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 \(\frac{4}{5}+\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}\)

Biểu thức trên đạt GTLN khi \(\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}\) đạt GTLN

                                        \(\Leftrightarrow\text{ }\left|3x+5\right|+\left|4y+5\right|+8\) nhỏ nhất

                                         \(\Rightarrow\text{ }\left|3x+5\right|+\left|4y+5\right|\) phải nhỏ nhất vì \(\text{ }\left|3x+5\right|\ge0\text{ và }\left|4y+5\right|\ge0\) nên khi cộng với 8 mới có GTNN

Ta có : \(\left|3x+5\right|\ge3x+5\) . Dấu " = " xảy ra khi \(3x+5\ge0\)  \(\Rightarrow\text{ }3x\ge-5\) \(\Rightarrow\text{ }x\ge-\frac{5}{3}\)

             \(\left|4y+5\right|\ge4y+5\).. Dấu " = " xảy ra khi \(4y+5\ge0\)   \(\Rightarrow\text{ }4y\ge-5\)  \(\Rightarrow\text{ }y\ge-\frac{5}{4}\)

Mà \(\left|3x+5\right|+\left|4y+5\right|\) nhỏ nhất \(\Rightarrow\text{ }x,y\text{ nhỏ nhất }\) 

Vậy \(x=-\frac{5}{3}\) , \(y=-\frac{5}{4}\)

\(\Rightarrow\text{ }\left|3x+5\right|+\left|4y+5\right|\ge\left(3x+5\right)+\left(4y+5\right)\)

\(\left|3x+5\right|+\left|4y+5\right|\ge\left(3x+4y\right)+10\)

Thay \(x=-\frac{5}{3}\) , \(y=-\frac{5}{4}\) vào vế phải của biểu thức ta được :

\(\left|3x+5\right|+\left|4y+5\right|\ge\left(3\cdot\frac{-5}{3}+4\cdot\frac{-5}{4}\right)+10\)

\(\left|3x+5\right|+\left|4y+5\right|\ge\left(-5+\left(-5\right)\right)+10\)

\(\left|3x+5\right|+\left|4y+5\right|\ge0\)

Vậy min \(\left|3x+5\right|+\left|4y+5\right|=0\)

\(\Rightarrow\text{ min }\left|3x+5\right|+\left|4y+5\right|+8=8\)

\(\Rightarrow\text{ }\frac{4}{5}+\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}\le\frac{4}{5}+\frac{20}{8}=\frac{33}{10}\)

\(\Rightarrow\text{ Max }\frac{4}{5}+\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}=\frac{33}{10}\)

Làm mẫu

a) Ta có: \(\left|3x+7\right|\ge0\)

\(\Leftrightarrow4\left|3x+7\right|\ge0\)

\(\Leftrightarrow4\left|3x+7\right|+3\ge3\)

\(\Leftrightarrow\frac{15}{4\left|3x+7\right|+3}\le5\)

\(\Leftrightarrow5+\frac{15}{4\left|3x+7\right|+3}\le10\)

Vậy GTLN của bt là 10\(\Leftrightarrow x=\frac{-7}{3}\)

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

câu 1

a)\(ĐKXĐ:x^3-8\ne0=>x\ne2\)

b)\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2-2x+4\right)}=\frac{3}{x-2}\left(#\right)\)

Thay \(x=\frac{4001}{2000}\)zô \(\left(#\right)\)ta được

\(\frac{3}{\frac{4001}{2000}-2}=\frac{3}{\frac{4001}{2000}-\frac{4000}{2000}}=\frac{3}{\frac{1}{2000}}=6000\)

c) Để phân thức trên có giá trị nguyên thì :

\(3⋮x-2\)

=>\(x-2\inƯ\left(3\right)=\left(\pm1\pm3\right)\)

=>\(x\in\left\{1,3,-1,5\right\}\)

zậy ....

câu 2)

a) \(8\left(3x-2\right)-14x=2\left(4-7x\right)+15x\)

=>\(24x-16-14x=8-14x+15x\)

=>\(24x-14x+14x-15x=8+16\)

=>\(9x=24=>x=\frac{24}{9}=\frac{8}{3}\)

b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

⇔\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)

⇔\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)

⇔\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)

⇔\(9x-10=0\)

hay 9x=10

⇔\(x=\frac{10}{9}\)

Vậy: \(x=\frac{10}{9}\)

c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)

⇔\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)

⇔\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)

⇔\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)

⇔\(6x-3-5x+10-3x-21=0\)

⇔\(-2x-14=0\)

⇔\(-2x=14\)

hay x=-7

Vậy: x=-7

d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)

⇔\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

⇔\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

⇔\(6x-18+7x-35-13x-4=0\)

⇔\(-21\ne0\)

Vậy: x∈∅

e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)

⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)

⇔\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)

⇔\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)

⇔\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)

⇔\(-12x+96=0\)

⇔\(-12x=-96\)

hay x=8

Vậy: x=8

yutyugubhujyikiu

6

a, để A_max khi\(\dfrac{15}{4\left|3x+7\right|+3}max\) khi:

\(\left\{{}\begin{matrix}4\left|3x+7\right|+3min\\4\left|3x+7\right|+3>0\end{matrix}\right.\)

mà\(4\left|3x+7\right|+3\ge3\)nên max A=10 khi x=\(\dfrac{-7}{3}\)

a: 4|3x+7|+3>=3

=>15/4|3x+7|+3<=5

=>A<=10

Dấu = xảy ra khi x=-7/3

b: 8|15x-21|+7>=7

=>21/8|15x-21|+7<=3

=>B<=3-1/3=8/3

Dấu = xảy ra khi x=7/5