tìm x , biết :
5 . | x - 3 | = 5/7
-3 . | 2x - 5 | = -3/4
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a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
a, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
a, 11/12 - ( 2/5 + x ) = 2/3
<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)
b, 2x . ( x - 1/7 ) = 0
<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
vậy x={\(0;\frac{1}{7}\)}
c, 3/4 + 1/4 : x = 2/5
<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)
vậy x=-5/7
a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)
\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)
\(\Leftrightarrow x=-\frac{3}{20}\)
b) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow4x=\frac{-20}{7}\)
\(\Leftrightarrow x=-\frac{5}{7}\)
a) -2(2x - 8) + 3(4 - 2x) = -72 - 5(3x - 7)
=> -4x + 18 + 12 - 6x = -72 - 15x + 35
=> -10x + 15x = -37 - 30
=> 5x = -37
=> x = -7,4
b) 3|2x2 - 7| = 33
=> |2x2 - 7| = 11
=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)
=> \(\orbr{\begin{cases}2x^2=18\\2x^2=-4\left(loại\right)\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Sửa đề : a) Tìm GTNN A
a) \(A=\left|x-5\right|+3\)có : \(\left|x-5\right|\ge0\Rightarrow\left|x-5\right|+3\ge0\)
\(\Leftrightarrow A\ge3\)dấu "=" xảy ra khi : \(\left|x-5\right|=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy GTNN A = 3 khi x = 5.
b) \(C=-\left|x+1\right|+5\)có : \(-\left|x+1\right|\le0\Rightarrow-\left|x+1\right|+5\le5\)
\(\Leftrightarrow C\le5\)dấu "=" xảy ra khi : \(-\left|x+1\right|=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy GTLN C = 5 khi x = -1.
\(D=5-\left|2x+3\right|\)có : \(-\left|2x+3\right|\le0\Rightarrow5-\left|2x+3\right|\le5\)
\(\Leftrightarrow D\le5\)dấu "=" xảy ra khi : \(-\left|2x+3\right|=0\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)
Vậy GTLN D = 5 khi x = -3/2.
c) \(\left|x-3\right|+\left|y+1\right|=0\)có \(\left|x-3\right|\ge0;\left|y+1\right|\ge0\Rightarrow\left|x-3\right|+\left|y+1\right|\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=0\\\left|y+1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}.\)