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\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.1-\left(x-1\right)^2.\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1-1\right)\left(x-1+1\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=2;x=0\end{cases}}}\)
Vậy: \(x\in\left\{1;2;0\right\}\)
Vì \(x=9\Rightarrow x+1=10\)
Thay x+1=10 vào biểu thức C ta dduojcw :
\(C=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-...-\left(x+1\right)x+10\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...-x^2-x+10\)
\(=-x+10\)
\(=-9+10\)
\(=1\)
A) 4x^2 - 3x -7 = 4x^2 + 4x - 7x - 7
=(x +1)(4x - 7) =0
=>x+1=0 <=> x=-1
hoac 4x-7=0 <=> x=7/4
Nhu cau sau lam tuong tu
a)\(27^6:9^3=\left(3^3\right)^6:\left(3^2\right)^3=3^{18}:3^6=3^{12}\)
b)\(24^n:2^{2n}=\left(2^3.3\right)^n:2^{2n}=2^{3n}.3^n:2^{2n}=2^{3n-2n}.3^n=2^n.3^n=6^n\)
c)\(32^4:8^6=\left(2^5\right)^4:\left(2^3\right)^6=2^{20}:2^{18}=2^2\)
a)Đặt \(A=8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
\(A< 8^9+8^9+8^9+8^9+8^9+8^9+8^9+8^9\)
\(A< 8\cdot8^9\)
\(A< 8^{10}< 9^{10}\)
\(\Rightarrow9^{10}>8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
a) \(8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
(8+7+6+5+4+3+2+1)9
369
Vậy369>99
\(\left(2^3\cdot9^4+9^3+45\right):\left(9^2\cdot10-9^2\right)\)
\(=\dfrac{9^3\cdot\left(2^3\cdot9+1\right)+45}{9^3}\)
\(=\dfrac{9^3\cdot73+45}{9^3}=\dfrac{5918}{81}\)