Tìm x ,biết :
(27x + 6 ) : - 11 = 9
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\(\left(27x+6\right):3-11=9\)
\(\Leftrightarrow27x+6=9\cdot11+3=102\)
hay \(x=\dfrac{32}{9}\)
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
\(\Leftrightarrow\) \(6x^5-12x^4-17x^4+34x^3-7x^3+14x^2+13x^2-26x-3x+\)6 =0
\(6x^5-29x^4+27x^3+27x^2-29x+6=0\)
\(\Leftrightarrow\left(6x^5-18x^4\right)+\left(-11x^4+33x^3\right)+\left(-6x^3+18x^2\right)+\left(9x^2-27x\right)+\left(-2x+6\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^4-11x^3-6x^2+9x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\left(6x^4-12x^3\right)+\left(x^3-2x^2\right)+\left(-4x^2+8x\right)+\left(x-2\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^3+x^2-4x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(\left(6x^3+6x^2\right)+\left(-5x^2-5x\right)+\left(x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(\left(6x^2-3x\right)+\left(-2x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow x=\left(3;2;-1;\frac{1}{2};\frac{1}{3}\right)\)
Bài 1 d)
\(1-2+3-4+5-6+...+2013\)
\(=1+\left(-2+3\right)+\left(-4+5\right)+...+\left(-2012+2013\right)\)
\(=1+1+1+...+1\left(1006s\right)\)
\(=1006.1=1006\)
Bài 6:
\(21,251+6,058+0,749+1,042\)
\(=\left(21,251+0,749\right)+\left(6,058+1,042\right)\)
\(=22+7,1\)
\(=29,1\)
___________________
\(1,53+5,309+12,47+5,691\)
\(=\left(1,53+12,47\right)+\left(5,309+5,691\right)\)
\(=14+11\)
\(=25\)
5:
a: =>x/17=5/17
=>x=5
b; =>6+x=7/11*33=21
=>x=15
c: \(\dfrac{12+x}{43-x}=\dfrac{2}{3}\)
=>3x+36=86-2x
=>5x=50
=>x=10
d: \(\dfrac{x}{5}< \dfrac{3}{7}\)
=>x<3/7*5
=>x<15/7
f: 15/26+x/16=46/52
=>x/16=23/26-15/26=8/26=4/13
=>x=4/13*16=64/13
a)\(29.\left(-13\right)-27.\left(-29\right)-14.\left(-29\right)\)
\(=13\left(-29\right)-27.\left(-29\right)-14.\left(-29\right)\)
\(=-29.\left(13-27-14\right)\)
\(=\left(-28\right).\left(-29\right)\)
\(=812\)
( 27x + 6 ) : - 11 = 9
\(\left(27x+6\right)=9.\left(-11\right)\)
\(\left(27x+6\right)=-99\)
\(27x=\left(-99\right)-6\)
\(27x=-105\)
\(x=\left(-105\right):27\)
\(x=-\frac{35}{9}\)
\(\left(27x+6\right):\left(-11\right)=9\)
\(27x+6=9.\left(-11\right)=-99\)
\(27x=\left(-99\right)-6=-105\)
\(x=\left(-105\right):27=\frac{-35}{9}\)