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a) \(=\left(\frac{-337}{100}:\frac{40}{100}\right).\frac{250}{100}=\frac{-337}{40}.\frac{40}{16}=-\frac{337}{16}\)
b) \(=\frac{-1}{8}.\left(-5,3\right).8=\frac{-1}{8}.8.\left(-5,3\right)=-1.\left(-5,3\right)=5,3\)
c) \(=10.\left(-7,9\right)=-79\)
d) \(=-\frac{3}{8}.\frac{13}{3}.\left(-8\right)=\left[-\frac{3}{8}.\left(-8\right)\right].\frac{13}{3}=3.\frac{13}{3}=13\)
a)(-3,37:0,4).2,5
=(-8,425).2,5
=-21,0625
b) (-0,125) . (-5,3) . 8
=(-0,125).8.(-5,3)
=(-1).(-5,3)
=5,3
c)(-2,5).(-4).(-7,9)
=10.(-7,9)
=-79
d)(-0,375).\(4\frac{1}{3}\).(-2)3
=(-0,375).(-2)3.\(4\frac{1}{3}\)
=3.\(\frac{13}{3}\)
=13
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\(11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
\(S=\left\{1,\dfrac{4}{11}\right\}\)
Đặt C(x)=0
\(\Leftrightarrow11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\11x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
Vậy: Nghiệm của đa thức \(C\left(x\right)=11x^2-15x+4\) là 1 và \(\dfrac{4}{11}\)
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Ta có: x+y+1=0
nên x+y=-1
Ta có: \(N=x^2\left(x+y\right)-y^2\left(x+y\right)+x^2-y^2+2\left(x+y\right)+3\)
\(=\left(x+y\right)\left(x^2-y^2\right)+\left(x^2-y^2\right)+2\left(x+y\right)+3\)
\(=\left(x^2-y^2\right)\left(x+y+1\right)+2\left(x+y\right)+3\)
\(=\left(x^2-y^2\right)\cdot0+2\cdot\left(-1\right)+3\)
=-2+3=1
\(=\left(-13,37+3,37\right)+\left(-24,58-5,42\right)\\ =-10-30=-40\)