó đa thức vô nghiệm.
Vậy...
b) \(y^2-y+1=\left(y^2-2.\frac{1}{2}y+\frac{1}{4}\right)+\frac{3}{4}=\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có :
\(\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
Do đó đa thức vô nghiệm.
Vậy ...
c) \(2y^2-2y+4\)
\(=2y^2-2y+\frac{1}{2}+\frac{7}{2}\)
\(=2\left(y^2-2.\frac{1}{2}.y+\frac{1}{4}\right)+\frac{7}{2}\)
\(=2\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\)
Ta có :
\(\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\ge\frac{7}{2}>0\)
Do đó đa thức vô nghiệm
Vậy...
d) \(3x^4+x^2+2\)
\(=2x^4+\left(x^4+2.\frac{1}{2}x^2+\frac{1}{4}\right)+3\)
\(=2\left(x^2\right)^2+\left(x^2+\frac{1}{2}\right)^2+3\)
Ta có :
\(\left(x^2\right)^2\ge0\)
\(\Rightarrow2\left(x^2\right)^2\ge0\)
\(\left(x^2+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(x^2\right)^2+\left(x^2+\frac{1}{2}\right)^2+3\ge3>0\)
Do đó đa thức vô nghiệm.
Vậy ...
e) \(x^2+x+1=\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có :
\(\left(x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
Do đó đa thức vô nghiệm.
Vậy ...
f) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-5\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=5\\x=1\end{cases}.}\)
g) \(x^3-x^2+2\)
\(=x^3-x^2+2x-2x+2\)
\(=\left(x^3-x\right)-\left(x^2-x\right)-2\left(x-1\right)\)
\(=x\left(x^2-1\right)-x\left(x-1\right)-2\left(x-1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-x\left(x-1\right)-2\left(x-1\right)\)
\(=\left[x\left(x+1\right)-x-2\right]\left(x-1\right)\)
\(=\left(x^2+x-x-2\right)\left(x-1\right)\)
\(=\left(x^2-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-2=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\in\left\{-\sqrt{2};\sqrt{2}\right\}\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x\in\left\{-\sqrt{2};\sqrt{2}\right\}\\x=1\end{cases}}.\)
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
a) Ta có : \(f\left(x\right)=x^2-10x+27=\left(x^2-10+25\right)+2=\left(x-5\right)^2+2\ge2>0\)
Vậy f(x) > 0 => Vô nghiệm.
b) Tương tự : \(g\left(x\right)=x^2+\frac{2}{3}x+\frac{4}{9}=\left(x^2+2.x.\frac{1}{3}+\frac{1}{9}\right)+\frac{4}{9}-\frac{1}{9}=\left(x+\frac{1}{3}\right)^2+\frac{1}{3}\ge\frac{1}{3}>0\)
Vậy g(x) > 0 => Vô nghiệm.