Chọn A

`b)1/3x+2<0`

Mk nghĩ là đáp án B

B

a) Ta có: \(2-x=2\left(x-2\right)^3\)

\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)

\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

b) Ta có: \(8x^3-72x=0\)

\(\Leftrightarrow8x\left(x^2-9\right)=0\)

\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

Vậy: S={0;3;-3}

c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)

\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)

\(\Leftrightarrow x-1.5=0\)

hay x=1,5

d) Ta có: \(2x^3+3x^2+3+2x=0\)

\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow2x=-3\)

hay \(x=-\dfrac{3}{2}\)

e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)

Vậy: S={0;1;-2}

f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)

Vậy: S={0;2;12}

a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c: Ta có: \(4\left(x-1\right)\left(x+5\right)-\left(x+5\right)\left(x+2\right)-3\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow4\left(x^2+4x-5\right)-\left(x^2+7x+10\right)-3\left(x^2+x-2\right)=0\)

\(\Leftrightarrow4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0\)

\(\Leftrightarrow6x=24\)

hay x=4

d: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-5+1=-4\)

hay \(x=\dfrac{2}{7}\)

Em hc bảng xét dáu chx ??

a) Ta có: (x-1)(x-4)>0

\(\Leftrightarrow\left[{}\begin{matrix}x-4>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\)

b) Ta có: (x-6)(x-7)<0

\(\Leftrightarrow\left\{{}\begin{matrix}x-6>0\\x-7< 0\end{matrix}\right.\Leftrightarrow6< x< 7\)

c) Ta có: \(\left(x-1\right)\left(x-2\right)\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2\le0\end{matrix}\right.\Leftrightarrow1\le x\le2\)

d) Ta có: \(\left(x-2\right)\left(x-\dfrac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2\ge0\\x-\dfrac{2}{3}\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le\dfrac{2}{3}\end{matrix}\right.\)

a) Ta có: (x-1)(x-4)>0

⇔[x−4>0x−1<0⇔[x>4x<1⇔[x−4>0x−1<0⇔[x>4x<1

b) Ta có: (x-6)(x-7)<0

⇔{x−6>0x−7<0⇔6<x<7⇔{x−6>0x−7<0⇔6<x<7

c) Ta có: (x−1)(x−2)≤0(x−1)(x−2)≤0

⇔{x−1≥0x−2≤0⇔1≤x≤2⇔{x−1≥0x−2≤0⇔1≤x≤2

d) Ta có: (x−2)(x−23)≥0(x−2)(x−23)≥0

⇔⎡⎣x−2≥0x−23≤0⇔⎡⎣x≥2x≤23

mng giúp với

pls gấp quá

a:Ta có: \(\left(x-2\right)\left(x-\dfrac{1}{2}\right)< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{2}>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 2\)

b: Ta có: \(\left(x+\dfrac{1}{3}\right)\left(x+1\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{1}{3}\\x< -1\end{matrix}\right.\)

a)<=>

A,=(x+y)(x-y)=x^2-y^2

x=(-1/2)^5:(1/2)^4=-1/2

x^2=1/4

y=8^2/(-2)^5=-2

y^2=4

A=1/4-4=-15/4