(81-1^2)(81-2^2)(81-3^2)....(81-9^2)

= (81-1^2)(81-2^2)(81-3^2)....(81-81)

= (81-1^2)(81-2^2)(81-3^2)....0

= 0

a. 96 : 8 : 2 = 6

96 : ( 8 x 2 ) = 6

96 : 8 : 2 = 96 : ( 8 x 2 )

b. 81 : 9 : 3 = 3

81 : ( 9 x 3 ) = 3

81 : 9 : 3 = 81 : ( 9x 3)

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

=14-x=1

tu hoc moi gioi

nói vậy thì những bài khó tự đi mà làm ngồi đó mà sĩ

Dễ thấy (\(\frac{3}{4}\)-81); (\(\frac{3^2}{5}\)-81); (\(\frac{3^3}{6}\)-81);... (\(\frac{3^{2007}}{2010}\)-81) có dạng (\(\frac{3^x}{3+x}\)-81) và x\(\varepsilon\){1;2;3;...2007}.

Nếu x=6 thì \(\frac{3^x}{3+x}\)-81=\(\frac{3^6}{3+6}\)-81=0

=>  (\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...​(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)=0

Mà |x-30|-6001=(\(\frac{3}{4}\)-81) (\(\frac{3}{4}\)-81)(\(\frac{3^3}{6}\)-81)...​(\(\frac{3^6}{3+6}\)-81)...(\(\frac{3^{2007}}{2010}\)-81)

=>|x-30|-6001=0

=>|x-30|=6001

=>x-30=6001 hoặc x-30=-6001

=>x=6031 hoặc x=-5971

-------------------The end----------------

\(\text{|x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\text{ |x - 30| - 6001 = }\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^6}{9}-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\left|x-30\right|- 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(3^4-3^4\right)...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow|x - 30| - 6001 = \left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...0...\left(\frac{3^{2007}}{2010}-81\right)\)

\(\Rightarrow\text{|x - 30| - 6001 = }0\)

\(\Rightarrow\left|x-30\right|=6001\) 

\(\Rightarrow x-30=6001\)hoặc \(x-30=-6001\)

\(\Rightarrow x=6031\)hoặc\(x=-5971\)

Vậy: x= 6031 hoặc x= -5971

(Nói thật thì mình mới lớp 7, đây có phải của lớp 8 không?)

a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)

\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)

\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)

\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)

\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)

b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)

\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)

\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)

\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)

\(\Leftrightarrow x=\frac{1}{-3}\)

c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)

\(\Leftrightarrow x=\frac{5}{6}\)

d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)

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

\(f\left(x\right)=4x^2+3x+1\)

\(g\left(x\right)=3x^2-2x+1.\)

a) \(h\left(x\right)=f\left(x\right)-g\left(x\right)\)

\(\Rightarrow h\left(x\right)=\left(4x^2+3x+1\right)-\left(3x^2-2x+1\right)\)

\(\Rightarrow h\left(x\right)=4x^2+3x+1-3x^2+2x-1\)

\(\Rightarrow h\left(x\right)=\left(4x^2-3x^2\right)+\left(3x+2x\right)+\left(1-1\right)\)

\(\Rightarrow h\left(x\right)=x^2+5x.\)

b) Ta có \(h\left(x\right)=x^2+5x.\)

Đặt \(x^2+5x=0\)

\(\Rightarrow x.\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x=0\) và \(x=-5\) là các nghiệm của đa thức \(h\left(x\right).\)

Chúc bạn học tốt!

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