\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

\(=\dfrac{\sqrt{x}.\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\dfrac{x-2}{\sqrt{x}}\)

\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)

\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)

\(=\dfrac{2\sqrt{3}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{\left|\sqrt{3}+1\right|}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)

a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)

\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-2}{\sqrt{x}}\)

b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)

Đặt \(\left|x-3\right|=t\left(t>0\right)\)

Ta có: \(A=t\left(2-t\right)=-t^2+2t=-\left(t-1\right)^2+1\le1\forall t\)

Dấu "=" xảy ra khi: \(t-1=0\Rightarrow t=1\Rightarrow\left|x-3\right|=1\)

\(\Rightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

Vậy GTLN của A là 1 khi x = 4 hoặc x = 2

xl mik nhầm phải là \(A=\left|x-3\right|\cdot\left(2-\left|3-x\right|\right)\)

\(\left|x-3\right|=\left|3-x\right|\) mà thì bạn làm giống trên thôi.

Giải PT hay gì thế

\(\left(x+3\right)^2+\left(x-2\right)^2=2x^2+2x+13\)

\(\Leftrightarrow x^2+6x+9+x^2-4x+4=2x^2+2x+13\)

\(\Leftrightarrow2x^2+2x+13=2x^2+2x+13\) Luôn đúng với mọi x

a ) \(\left(3\times x-15\right)^7=0.\)

\(3\times x-15=0\)

\(3\times x=15\)

\(x=5\)

b ) \(10-\left\{\left[\left(x\div3+17\right)\div10+3\times2^4\right]\div10\right\}=5\)

\(10-\left\{\left[\left(x\div3+17\right)\div10+3\times16\right]\div10\right\}=5\)

\(10-\left\{\left[\left(x\div3+17\right)\div10+48\right]\div10\right\}=5\)

\(\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\left(x\div3+17\right)\div10+48=50\)

\(\left(x\div3+17\right)\div10=2\)

\(x\div3+17=20\)

\(x\div3=3\)

\(x=9\)

a) Ta có: 

VT = |x + 1| + |x + 2| + |2x - 3| \(\ge\)|x + 1 + x + 2| + |3 - 2x| =  |2x + 3| + |3 - 2x| \(\ge\)|2x + 3 + 3 - 2x| = 6

VP = 6

Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(x+2\right)\ge0\\\left(2x+3\right)\left(3-2x\right)\ge0\end{cases}}\)  => \(\orbr{\begin{cases}x\ge-1\\x\le-2\end{cases}}\)và \(-\frac{3}{2}\le x\le\frac{3}{2}\)

<=> \(-1\le x\le\frac{3}{2}\)

b) Ta có: VT = |x + 1| + |x - 2| + |x - 3| + |x - 5| = (|x + 1| + |5 - x|) + (|x - 2| + |3 - x|) \(\ge\)|x + 1 + 5 - x| + |x - 2 + 3 - x| = |6| + |1| = 7

VP = 7

Dấu "=" xảy ra<=> \(\hept{\begin{cases}\left(x+1\right)\left(5-x\right)\ge0\\\left(x-2\right)\left(3-x\right)\ge0\end{cases}}\) <=> \(\hept{\begin{cases}-1\le x\le5\\2\le x\le3\end{cases}}\) <=> \(2\le x\le3\)

\(P=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^4+...+\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\)

\(=\left(\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2\right)+\left(\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{4}\right)^4\right)+...+\left(\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\right)\)

\(=\dfrac{1}{3}.\dfrac{2}{3}+\left(\dfrac{1}{3}\right)^3.\dfrac{2}{3}+...+\left(\dfrac{1}{3}\right)^{19}.\dfrac{2}{3}\)

\(=\dfrac{2}{3}.\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{19}\right]\)