(2/5)^6x(25/4)^2
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\(a,\left(x^2-25\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\left(x-5\right)\left(x+5\right)+\left(x-5\right)\left(2x-11\right)=0\)
\(\left(x-5\right)\left(x+5+2x-11\right)=0\)
\(\left(x-5\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x-6=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
\(b,\left(x^2-6x+9\right)-4=0\)
\(x^2-6x+9-4=0\)
\(x^2+6x+5=0\)
\(x^2+x+5x+5=0\)
\(x\left(x+1\right)+5\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)
Chúc bạn hok tốt
1. \(\left(x^2-25\right)+\left(x-5\right).\left(2x-11\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right).\left(x+5\right)+\left(x-5\right).\left(2x-11\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right).\left(x+5+2x-11\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right).\left(3x-6\right)=0\)
\(\Leftrightarrow\)\(3.\left(x-5\right).\left(x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
Giải:
1) \(\left(x-6\right)\left(x^2+6x+36\right)-\left(x+4\right)^3=\left(x-2\right)^3+\left(x+5\right)\left(x^2-10x+25\right)-\left(2x^3+6x^2\right)\)
\(\Leftrightarrow x^3-216-\left(x^3+12x^2+48x+64\right)=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow x^3-216-x^3-12x^2-48x-64=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow-280-12x^2-48x=-12x^2+12x+117\)
\(\Leftrightarrow-280-48x-12x-117=0\)
\(\Leftrightarrow-397-60x=0\)
\(\Leftrightarrow-60x=397\)
\(\Leftrightarrow x=-\dfrac{397}{60}\)
Vậy ...
2) \(\left(2x+3\right)^3-\left(2x+5\right)\left(4x^2-10x+25\right)=\left(6x-1\right)^2-\left(x-2\right)\left(x^2+2x+4\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-\left(8x^3+125\right)=36x^2-12x+1-\left(x^3-8\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3-125=36x^2-12x+1-x^3+8+x^3\)
\(\Leftrightarrow54x-98=-12x+9\)
\(\Leftrightarrow54x+12x=9+98\)
\(\Leftrightarrow66x=107\)
\(\Leftrightarrow x=\dfrac{107}{66}\)
Vậy ...
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
a: 9-25=7-x-(25+7)
=>7-x-25-7=-16
=>-x-25=-16
=>x+25=16
hay x=-9
b: \(10+2\left|x\right|=2\cdot\left(3^2-1\right)\)
\(\Leftrightarrow2\left|x\right|=2\cdot8-10=6\)
=>x=3 hoặc x=-3
c: -6x=18
nên x=18:(-6)=-3
(2/5)^6x(25/4)^2
=64/15625 x 625/16
= 4/25
Bài làm
\(\left(\frac{2}{5}\right)^6x\left(\frac{25}{4}\right)^2=\frac{4}{25}x\left(\frac{625}{16}\right)=\frac{25}{4}x\)