a) \(\left(x-3\right)^2+2x-6=0\)

\(\Leftrightarrow x^2-6x+9+2x-6=0\)

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

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

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\) (ĐKXĐ: \(x\ne\pm3\))

\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x+3}\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2+6x+9-48=x^2-6x+9\)

\(\Leftrightarrow x^2-x^2+6x+6x+9-9-48=0\)

\(\Leftrightarrow12x-48=0\)

\(\Leftrightarrow12x=48\)

\(\Leftrightarrow x=\dfrac{48}{12}\)

\(\Leftrightarrow x=4\left(tm\right)\)

a: (x-3)^2+2x-6=0

=>(x-3)^2+2(x-3)=0

=>(x-3)(x-3+2)=0

=>(x-3)(x-1)=0

=>x=3 hoặc x=1

b:

ĐKXĐ: x<>3; x<>-3

 \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\)

=>\(\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{48}{\left(x-3\right)\cdot\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)^2}\)

=>(x+3)^2-48=(x-3)^2

=>x^2+6x+9-48=x^2-6x+9

=>6x-39=-6x+9

=>12x=48

=>x=4(nhận)

\(a,\left(x-3\right)^2-2x+6=0\\ \Leftrightarrow x^2-6x+9-2x+6=0\\ \Leftrightarrow x^2-8x+15=0\\ \Leftrightarrow x^2-3x-5x+15=0\\ \Leftrightarrow x\left(x-3\right)-5\left(x-3\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\\ Vậy:S=\left\{5;3\right\}\\ b,\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{\left(x+3\right)^2-48-\left(x-3\right)^2}{x^2-9}=0\\ \Leftrightarrow\left[\left(x+3-x+3\right)\left(x+3+x-3\right)\right]-48=0\\ \Leftrightarrow6.2x-48=0\\ \Leftrightarrow12x=48\\ \Leftrightarrow x=4\left(TM\right)\\ Vậy:S=\left\{4\right\}\)

a: \(P\left(x\right)=3x^2-4x+7\)

\(Q\left(x\right)=5x^3-x^2+4x-3\)

b: \(P\left(x\right)-Q\left(x\right)=3x^2-4x+7-5x^3+x^2-4x+3\)

\(=-5x^3+4x^2-8x+10\)

a: \(\dfrac{x}{2}+\dfrac{1-x}{3}>0\)

=>3x+2(1-x)>0

=>3x+2-2x>0

=>x+2>0

=>x>-2

b: (x-9)^2-x(x+9)<0

=>x^2-18x+81-x^2-9x<0

=>-27x+81<0

=>-27x<-81

=>x>3

S = (2+2^2)+(2^3+2^4)+...+(2^99+2^100)

= 2(1+2)+2^3(1+2)+...+2^99(1+2)

= 2.3+2^3 .3+...+2^99 .3

=3(2+2^3+...+2^99) chia hết cho 3

Hay S là bội của 3

Lời giải:

$(\sqrt{2}-\sqrt{3})^2=(\sqrt{2})^2-2.\sqrt{2}.\sqrt{3}+(\sqrt{3})^2$
$=2-2\sqrt{6}+3=5-2\sqrt{6}$

Bạn chú ý lần sau viết đề bằng công thức toán và viết đúng chính tả.

so sánh hả bn

34%:51/16-3 = -217/75

7/9x6,5-(0,4) mũ 2 = 2203/450

suy ra 2203/450 lớn hơn lúc này ta có 7/9x6,5-(0,4)mũ 2 lớn hơn 34%:51/16-3

Ko phải đâu tớ quên ko nhắc 3 và 7/9 là hỗn số. Cậu ghi lại câu trả lại đc ko. Tớ cảm ơn

a.

\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

b.

\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)

c.

\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d.

\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

e.

\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)

f.

\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)