\(a,f\left(x\right)=x^2-10x+27\)

\(\Rightarrow f\left(x\right)=x^2-5x-5x+25+2\)

\(\Rightarrow f\left(x\right)=x\left(x-5\right)-5\left(x-5\right)+2\)

\(\Rightarrow f\left(x\right)=\left(x-5\right)^2+2\ge2>0\)  (Vì \(\left(x-5\right)^2\ge0\)  \(Vx\) )

Vậy đa thức f(x) vô nghiệm

\(b,g\left(x\right)=x^2+\frac{2}{3}x+\frac{4}{9}\)

\(\Rightarrow g\left(x\right)=x^2+\frac{1}{3}x+\frac{1}{3}x+\frac{1}{9}+\frac{3}{9}\)

\(\Rightarrow g\left(x\right)=x\left(x+\frac{1}{3}\right)+\frac{1}{3}\left(x+\frac{1}{3}\right)+\frac{1}{3}\)

\(\Rightarrow g\left(x\right)=\left(x+\frac{1}{3}\right)^2+\frac{1}{3}\ge\frac{1}{3}>0\)  (Vì \(\left(x+\frac{1}{3}\right)^2\ge0\)  \(Vx\) )

Vậy đa thức g(x) vô nghiệm

TH1:a=0; b=0

TH2:a=2; b=2

nếu ab=a.b 

\(M=\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+\frac{20}{832}\)

\(\Rightarrow\frac{1}{M}=\frac{112}{20}+\frac{280}{20}+\frac{520}{20}+\frac{832}{20}\)

\(\Rightarrow\frac{1}{M}=\frac{1744}{20}=\frac{436}{5}\)

\(\Rightarrow M=\frac{5}{436}\)

\(a,\left(x+3\right)^2=\left(x-2\right)\left(x+4\right)\)

\(\Rightarrow\left(x+3\right)^2=x^2+2x-8\)    \(\Rightarrow\left(x+3\right)^2=x^2+2x+1-9\)

\(\Rightarrow\left(x+3\right)^2=\left(x+1\right)^2-9\)   \(\Rightarrow\left(x+1\right)^2-\left(x+3\right)^2=9\)

\(\Rightarrow\left(x+1+x+3\right)\left(x+1-x-3\right)=9\)

\(\Rightarrow\left(2x+4\right)\left(-2\right)=9\)   \(\Rightarrow2x+4=-\frac{9}{2}\)

\(\Rightarrow2x=-\frac{17}{2}\)    \(\Rightarrow x=-\frac{17}{4}\)

\(b,\left(x+4\right)^2=2x^2+16\)

\(\Rightarrow\left(x+4\right)^2=x^2+x^2+4^2\)

\(\Rightarrow\left(x+4\right)^2=x^2+\left(x+4\right)^2-8x\)

\(\Rightarrow\left(x+4\right)^2=x\left(x-8\right)+\left(x+4\right)^2\)

\(\Rightarrow x\left(x-8\right)=0\)   \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=0\\x-8=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=0\\x=8\end{array}\right.\)

a, \(3-17+x=289-36-289\)

\(\Rightarrow x-14=-36\)

\(\Rightarrow x=-22\)

\(b,x-73=\left(-35\right)+\left|-55\right|\)

\(\Rightarrow x-73=\left(-35\right)+55\)

\(\Rightarrow x-73=20\)

\(\Rightarrow x=93\)

\(c,\left|x\right|-3=2\)

\(\Rightarrow\left|x\right|=5\)

\(\Rightarrow x=\pm5\)

\(d,\left|x+1\right|=5\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+1=5\\x+1=-5\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=4\\x=-6\end{array}\right.\)

\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)

\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)

\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)

\(\Rightarrow-14x-60=0\)

\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)

\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)

\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)

\(\Rightarrow-x^2+12x-32=7x-x^2-10\)

\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)

\(\Rightarrow5x-22=0\)

\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)

a) (x - 1) (x² + x + 1) - (x + 1) (x² - x + 1) + 2(x - 1) (x + 1) - 2(x + 2)²

= x³ - 1 -  x³ - 1 + 2(x² - 1) - 2(x² + 4x + 4)

= -2 + 2x² -  2 - 2x² - 8x - 8

= -12  

Gọi số học sinh giỏi lớp 6a là a, 6b là b

Theo bài ra ta có: \(a=\frac{2}{3}b\) 

\(a-3=\frac{3}{7}\left(b+3\right)\Rightarrow\frac{2}{3}b-3=\frac{3}{7}b+\frac{9}{7}\)

\(\Rightarrow\frac{2}{3}b-\frac{3}{7}b=3+\frac{9}{7}\Rightarrow\frac{5}{21}b=\frac{30}{7}\)

\(\Rightarrow b=\frac{30}{7}:\frac{5}{21}=\frac{30}{7}.\frac{21}{5}=18\)

\(\Rightarrow a=18.\frac{2}{3}=12\)

Vậy số học sinh giỏi lớp 6a là 12 em, 6b là 18 em