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\(\left(-12\right)^2.x=56+10.13.10\)
\(\left(-12\right)^2.x=56+1300\)
\(\left(-12\right)^2.x=1356\)
\(-12.-12.x=1356\)
\(144.x=1356\)
\(x=1356:144\)
\(x=9,4166666666...\)
(-122) . x = 56 + 10.13.x
144.x = 56 + 130.x
144.x - 130.x = 56
14x = 56
x = 56 : 14
x = 4
Vậy x = 4
(-12)².x=56+10.13.x
<=>144.x=56+130.x
<=>144x-130x=56
<=>14x=56<=>x=4
Vậy x=4
Nhớ tick
\(p=\left|x-1\right|+5\)
\(\left|x-1\right|\ge0\)
\(\Rightarrow\left|x-1\right|+5\ge5\)
Dấu ''='' xảy ra khi \(\left|x-1\right|=0\)
\(x-1=0\)
\(x=1\)
Vậy \(MinP=5\Leftrightarrow x=1\)
Ta có: |x-1| \(\ge\) 0
<=> |x-1| + 5 \(\ge\) 5
Dấu "=" xảy ra khi x = 1
Vậy MinP = 5 khi x = 1
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
\(\Leftrightarrow5\left(7-8x\right)=6\left(-4+2x\right)\)
\(\Leftrightarrow35-40x=-24+12x\)
\(\Leftrightarrow35+24=40x+12x\)
\(\Leftrightarrow59=52x\)
\(\Rightarrow x=\frac{59}{52}\)
\(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
\(\Leftrightarrow1-3:x=8\)
\(\Leftrightarrow-3:x=7\)
\(\Leftrightarrow x=-\frac{7}{3}\)
ta có : 144.x=56+130.x
144.x-130.x=56
(144-130).x=56
14.x=56suy ra x=4
144x=56+130x
=>144x-130x=56
=>14x=56
=>x=4