10x+56-8=102+12x
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12x - 10x = 5^8 : 5^4 + 201^0
2x = 5^4 + 1
2x = 625 + 1
2x = 626
x = 626 : 2
x = 313
\(12x-10x=5^8:5^4+201^0\)
\(\Rightarrow x\left(12-10\right)=5^4+1\)
\(\Rightarrow2x=625+1=626\)
\(\Rightarrow x=626:2=313\)
Vậy x = 313
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a:
ĐKXĐ: \(x^2+3x>=0\)
=>x(x+3)>=0
=>\(\left[{}\begin{matrix}x>=0\\x< =-3\end{matrix}\right.\)
\(\sqrt{16}-\sqrt{x^2+3x}=0\)
=>\(\sqrt{x^2+3x}=\sqrt{16}\)
=>x^2+3x=16
=>x^2+3x-16=0
\(\text{Δ}=3^2-4\cdot1\cdot\left(-16\right)=9+64=73>0\)
Do đó: Phương trình có 2 nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{73}}{2}\\x_2=\dfrac{-3+\sqrt{73}}{2}\end{matrix}\right.\)
b:
ĐKXĐ: \(x\in R\)
\(3x-1-\sqrt{4x^2-12x+9}=0\)
=>\(\sqrt{\left(2x-3\right)^2}=3x-1\)
=>\(\left\{{}\begin{matrix}3x-1>=0\\\left(3x-1\right)^2=\left(2x-3\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(3x-1-2x+3\right)\left(3x-1+2x-3\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(x+2\right)\left(5x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{4}{5}\left(nhận\right)\end{matrix}\right.\)
c:
ĐKXĐ: \(\left\{{}\begin{matrix}x^2-6x+8>=0\\2x^2-10x+11>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\x< =2\end{matrix}\right.\\\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=\dfrac{5+\sqrt{3}}{2}\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=4\end{matrix}\right.\)
\(\sqrt{2x^2-10x+11}=\sqrt{x^2-6x+8}\)
\(\Leftrightarrow2x^2-10x+11=x^2-6x+8\)
=>\(x^2-4x+3=0\)
=>(x-1)(x-3)=0
=>x=3(loại) hoặc x=1(nhận)
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\(4x^2-4x-5=4x^2-4x+1-6=\left(2x-1\right)^2-6\ge-6\)
\(Min=-6\Leftrightarrow x=\dfrac{1}{2}\)
\(4x^2+12x+10=4\left(x^2+3x+\dfrac{9}{4}\right)+1=4\left(x+\dfrac{3}{2}\right)^2+1\ge1\)
\(Min=1\Leftrightarrow x=-\dfrac{3}{2}\)
\(4x^2-12x-5=4\left(x^2-3x+\dfrac{9}{4}\right)-14=4\left(x-\dfrac{3}{2}\right)^2-14\ge-14\)
\(Min=-14\Leftrightarrow x=\dfrac{3}{2}\)
\(9x^2+12x+8=\left(9x^2+12x+4\right)+4=\left(3x+2\right)^2+4\ge4\)
\(Min=4\Leftrightarrow x=-\dfrac{2}{3}\)
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a: \(=2\left(x^2+6x+4\right)\)
\(=2\left(x^2+6x+9-5\right)\)
\(=2\left(x+3\right)^2-10>=-10\)
Dấu = xảy ra khi x=-3
b: \(=-\left(x^2-10x-3\right)\)
\(=-\left(x^2-10x+25-28\right)\)
\(=-\left(x-5\right)^2+28< =28\)
Dấu = xảy ra khi x=5
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a: \(=\left(4x-1\right)^3\)
b: \(=1000x^3-1-10x\left(100x^2-1\right)\)
\(=-1+10x\)