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a. \(3\dfrac{1}{3}x+16=13,25\)
=> x + 16 = 13,25
=> x = 13,25 - 16
=> x = \(-\dfrac{11}{4}\)
b. x - 43 = (57 - x) - 50
=> 2x = 57 - 50 + 43
=> 2x = 7 + 43
=> 2x = 50
=> x = 50 : 2
=> x = 25
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(=>11-3x+1=\frac{9}{2}-5+3,5x\)
\(=>-3x+12=3,5x-\frac{1}{2}\)
\(=>-3x-3,5x=-\frac{1}{2}-12\)
\(=>-6,5x=-12,5\)
\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)
Ủng hộ nha
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(11-3x+1=\frac{9}{2}-5+3,5x\)
\(12-3x=-\left(0,5\right)+3,5x\)
\(12,5-3x=3,5x\)
\(12,5=6,5x\)
\(x=12,5:6,5=\frac{25}{13}\)
<=> 3 - 1/6 + x = 2/3
<=> 17/6 + x = 2/3
<=> x = 2/3 - 17/6
=> x = -13/6
Vậy x = -13/6
\(3-\left(\frac{1}{6}-x\right)=\frac{2}{3}\)
\(\Rightarrow\frac{1}{6}-x=3-\frac{2}{3}\)
\(\Rightarrow\frac{1}{6}-x=\frac{7}{3}\)
\(\Rightarrow x=\frac{1}{6}-\frac{7}{3}\)
\(\Rightarrow x=\frac{-13}{6}\)
a) Ta có: \(\left(x-3\right)\left(x-5\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-3< 0\\x-5>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-3>0\\x-5< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 3\\x>5\end{cases}}\) (vô lý) hoặc \(\hept{\begin{cases}x>3\\x< 5\end{cases}}\)(thỏa mãn).
Vậy 3 < x < 5 thì (x-3)(x-5) <0.
b) \(-6x-\left(-7\right)=25\)
\(\Rightarrow-6x=25-7\)
\(\Rightarrow-6x=18\Rightarrow x=\frac{18}{-6}=-3\)
Vậy x = -3.
c) \(46-\left(x-11\right)=-48\)
\(\Rightarrow46-x+11=-48\)
\(\Rightarrow46+11+48=x\Rightarrow x=105\).
d) \(\left(x+15\right)\left(x-2\right)=0\)
\(\Rightarrow\)x + 15 = 0 hoặc x - 2 = 0
\(\Rightarrow x=-15\)hoặc \(x=2\).
e) \(3\left(4-x\right)-2\left(x-5\right)=12\)
\(\Rightarrow12-3x-2x+10=12\)
\(\Rightarrow-3x-2x=12-10-12\)
\(\Rightarrow-5x=-10\Rightarrow x=2\).
Chúc bn hc tốt!
c,\(43+x=2.5^2-\left(x-57\right)\)
\(< =>43+x=50-x+57\)
\(< =>2x=50+57-43\)
\(< =>x=\frac{107-43}{2}=32\)
d,\(-3.2^2\left(x-5\right)+7\left(3-x\right)=5\)
\(< =>-12.\left(x-5\right)+7.\left(3-x\right)=5\)
\(< =>-12x+60+21-7x=5\)
\(< =>-19x=5-81=-76\)
\(< =>x=-\frac{76}{-19}=4\)
Bài 2:
a) \(A=\left|x-3\right|+10\)
Vì \(\left|x-3\right|\ge0\forall x\)\(\Rightarrow\left|x-3\right|+10\ge10\forall x\)
hay \(A\ge10\)
Dấu " = " xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)
Vậy \(minA=10\Leftrightarrow x=3\)
b) \(B=-7+\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow-7+\left(x-1\right)^2\ge-7\forall x\)
hay \(B\ge-7\)
Dấu " = " xảy ra \(\Leftrightarrow x-1=0\)\(\Leftrightarrow x=1\)
Vậy \(minB=-7\Leftrightarrow x=1\)
Tính nhanh : \(\left(59-5^{30}\right)-\left(59+3^3-5^{30}\right)\)
Giúp mình nha , cảm ơn minna nhiều
\(\left(59-5^{30}\right)-\left(59+3^3-5^{30}\right)=59-5^{30}-59-3^3+5^{30}\)
\(=\left(59-59\right)-\left(5^{30}-5^{30}\right)-3^3\)
\(=0-3^3\)
\(=-27\)
<=> x - 43 = 57 - x - 50
<=> x - 43 = 57 - 50 - x
<=> x - 43 = 7 - x
<=> x + x = 7 + 43
<=> 2x = 50
=> x = 25
Vậy x = -18
x-43=(57-x)-50
=>x-43=57-x-50
=>x-43=7-x
=>x+x=7+43
=>2x=50
=>x=50:2
=>x=25