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2x-3y+5z=1 hoặc =-1
TH1: \(\dfrac{x}{y}\)=\(\dfrac{3}{2}\)=>\(\dfrac{x}{3}\)=\(\dfrac{y}{2}\)=>\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)
\(\dfrac{y}{z}\)=\(\dfrac{5}{7}\)=>\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)=>\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)
\(\Rightarrow\)\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)=>\(\dfrac{2x}{30}\)=\(\dfrac{3y}{30}\)=\(\dfrac{5z}{70}\)
Áp dụng tính chát dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-3y+5z}{30-30+70}\)=\(\dfrac{1}{70}\)
=>x=1.15:7=\(\dfrac{3}{14}\)
y=\(\dfrac{1}{7}\)
z=\(\dfrac{1}{5}\)
TH2:............=-1 tự tính nhé làm tương tựmình còn phải ôn bài
a) C = 20013 - |5−2x|
do \(-\left|5-2x\right|\le0\forall x\)
=> 20013-\(\left|5-2x\right|\le20013\)
=>A≤20013
=> GTLN C =20013 khi 5-2x=0
=> 2x=5
=> x=\(\dfrac{5}{2}\)
vậy GTLN C = 20013 khi x=\(\dfrac{5}{2}\)
b) D = 7 - \(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\)
do \(-\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le0\forall x\)
=> 7-\(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le7\)
=> D≤7
=> GTLN D =7 khi \(\dfrac{2}{3}+\dfrac{1}{4}x=0\)
=> x=-\(\dfrac{8}{3}\)
\(\left|x+\dfrac{1}{2}\right|-\dfrac{2}{5}=0\)
\(\Rightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{2}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\pm\dfrac{2}{5}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{5}\\x+\dfrac{1}{2}=-\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
Vậy..............
a: \(B\left(x\right)-A\left(x\right)=-3x+9-x^3+3x^2-7x+5\)
\(=-x^3+3x^2-10x+14\)
b: B(x)=-3x+9
Bậc là 1
Đặt B(x)=0
=>-3x+9=0
=>-3x=-9
hay x=3
c: B(3/5)=-9/5+9=7,2
\(y=\dfrac{2x+1}{x-1}+5=\dfrac{2x-2+3}{x-1}+5=\dfrac{2x-2}{x-1}+\dfrac{3}{x-1}+5=7+\dfrac{3}{x-1}\)
Để \(max_y\) thì \(\dfrac{3}{x-1}\) nhỏ nhất và \(x-1>0\Leftrightarrow x-1=1\Leftrightarrow x=2\)
Khi đó \(max_y=\dfrac{2.2+1}{2-1}+5=10\)
\(\left(2x-5\right)^2=\left(x-\dfrac{5}{2}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=x-\dfrac{5}{2}\\2x-5=\dfrac{5}{2}-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\3x=\dfrac{15}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{5}{2}\)
Vậy x = \(\dfrac{5}{2}\)
\(\left(2x-5\right)^2=\left(x-\dfrac{5}{2}\right)^2\)
\(\Leftrightarrow2x-5=x-\dfrac{5}{2}\)
\(\Leftrightarrow2x-x=-\dfrac{5}{2}+5\)
\(\Leftrightarrow x=\dfrac{5}{2}\)