x x 1/2 + 1/3 = 5,6
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1: (3x+2)(x+2)(2x-1)
=(3x^2+6x+2x+4)(2x-1)
=(3x^2+8x+4)(2x-1)
=6x^3-3x^2+16x^2-8x+8x-4
=6x^3+13x^2-4
2: (5x+1)(x-1)+3x(2x+2)
=5x^2-5x+x-1+6x^2+6x
=11x^2+10x-1
3: 4x(2x+1)(x-1)+(x+5)(x-3)
=4x(2x^2-2x+x-1)+x^2+2x-15
=8x^3-4x^2-4x+x^2+2x-15
=8x^3-3x^2-2x-15
4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)
=(2x-1)(x^2-4)+3x^2-4x+1
=2x^3-8x-x^2+4+3x^2-4x+1
=2x^3+2x^2-12x+5
`x(x + 3)^2 - 3x = (x + 2)^3 + 1`
`<=> x(x^2 + 6x + 9) - 3x = x^3 + 12x + 6x^2 + 8 + 1`
`<=> x^3 + 6x^2 + 9x - 3x - x^3 - 12x - 6x^2 - 9 = 0`
`<=> -6x - 9 = 0`
`<=> -6x = 9`
`=> x = -3/2`
Vậy `S = {-3/2}`
`#3107.101107`
\(\dfrac{1}{5}+\dfrac{2}{3}\times x=\dfrac{1}{3}\)
\(\dfrac{2}{3}\times x=\dfrac{1}{3}-\dfrac{1}{5}\)
\(\dfrac{2}{3}\times x=\dfrac{2}{15}\)
\(x=\dfrac{2}{15}\div\dfrac{2}{3}\)
\(x=\dfrac{1}{5}\)
Vậy, \(x=\dfrac{1}{5}.\)
1/5 + 2/3 . x = 1/3
2/3 . x = 1/3 - 1/5
2/3 . x = 2/15
x = 2/15 : 2/3
x = 1/5
1/2* x+2/3=9/2
1/2 * x = 9/2 - 2/3
1/2 * x= 23/6
x= 23/6 : 1/2
x= 23/6 x 2= 23/3
___
1/2*x-1/3=2/3
1/2*x = 2/3 + 1/3
1/2 * x= 1
x= 1: 1/2
x= 2
____
1/4+3/4:x=3
3/4 : x = 3 - 1/4
3/4 : x= 11/4
x= 11/4 : 3/4
x= 11/3
\(\dfrac{1}{2}\)\(\times\)\(x\) + \(\dfrac{2}{3}\) = \(\dfrac{9}{2}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) = \(\dfrac{9}{2}\) - \(\dfrac{2}{3}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) = \(\dfrac{23}{6}\)
\(x\) = \(\dfrac{23}{6}\):\(\dfrac{1}{2}\)
\(x\) = \(\dfrac{23}{3}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\)
\(\dfrac{1}{2}\)\(\times\)\(x\) = \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\times\)\(x\) = 1
\(x\) = 1 : \(\dfrac{1}{2}\)
\(x\) = 2
\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\): \(x\) = 3
\(\dfrac{3}{4}\): \(x\) = 3 - \(\dfrac{1}{4}\)
\(\dfrac{3}{4}\):\(x\) = \(\dfrac{11}{4}\)
\(x\) = \(\dfrac{3}{4}\): \(\dfrac{11}{4}\)
\(x\) = \(\dfrac{3}{11}\)
Lời giải:
$A=(x-2)^2+|x-1|+5$
Nếu $x\geq 1$ thì:
$A=(x-2)^2+x-1+5=x^2-4x+4+x-1+5=x^2-3x+8=(x-\frac{3}{2})^2+\frac{23}{4}\geq \frac{23}{4}(*)$
Nếu $x< 1$:
$A=(x-2)^2+1-x+5=x^2-5x+10=(x-1)(x-4)+6> 6(**)$
Từ $(*); (**)\Rightarrow A_{\min}=\frac{23}{4}$ khi $x=\frac{3}{2}$
Lời giải:
\(B=2(x+1)^2-|x+3|-11\)
Nếu $x\geq -3$ thì:
\(B=2(x+1)^2-(x+3)-11=2x^2+3x-12=2(x+\frac{3}{4})^2-\frac{105}{8}\)
\(\geq \frac{-105}{8}\) (1)
Nếu $x< -3$
$B=2(x+1)^2+(x+3)-11=2x^2+5x-6=(x+3)(2x+1)-9> -9$ (2)
Từ $(1); (2)\Rightarrow B_{\min}=\frac{-105}{8}$ khi $x+\frac{3}{4}=0\Leftrightarrow x=\frac{-3}{4}$
`x xx1/4+x xx1/3=2`
`=> x xx (1/4+1/3)=2`
`=> x xx (3/12+4/12)=2`
`=> x xx 7/12=2`
`=> x=2:7/12`
`=>x=2xx12/7`
`=>x=24/7`
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
=>1/2x=5,6-1/3=79/15
=>x=79/15*2=158/15