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( 5/4x - 2/3 )^4 = 57/16 + 3/2
=> ( 5/4x - 2/3 )^4 = 81/16
=> ( 5/4x - 2/3 )^4 = 3^4/2^4
=> 5/4x - 2/3 = 3/2
=> 5/3x = 13/6
=> x = 13/10
Vậy x = 13/10
\(\left(\frac{5}{4}x-\frac{2}{3}\right)^4-\frac{3}{2}=\frac{57}{16}\)
\(\left(\frac{5}{4}x-\frac{2}{3}\right)^4-\frac{3}{2}=\frac{57}{16}+\frac{3}{2}\)
\(\left(\frac{5}{4}x-\frac{2}{3}\right)=\frac{81}{16}\)
Ta xét 2th:
Th1: \(\frac{5}{4}x-\frac{2}{3}=\left(\frac{81}{16}\right)^{\frac{1}{4}}\)
\(\Rightarrow x=\frac{26}{15}\)
Th2: \(\frac{5}{4}x-\frac{2}{3}=-\left(\frac{81}{16}\right)^{\frac{1}{4}}\)
\(\Rightarrow x=-\frac{2}{3}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{26}{15}\\x=-\frac{2}{3}\end{cases}}\)
để G lớn nhất thì :
\(|4x-1|\);\(|xy-3|\)nhỏ nhất và bằng 0
=> 4x-1 =0 1/4 . y-3=0
x= (1+0) :4 y=(0+3) :1/4
x=1/4 =>x=1/4 y=12
=> G= 3 - \(|4x-1|\)- \(|xy-3|\)
=> G=3-0-0=3
vậy maxG=3
(2,5x - 4/3)2 - 5/6 = -7/18
=> (2,5x - 4/3)2 = -7/18 + 5/6
=> (2,5x - 4/3)2 = 4/9
=> (2,5x - 4/3)2 = (2/3)2
=> \(\orbr{\begin{cases}2,5x-\frac{4}{3}=\frac{2}{3}\\2,5x-\frac{4}{3}=-\frac{2}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)
\(\left(2,5x-\frac{4}{3}\right)^2-\frac{5}{6}=-\frac{7}{18}\)
\(\left(2,5x-\frac{4}{3}\right)^2=-\frac{7}{18}+\frac{5}{6}\)
\(\left(2,5x-\frac{4}{3}\right)^2=\frac{4}{9}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{4}{15}\end{cases}}\)
\(\left|0,75-\frac{5}{4}x\right|=\left|-\frac{2}{3}x+\frac{5}{6}\right|\)
\(\Rightarrow\left|0,75-\frac{5}{4}x\right|-\left|-\frac{2}{3}x+\frac{5}{6}\right|=0\)
Mà GTTĐ luôn lớn hơn hoặc bằng 0
=> +) \(0,75-\frac{5}{4}x=0\)
\(\frac{5}{4}x=0,75\)
\(x=\frac{3}{5}\)
+) \(-\frac{2}{3}x+\frac{5}{6}=0\)
\(-\frac{2}{3}x=-\frac{5}{6}\)
\(x=\frac{5}{4}\)
Vậy,.........
\(|\)0,75 - 5/4x\(|\)= \(|\)-2/3x +5/6 \(|\)
\(\Leftrightarrow\)0,75-5/4x = -2/3x+5/6 hoặc 0,75-5/4x=-(-2/3x+5/6)
TH1: 0,75-5/4x = -2/3x+5/6
\(\Leftrightarrow\)-5/4x+2/3x= 5/6-0,75
\(\Leftrightarrow\)-7/12x= 1/12
\(\Leftrightarrow\)x = 1/12 \(\div\)(-7/12)
\(\Leftrightarrow\)x = -1/7
TH2: 0,75-5/4x=-(-2/3x+5/6)
\(\Leftrightarrow\)0,75 - 5/4x= 2/3x-5/6
\(\Leftrightarrow\)-5/4x-2/3x = -5/6-0.75
\(\Leftrightarrow\)-23/12x= -19/12
\(\Leftrightarrow\)x=-(19/12)\(\div\)(-23/12)
\(\Leftrightarrow\)x=19/23
Vậy, x =-1/7 hoặc x= 19/23
a) Ta có : ( x + 1 ).( 3 - x ) > 0
Th1 : \(\hept{\begin{cases}x+1>0\\3-x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x>3\end{cases}\Rightarrow}x>3}\)
Th2 : \(\hept{\begin{cases}x+1< 0\\3-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x< 3\end{cases}\Rightarrow}x< -1}\)
|5x + 3| - x = 7
|5x + 3| = 7 + x
TH1: 5x + 3 = -7 - x
5x + x = -7 - 3
6x = -10
\(x=\frac{-5}{3}\)
TH2: 5x + 3 = 7 + x
5x - x = 7 - 3
4x = 4
x = 1
\(\frac{7}{12}-0,75:\left(4x-3\right)^2=\frac{-3}{2}\)
\(\frac{3}{4}:\left(4x-3\right)^2=\frac{25}{12}\)
\(\left(4x-3\right)^2=\frac{9}{25}=\left(\frac{\pm3}{5}\right)^2\)
+) 4x - 3 = 3/5
4x = 18/5
x = 9/10
+) 4x - 3 = -3/5
4x = 12/5
x = 3/5
Vậy,.........
\(\frac{7}{12}-0,75:\left(4x-3\right)^2=-\frac{3}{2}\)
\(\Leftrightarrow\frac{7}{12}.12\left(4x-3\right)^2-\frac{0,75}{\left(4x-3\right)^2}.12\left(4x-3\right)^2=-\frac{3}{2}.12\left(4x-3\right)^2\)
\(\Leftrightarrow7\left(4x-3\right)^2-9=-18\left(4x-3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{10}\\x=\frac{3}{5}\end{cases}}\)