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2.| 2x - 3 | = \(\frac{1}{2}\)
| 2x - 3 | = \(\frac{1}{2}:2\)
| 2x - 3 | = \(\frac{1}{4}\)
Th 1 : 2x - 3 = \(\frac{1}{4}\)
2x = \(\frac{1}{4}+3\)
2x = \(\frac{13}{4}\)
x = \(\frac{13}{4}:2\)
x = \(\frac{13}{8}\)
2 . | 2x - 3 | = 1/2
<=> | 2x - 3 | = 1/4
<=> 2x - 3 = 1/4
hoặc 2x - 3 = -1/4
<=> x = 13/8
hoặc x = 11/8
7,5 - 3 . | 5- xx | = - 4,5
<=> - 3 | 5 - x | = -12
<=> | 5 - x | = 4
<=> 5 - x = 4
hoặc 5 -x = -4
<=> x = 1 hoặc x = 9
\(\dfrac{-12}{25}.\left(\dfrac{3}{4}-x+\dfrac{6}{-11}-\dfrac{5}{6}\right)=0\)
\(\dfrac{3}{4}-x+\dfrac{-6}{11}-\dfrac{5}{6}=0\)
\(\dfrac{3}{4}-x+\dfrac{-91}{66}=0\)
\(\dfrac{3}{4}-x=0-\left(\dfrac{-91}{66}\right)\)
\(\dfrac{3}{4}-x=\dfrac{91}{66}\)
\(x=\dfrac{-83}{132}\)
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
\(P\left(x\right)=4x^4+2x^2-8x+\dfrac{1}{2}\)
\(Q\left(x\right)=-x^4-5x^2-8x-\dfrac{3}{4}\)
a: \(R\left(x\right)=P\left(x\right)-Q\left(x\right)=3x^4+7x^2+\dfrac{5}{4}\)
b: \(R\left(x\right)=3x^4+7x^2+\dfrac{5}{4}\ge\dfrac{5}{4}\forall x\)
nên R(X) không có nghiệm
`x^2 +3(x-1/2)=x^2+3`
`=>x^2+3x-3/2 =x^2+3`
`=> x^2 +3x-x^2=3+3/2`
`=> 3x=6/2+3/2`
`=>3x= 9/2`
`=>x= 9/2 : 3`
`=>x= 9/6= 3/2`
Vậy `x=3/2`
\(D=\frac{2x+1}{x-3}=\frac{2x-6}{x-3}+\frac{7}{x-3}=2+\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Leftrightarrow x\in\left\{-4;2;4;10\right\}\)
D= \(\frac{2x+1}{x-3}=2+\frac{7}{x-3}\)
để D dương thì x-3 là uocs của 7=(-1,1,-7,7)
xét từng TH:
x-3=-1=> x=2
x-3=1=>x=4
x-3=-7=>x=-4
x-3=7=>x=10
các giá trị x là 2,4,-4,10
\(X-\frac{3}{2}\times\left(-1\right)=\frac{4}{3}\)
\(X-\frac{3}{2}=\frac{4}{3}\div\left(-1\right)\)
\(X-\frac{3}{2}=\frac{4}{3}\div\left(-\frac{3}{3}\right)\)
\(X-\frac{3}{2}=\frac{4}{3}\times\left(-\frac{3}{3}\right)\)
\(X-\frac{3}{2}=-\frac{12}{9}=-\frac{4}{3}\)
\(X=-\frac{4}{3}+\frac{3}{2}\)
\(X=-\frac{8}{6}+\frac{9}{6}\)
\(X=\frac{-8+9}{6}=\frac{1}{6}\)
(sai thì thôi)
\(x-\frac{3}{2}.\left(-1\right)=\frac{4}{3}\)
\(x-\frac{3}{2}=\frac{4}{3}.\left(-1\right)\)
\(x-\frac{3}{2}=-\frac{4}{3}\)
\(x=-\frac{4}{3}+\frac{3}{2}\)
\(x=-\frac{8}{6}+\frac{9}{6}\)
\(x=\frac{1}{6}\)