a) \(x^2+x=0\)

\(\Rightarrow x\left(x+1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x+1=0\Rightarrow x=-1\)

Vậy nghiệm của đa thức \(x^2+x\) là \(x=0;x=-1\)

b) Làm tương tự như trên.

Đ/s: \(x=0;x=1\)

\(2x\left(3x+1\right)+3x\left(4-2x\right)=7\)

\(\Rightarrow6x^2+2x+12x-6x^2=7\)

\(\Rightarrow\left(6x^2-6x^2\right)+\left(2x+12x\right)=7\)

\(\Rightarrow14x=7\)

\(\Rightarrow x=\dfrac{1}{2}\)

Ta có: \(a+b+c=0\Rightarrow c=-\left(a+b\right)=-a-b\)

Đặt \(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow f\left(x\right)=ax^2+bx-a-b\)

\(\Rightarrow f\left(x\right)=ax^2-ax+ax+bx-a-b\)

\(\Rightarrow f\left(x\right)=\left(ax^2-ax\right)+\left(ax-a\right)+\left(bx-b\right)\)

\(\Rightarrow f\left(x\right)=ax\left(x-1\right)+a\left(x-1\right)+b\left(x-1\right)\)

\(\Rightarrow f\left(x\right)=\left(x-1\right).\left(ax+a+b\right)\)

\(f\left(x\right)=0\Rightarrow\left(x-1\right).\left(ax+a+b\right)=0\)

\(\Rightarrow x-1=0\) hoặc \(ax+a+b=0\)

+) \(x-1=0\Rightarrow x=1\)

+) \(ax+a+b=0\)

\(\Rightarrow a\left(x+1\right)=-b\)

\(\Rightarrow x=\dfrac{-b}{a}-1\)

\(\Rightarrow f\left(x\right)\) sẽ có 1 nghiệm là \(x=1\)

\(\Rightarrowđpcm\)

Xét \(\Delta ABD\) và \(\Delta ACD\) có:

\(AB=AC\left(gt\right)\)

\(BD=CD\left(gt\right)\)

\(AD\) là cạnh chung

\(\Rightarrow\Delta ABD=\Delta ACD\left(c.c.c\right)\)

\(\Rightarrow\widehat{BAD}=\widehat{CAD}\) (2 góc tương ứng)

Gọi giao điểm của \(AD\) và \(BC\) là \(I\)

Xét \(\Delta ABI\) và \(\Delta ACI\) có:

\(AB=AC\left(gt\right)\)

\(AI\) là cạnh chung

\(\widehat{BAI}=\widehat{CAI}\left(cmt\right)\)

\(\Rightarrow\Delta ABI=\Delta ACI\left(c.g.c\right)\)

\(\Rightarrow\widehat{AIB}=\widehat{AIC}\) (2 góc tương ứng)

Mà \(\widehat{AIB}+\widehat{AIC}=180^0\) (2 góc kề bù)

\(\Rightarrow\widehat{AIB}+\widehat{AIB}=180^0\)

\(\Rightarrow2.\widehat{AIB}=180^0\Rightarrow\widehat{AIB}=90^0\)

\(\Rightarrow AD\perp BC\left(đpcm\right)\)

\(\sqrt{x^2}.\left(x^2-1\right)=0\)

\(\Rightarrow\sqrt{x^2}=0\) hoặc \(x^2-1=0\)

+) \(\sqrt{x^2}=0\Rightarrow x=0\)

+) \(x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\)

Vậy \(x\in\left\{-1;0;1\right\}\)

+) Với \(x=3\) ta được:

\(P\left(x\right)=3^2-6.3+9\)

\(\Rightarrow P\left(x\right)=9-18+9\)

\(\Rightarrow P\left(x\right)=\left(9+9\right)-18\)

\(\Rightarrow P\left(x\right)=18-18=0\)

Vậy \(P\left(x\right)=0\) tại \(x=3\)

+) Với \(x=-3\) ta được:

\(P\left(x\right)=\left(-3\right)^2-6.\left(-3\right)+9\)

\(\Rightarrow P\left(x\right)=9-\left(-18\right)+9\)

\(\Rightarrow P\left(x\right)=9+18+9\)

\(\Rightarrow P\left(x\right)=36\)

Vậy \(P\left(x\right)=36\) tại \(x=-3\)

\(\frac{7}{12}-\left(\frac{5}{12}-\frac{5}{6}\right)\)

\(=\frac{7}{12}-\frac{5}{12}+\frac{5}{6}\)

\(=\frac{2}{12}+\frac{5}{6}\)

\(=\frac{1}{6}+\frac{5}{6}\)

\(=\frac{6}{6}=1\)

Ta có:

\(\overline{X}=\frac{5.2+6.5+9.n+10.1}{2+5+n+1}=6,8\)

\(\Rightarrow\overline{X}=\frac{10+30+9n+10}{8+n}=6,8\)

\(\Rightarrow\overline{X}=\frac{50+9n}{8+n}=6,8\)

\(\Rightarrow6,8.\left(8+n\right)=50+9n\)

\(\Rightarrow54,4+6,8n=50+9n\)

\(\Rightarrow54,4-50=9n-6,8n\)

\(\Rightarrow4,4=2,2n\)

\(\Rightarrow n=2\)

Ta có:

\(x\left(y+z\right)=32\Rightarrow xy+xz=32\)

\(y\left(x+z\right)=27\Rightarrow xy+yz=27\)

\(z\left(y+x\right)=yz+xz=35\)

\(\Rightarrow xy+xz+xy+yz+yz+xz=32+27+35\)

\(\Rightarrow2\left(xy+yz+zx\right)=94\)

\(\Rightarrow xy+yz+zx=47\)

Mà \(xy+yz=27\)

\(\Rightarrow27+zx=47\)

\(\Rightarrow zx=20\)

Tương tự ta được : \(xy=12\) ; \(yz=15\)

\(\Rightarrow zx.xy=20.12\)

\(\Rightarrow x^2.yz=240\)

Mà \(yz=15\)

\(\Rightarrow x^2=240:15\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=\pm4\)

+)Nếu \(x=4\Rightarrow xyz=4.yz=4.15=60\)

+)Nếu \(x=-4\Rightarrow xyz=-4.yz=-4.15=-60\)

Vậy \(xyz=60;xyz=-60\)