\(\frac{100-y+0,5x2}{2}-0,5=0\)
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Chọn A
y ' = 2.4 x 3 − 3.3 x 2 + 0 , 5.2 x − 3 2 = 8 x 3 − 9 x 2 + x − 3 2
Tiếp tục:\(-A=\frac{x^3+y^3+z^3}{2xyz}\)
thay(1) vào A ta có
\(-A=\frac{y^3+z^3-\left(y+z\right)^3}{2xyz}=\frac{y^3+z^3-y^3-z^3-3yz\left(y+z\right)}{2xyz}\)
\(-A=\frac{3xyz}{2xyz}=\frac{3}{2}\Rightarrow A=\frac{-3}{2}\)
P/s tham khảo bài mình nhé nhớ
ta có:\(x+y+z=0\) \(\Rightarrow x=-\left(y+z\right)\)
\(\Rightarrow x^3=-\left(y+z\right)^3\left(1\right)\)\(;x^2=\left(y+z\right)^2\)
\(\Rightarrow y^2+z^2-x^2=-2yz\)
CMTT:\(z^2+x^2-y^2=-2xz;x^2+y^2-z^2=-2xy\)
thay vào A ta có:
\(A=\frac{-x^2}{2yz}+\frac{-y^2}{2xz}+\frac{-z^2}{2xy}\)
`-0,5x^2(-yz).(-3xy^3z)`
`=-1,5x^{2+1}.y^{1+3}.z^{1+1}`
`=-1,5x^3y^4z^2`
\(=\frac{75}{100}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\left(-\frac{1}{3}\right)^2\)
\(=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}-\frac{1}{9}\)
\(=\frac{-3}{4}+\frac{6}{5}-\frac{1}{9}\)
\(=\frac{9}{20}-\frac{1}{9}\)
\(=\frac{61}{180}\)
\(\frac{75}{100}-1\frac{1}{2}+0,5:\frac{5}{12}-\left(-\frac{1}{2}\right)^3\)
= \(\frac{3}{4}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\frac{-1}{8}\)
= \(\frac{3}{4}-\frac{6}{4}+\frac{1}{2}\cdot\frac{12}{5}-\frac{-1}{8}\)
= \(-\frac{3}{4}+\frac{6}{5}-\frac{-1}{8}=\frac{-30}{40}+\frac{48}{40}+\frac{-6}{40}=\frac{12}{40}=\frac{3}{10}\)
CHÚC BN HK TỐT #
\(=-\frac{3}{4}+\frac{6}{5}-\left(\frac{-1}{8}\right)\))
=\(-\frac{3}{4}+\frac{6}{5}+\frac{1}{8}\)
=\(\frac{9}{20}+\frac{1}{8}\)
=\(\frac{23}{40}\)
Vậy..........
\(\frac{75}{100}-1\frac{1}{2}+0,5:\frac{5}{12}-\left(-\frac{1}{2}\right)^3\)
= \(\frac{3}{4}-\frac{3}{2}+\frac{6}{5}+\frac{1}{8}\)
= \(\frac{30}{40}-\frac{60}{40}+\frac{48}{40}+\frac{5}{40}\)
= \(\frac{23}{40}\)