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Có: \(\left(\left|x\right|-\dfrac{1}{8}\right).\left(-\dfrac{1}{5}\right)^5=\left(-\dfrac{1}{8}\right)^7\)
<=> \(\left|x\right|.\left(-\dfrac{1}{5}\right)^5-\dfrac{1}{8}.\left(-\dfrac{1}{5}\right)^5=\left(-\dfrac{1}{8}\right)^7\)
<=> \(\left|x\right|.\left(-\dfrac{1}{5}\right)^5+\dfrac{1}{8}.\left(\dfrac{1}{5}\right)^5=\left(-\dfrac{1}{8}\right)^7\)
<=> \(\left|x\right|.\dfrac{-1}{3125}=-\dfrac{1}{8^7}-\dfrac{1}{8}.\dfrac{1}{3125}\)
<=> \(\left|x\right|=\dfrac{\dfrac{-1.3125}{8^7.3125}-\dfrac{1}{8.3125}}{-\dfrac{1}{3125}}=\dfrac{\dfrac{-3125}{8^7}.\dfrac{1}{3125}-\dfrac{1}{8}.\dfrac{1}{3125}}{-\dfrac{1}{3125}}=\dfrac{\dfrac{-1}{3125}\left(\dfrac{3125}{8^7}+\dfrac{1}{8}\right)}{-\dfrac{1}{3125}}\)
<=> \(\left|x\right|=\dfrac{3125}{8^7}+\dfrac{8^6}{8^7}=\dfrac{265269}{2097152}\)
=> x\(\in\left\{\dfrac{265269}{2097152};\dfrac{-265269}{2097152}\right\}\)
Câu 2:
a: THeo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}-a+b=-11\\2a+b=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-9\end{matrix}\right.\)
b: Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}a\cdot0+b\cdot0+c=5\\4a-2b+c=21\\a-b+c=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=5\\4a-2b=16\\a-b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=5\\a=3\\b=-2\end{matrix}\right.\)
1, f(x)=x3-x2+x-1
\(f\left(x\right)=x^2\left(x-1\right)+\left(x-1\right)=\left(x^2+1\right)\left(x-1\right)\)
f(x) = 0 \(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+1=0\\x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2=-1\\x=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\in\phi\\x=1\end{array}\right.\)
\(\Leftrightarrow x=1\)
4, P(x)=x2+5x
\(\Rightarrow P\left(x\right)=x\left(x+5\right)\)
\(P\left(x\right)=0\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x=0\\x+5=0\Rightarrow x=-5\end{array}\right.\)
6, Q(x)=3x2-4x
\(\Rightarrow Q\left(x\right)=x\left(3x-4\right)\)
7, H(x)= 5x5+10x
đề ? f91) --> f(1) hả
f(1) =2+a+4=a+6
g(2)=4-10-b =-6-b
g(5) =25-25-b =-b
.............
f(1) =g(2)=g(5)
=>xem lại đề b không tồn tại
\(a,G\left(-11\right)=6.\left(-11\right)^3-5.\left(-11\right)^2+10.\left(-11\right)\)
\(G\left(-11\right)=6.\left(-1331\right)-5.121+\left(-110\right)\)
\(G\left(-11\right)=-7986-605+\left(-110\right)\)
\(G\left(-11\right)=-8701\)