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chỉnh đề B
\(B=x^5-15x^4+16x^3-29x^2+13x\)
\(=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3+\left(2x+1\right)x^2+\left(x-1\right)x\)
\(=x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x\)
\(=-x=-14\)
\(a.\dfrac{-3}{8}-\dfrac{13}{65}+\dfrac{3}{8}=\left(\dfrac{3}{8}-\dfrac{3}{8}\right)-\dfrac{13}{65}=-\dfrac{13}{65}\)
\(b.\left(\dfrac{-13}{7}-\dfrac{4}{9}\right)-\left(-\dfrac{10}{7}-\dfrac{4}{9}\right)=\dfrac{-13}{7}-\dfrac{4}{9}+\dfrac{10}{7}+\dfrac{4}{9}\\ =\left(\dfrac{-13}{7}+\dfrac{10}{7}\right)+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)=-\dfrac{3}{7}\)
\(c.17\dfrac{1}{3}\cdot\left(\dfrac{-3}{7}\right)+3\dfrac{2}{3}\cdot\left(\dfrac{-3}{7}\right)=\dfrac{-3}{7}\cdot\left(17\dfrac{1}{3}+3\dfrac{2}{3}\right)\\ =\dfrac{-3}{7}\cdot\left(\dfrac{52}{3}+\dfrac{11}{3}\right)=\dfrac{-3}{7}\cdot21=-9\)
D = \(x^{10}-25x^9+25x^8-25x^7+...+25x^2-25x+25\)với x = 24
thiếu 1 câu
A= x5−5x4+5x3−5x2+5x−1x5−5x4+5x3−5x2+5x−1 với x = 4
= x5−(x+1)x4+(x+1)x3−(x+1)x2+(x+1)x−1
= x5−x5−x4+x4+x3−x3+x2−x2+x−1
=x−1=4−1=3
Tương tự với các câu B,C,D
Lời giải:
a.
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x}{2}=\frac{y}{\frac{3}{2}}=\frac{z}{\frac{4}{3}}=\frac{x-y}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=60\\ y=45\\ z=40\end{matrix}\right.\)
b)
Từ đkđb suy ra \(\frac{10x}{1}=\frac{5y}{\frac{1}{3}}=\frac{z}{\frac{1}{6}}=\frac{10x-5y+z}{1-\frac{1}{3}+\frac{1}{6}}=\frac{25}{\frac{5}{6}}=30\)
\(\Rightarrow \left\{\begin{matrix} x=3\\ y=2\\ z=5\end{matrix}\right.\)
a: 2x-3/2+3/4=-4
=>2x-3/4=-4
=>2x=-13/4
hay x=-13/8
b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)
=>2/3x+3/5=12/145
=>2/3x=-15/29
hay x=-45/58
c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)
=>1/2x-3/5x+13/5=-7/10x-7/5
=>-1/10x+7/10x=-7/5-13/5
=>3/5x=-2
hay x=-2:3/5=-10/3
