\(\Leftrightarrow\dfrac{3}{x-2}>0\)

=>x-2>0

hay x>2

4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)

=> (a+b)^2=(a-b)^2+4ab

(x – 3)(2x + 1) = 0

x = 3 hay x = -1/2

\(Q\left(x\right)-P\left(x\right)=0\)

\(\Leftrightarrow\left(-6x^2+x^3-8+12\right)-\left(x^3-3x^2+6x-8\right)=0\)

\(\Leftrightarrow\left(-6x^2+x^3+4\right)-\left(x^3-3x^2+6x-8\right)=0\)

\(\Leftrightarrow-6x^2+x^3+4-x^3+3x^2-6x+8=0\)

\(\Leftrightarrow-3x^2-6x+12=0\)

\(\Leftrightarrow-3\left(x^2+2x-4\right)=0\)

\(\Leftrightarrow x^2+2x-4=0\)

\(\Leftrightarrow x^2+2x+1=5\)

\(\Leftrightarrow\left(x+1\right)^2=5\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=\sqrt{5}\\x+1=-\sqrt{5}\end{cases}}\Leftrightarrow x=\pm\sqrt{5}-1\)

\(P\left(x\right)-Q\left(x\right)=\left(x^3-3x^2+6x-8\right)-\left(-6x^2+x^3-8+12\right)\)

\(P\left(x\right)-Q\left(x\right)=\left(x^3-3x^2+6x-8\right)-\left(-6x^2+x^3+4\right)\)

\(P\left(x\right)-Q\left(x\right)=x^3-3x^2+6x-8+6x^2-x^3-4\)

\(P\left(x\right)-Q\left(x\right)=3x^2+6x-4\)

Ta cần phân tích \(3x^2+6x-4\) thành nhân tử

Ta có:\(P\left(x\right)-Q\left(x\right)=-\frac{1}{3}\left(-9x^2-18x+12\right)\)

\(=-\frac{1}{3}\left[21-\left(9x^2+18x+9\right)\right]\)

\(=-\frac{1}{3}\left[21-\left(3x+3\right)^2\right]\)

\(=-\frac{1}{3}\left(\sqrt{21}-3x-3\right)\left(\sqrt{21}+3x+3\right)\)

\(\Rightarrow x=\frac{\sqrt{21}-3}{3};x=\frac{-\sqrt{21}-3}{3}\)

Ớ lộn \(P\left(x\right)-Q\left(x\right)\) rồi.bạn thông cảm nha.tham khảo bài bạn NCTK ý

a) ( x-2) ( y+1) =7

=> x-2 \(\in\)Ư(7)= { 1,7}

Nếu x-2 = 1 => x= 1+2 => x= 3

Nếu x-2= 7 => x= 7+2 => x= 9

Nếu x= 3 thì ( x-2) ( y+1) = ( 3-2)(y+1)=7

=>  y+1 =7 => y= 7-1 => y = 6

Nếu x = 9 thì ( x- 2 )( y+1)= 7 => ( 9-2) ( y+1) =7

=> 7( y+1) =7 => y+1= 7:7 => y+1 = 1 => y= 1-1 => y=0

Vậy...

Trình bày có chỗ nào sao mong mn sửa hộ nhaaa

b) ( 2x-1)( x+3) =6

=> ( 2x-1) \(\in\)Ư( 6) = { 1: 2: 3: 6}

Mà 2x-1 là số lẻ nên 2x-1 \(\in\){ 1; 3}

Nếu 2x-1 = 1 thì 2x= 1+1 => 2x= 2 => x= 2:2 => x= 1

Nếu 2x-1 = 3 thì 2x= 1+3 => 2x=4 => x= 4:2 => x= 2

Phần còn lại làm như phần a nha :33333

1/4×2/6×3/8×4/10×...×14/30×15/32=1/2^x

<=>1/(2×2)×2/(2×3)×...×14/(2×15)×15/2^5=1/2^x

<=>1/2×1/2×...×1/2×1/(2^5)=1/2^x

<=>1/2^19=1/2^x=>x=19

Đề mình không ghi lại nhé.

^{\(\Rightarrow\frac{1\times2\times3\times4\times...\times14\times15}{4\times6\times10\times...\times30\times32}=\frac{1}{2^x}\)}\(\frac{1}{2^x}\)

\(\Rightarrow\frac{1\times2\times3\times4\times...\times14\times15}{2\times4\times6\times8\times10\times...\times30\times32}\)\(=\frac{1}{2^{x+1}}\)

\(\Rightarrow\frac{1}{2^{15}\times32}=\)\(\frac{1}{2^{x+1}}\)

\(\Rightarrow2^{15}\times2^5=2^{x+1}\)

\(\Rightarrow2^{20}=2^{x+1}\)

\(\Rightarrow x+1=20\Rightarrow x=19\)

Vậy \(x=1\)

Học tốt nhaaa!

Vậy x = 19 nha cậu ơi! Lỗi kỹ thuật xíu!

Ta thấy : \(\left(x-y^2+z\right)^2\ge0\forall x,y,z\)

\(\left(y-2\right)^2\ge0\forall y\)

\(\left(z+3\right)^2\ge0\forall z\)

Do đó : \(\left(x-y^2+z\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\forall x,y,z\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-y^2+z\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x-y^2+z=0\\y-2=0\\z+3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-2^2+\left(-3\right)=0\\y=2\\z=-3\end{cases}}\)  \(\Leftrightarrow\hept{\begin{cases}x=7\\y=2\\z=-3\end{cases}}\)

Vậy : \(\left(x,y,z\right)=\left(7,2,-3\right)\)

CẢM ƠN BN ĐẠT NHIỀU!!!!!!