1/6*(2X-3)=-1/2*(X-1/4)-2/3 LÀM NHANH CHO MIK NHÉ
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Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
Trả lời:
1, \(\left(x-2\right)^3-\left(x+1\right)\left(x^2-x+1\right)+6\left(x-1\right)^2\)
\(=x^3-6x^2+12x-8-\left(x^3+1\right)+6\left(x^2-2x+1\right)\)
\(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)
\(=-2\)
2, \(-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)
\(=\)\(-x\left(x^2+4x+4\right)+4x^2+4x+1+x^3+27-1\)
\(=-x^3-4x^2-4x+4x^2+4x+1+x^3+27-1\)
\(=27\)
a) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow x+\frac{1}{2}=0\) \(\Rightarrow\frac{2}{3}-2x=0\)
\(x=\frac{-1}{2}\) \(2x=\frac{2}{3}\)
\(x=\frac{2}{3}:2\)
\(x=\frac{1}{3}\)
KL: x = -1/2 hoặc x= 1/3
b) \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=\frac{11}{6}\)
\(\left|2x-\frac{1}{3}\right|=1\)
TH1: \(2x-\frac{1}{3}=1\)
\(2x=\frac{4}{3}\)
\(x=\frac{2}{3}\)
TH2: \(2x-\frac{1}{3}=-1\)
\(2x=\frac{-2}{3}\)
\(x=\frac{-2}{3}:2\)
\(x=\frac{-1}{3}\)
KL: x =.........
Học tốt nhé bn!!!
a) \(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x\right|=\frac{29}{12}\)
\(\orbr{\begin{cases}x=\frac{29}{12}\\x=-\frac{29}{12}\end{cases}}\)
15/2 - ( x * 3 ) = 0,2
x * 3 = 15/2 - 0,2
x * 3 = 37,5
x = 37,5 : 3
x = 12,5
2/3 : x - 6 = 1/2
2/3 : x = 1/2 + 6
2/3 : x = 13/2
x = 2/3 : 13/2
x = 4/39
a) \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1\right)-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x.\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\)
\(\Leftrightarrow26x+28=54\Leftrightarrow26x=54-28\Leftrightarrow26x=26\Leftrightarrow x=1\)
Vậy nghiệm của phương trình là x=1
b) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+6.\left(x^2+2x+1\right)+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
\(\Leftrightarrow27x+12x+6=-33\Leftrightarrow39x=-33-6\Leftrightarrow39x=-39\Leftrightarrow x=-1\)
Vậy nghiệm của phương trình là x = -1
Trần Anh: Hí hí =)) ÀI LỚP DIU CHIU CHIU CHÍU :3 CẢM ƠN PẠN NHIỀU NHÁ ;) ;) ;)
1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)
\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)
\(=x^3.\left(x+2\right)+4x.\left(x+2\right)-5.\left(x+2\right)=\left(x+2\right)\left(x^3+4x-5\right)\)
\(=\left(x+2\right)\left(x^3-x^2+x^2-x+5x-5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)
2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)
Đặt \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)
\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)
Thay a , ta có :
\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
NHANH LÊN MIK ĐANG CẦN GẤP NHA CÁC BAN
\(\frac{1}{6}\)(2x-3) = \(\frac{-1}{2}\)(x-\(\frac{1}{4}\))-\(\frac{2}{3}\)
\(\frac{1}{3}\)x - \(\frac{1}{2}\)=\(\frac{-1x}{2}\)-\(\frac{-1}{8}\)-\(\frac{2}{3}\)
\(\frac{1x}{3}\)-\(\frac{1}{2}\)=\(\frac{-1x}{2}\) - \(\frac{19}{24}\)
\(\frac{1x}{3}\) - \(\frac{1}{2}\) - \(\frac{-1x}{2}\) - \(\frac{19}{24}\) =0
\(\frac{5x}{6}\) - \(\frac{7}{24}\)=0
\(\frac{5x}{6}\) = \(\frac{7}{24}\)
x = \(\frac{7}{20}\)