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A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=\frac{-2x}{3}+3x\left(\frac{15x+20-126}{90}\right)-\frac{5x^2-20x}{10}\)
\(M=\frac{-2x}{3}+3x.\frac{15x-106}{90}-\frac{5.\left(x^2-4x\right)}{10}\)
\(M=\frac{-2x}{3}+\frac{45x^2-318x}{90}-\frac{x^2-4x}{2}\)
Khi \(B=-\frac{3}{5}\)ta có :
\(B=\left|x-\frac{1}{7}\right|-\left|x+\frac{3}{5}\right|+\frac{4}{5}\)
\(B=\left|-\frac{3}{5}-\frac{1}{7}\right|-\left|-\frac{3}{5}+\frac{3}{5}\right|+\frac{4}{5}\)
\(B=-\frac{26}{35}-0+\frac{4}{5}\)
\(B=-\frac{26}{35}+\frac{4}{5}\)
\(B=\frac{2}{35}\)
\(B=\left|\frac{-3}{5}-\frac{1}{7}\right|-\left|\frac{-3}{5}+\frac{3}{5}\right|+\frac{4}{5}=\frac{26}{35}+\frac{4}{5}=\frac{2}{35}\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left[\frac{x}{6}-\left(-\frac{2}{9}\right)-\frac{7}{5}\right]-\frac{5x}{4}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{5x\left(x-4\right)}{10}\)
\(M=-\frac{2x}{3}+3x\left(\frac{x}{6}-\frac{53}{45}\right)-\frac{x\left(x-4\right)}{2}\)
\(M=-\frac{2x}{3}+\frac{x^2}{2}-\frac{53x}{15}-\frac{x\left(x-4\right)}{2}\)
\(M=\left(-\frac{2x}{3}-\frac{53x}{15}\right)+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)
\(M=-\frac{21x}{5}+\frac{x^2}{2}-\frac{x\left(x-4\right)}{2}\)
\(M=\frac{-2.21x+5x^2-5x\left(x-4\right)}{10}\)
\(M=\frac{-42x+5x^2-5x\left(x-4\right)}{10}\)
\(M=\frac{-x\left[42-5x+5\left(x-4\right)\right]}{10}\)
\(M=\frac{-x\left(42-5x+5x-20\right)}{10}\)
\(M=\frac{-x\left(42-20\right)}{10}\)
\(M=\frac{-x.22}{10}\)
\(M=-\frac{22x}{10}\)
\(M=-\frac{11x}{5}\)
\(b)\) Ta có :
\(C=\left|x+1\right|+\left|x-3\right|\)
\(C=\left|x+1\right|+\left|3-x\right|\ge\left|x+1+3-x\right|=\left|4\right|=4\)
Dấu "=" xảy ra khi và chỉ khi \(\left(x+1\right)\left(3-x\right)\ge0\)
Trường hợp 1 :
\(\hept{\begin{cases}x+1\ge0\\3-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\le3\end{cases}\Leftrightarrow}-1\le x\le3}\)
Trường hợp 2 :
\(\hept{\begin{cases}x+1\le0\\3-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le-1\\x\ge3\end{cases}}}\) ( loại )
Vậy \(C=4\) khi \(-1\le x\le3\)
Chúc bạn học tốt ~