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

Điều kiện : \(xy\ge0\) hoặc \(xy\le0\) ; \(xy\ne1\); \(x\ge0\);\(y\ge0\)

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

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

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

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

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

\(P=\dfrac{2\sqrt{x}}{x+1}\)

b) ta có :\(x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{4-2\sqrt{3}}{4-3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

thay \(x=\left(\sqrt{3}-1\right)^2\) vào biểu thức P
ta được : \(P=\dfrac{2\sqrt{\left(\sqrt{3}-1\right)^2}}{\left(\sqrt{3}-1\right)^2+1}\)

\(P=\dfrac{2\left|\sqrt{3}-1\right|}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}\)

\(P=\dfrac{\left(2\sqrt{3}-2\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}=\dfrac{10\sqrt{3}+12-10-4\sqrt{3}}{25-12}\)

\(P=\dfrac{6\sqrt{3}+2}{13}\)

c) để P\(\le\)1 thì \(\dfrac{2\sqrt{x}}{x+1}\le1\)

\(\Leftrightarrow\dfrac{2\sqrt{x}}{x+1}-1\le0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-x-1}{x+1}\le0\)

\(\Leftrightarrow\dfrac{-\left(x-2\sqrt{x}+1\right)}{x+1}\le0\)

\(\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x+1}\le0\)

Vì \(-\left(x-1\right)^2\le0\) nên x + 1 \(\ge\) 0

\(\Leftrightarrow\) x \(\ge\) -1
đúng thì cho xin 1 like nha

\(\sqrt{2x+1}-\sqrt{3x}=x-1\)

Điều kiện : x\(\ge0\)

\(\Leftrightarrow\sqrt{2x+1}=x-1+\sqrt{3x}\)

\(\Leftrightarrow\left(\sqrt{2x+1}\right)^2=\left(x-1+\sqrt{3x}\right)^2\)

\(\Leftrightarrow2x+1=\left(x-1\right)^2+2\left(x-1\right)\sqrt{3x}+3x\)

\(\Leftrightarrow2x+1=x^2-2x+1+2\left(x-1\right)\sqrt{3x}+3x\)

\(\Leftrightarrow2x+1-x^2-x-x-2\left(x-1\right)\sqrt{3x}=0\)

\(\Leftrightarrow-x^2+x-2\left(x-1\right)\sqrt{3x}=0\)

\(\Leftrightarrow-x\left(x-1\right)-2\left(x-1\right)\sqrt{3x}=0\)

\(\Leftrightarrow\left(x-1\right)\left(-x-2\sqrt{3x}\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x\in\varnothing\end{matrix}\right.\) Vậy pt tập nghiệm S={1;0}

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

\(N=1:\left(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\sqrt{x}+1}{x+1+\sqrt{x}}-\dfrac{\sqrt{x}+1}{x-1}\right)\)

Điều kiện:x\(\ge\)0,x\(\ne\)1

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

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

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

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

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

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

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(C=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\div\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)

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

\(=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}-1}\div\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{1}{\sqrt{x}-1}\div\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\)

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

a)\(\dfrac{2}{x^2-1}+\dfrac{1}{x+1}=2\) Điều kiện:x#1,-1

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

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

\(\Leftrightarrow\dfrac{1}{x-1}=2\)

\(\Leftrightarrow1=2\left(x-1\right)\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

b)\(1-\dfrac{12}{x^2-4}=\dfrac{3}{x+2}\) Điều kiện:x#2,-2

\(\Leftrightarrow\dfrac{x^2-4-12}{x^2-4}=\dfrac{3}{x+2}\)

\(\Leftrightarrow x^2-16=3\left(x-2\right)\)

\(\Leftrightarrow x^2-16-3x+6=0\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

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

Vậy \(S=\left\{5\right\}\)

Bài 1:tìm x,biết
a)x³-6x²-x+30=0
\(\Leftrightarrow\text{x³ + 2x² - 8x² - 16x + 15x + 30 = 0 }\)
\(\Leftrightarrow\text{x² ( x + 2 ) - 8x ( x + 2 ) + 15 ( x + 2 ) = 0 }\)
\(\Leftrightarrow\text{( x + 2 )( x² - 8x + 15 ) = 0 }\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+16-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[\left(x-4\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-4-1\right)\left(x-4+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\\x=3\end{matrix}\right.\)
b)x²+16=10x
\(\Leftrightarrow x^2-10x+16=0\)
\(\Leftrightarrow x^2-2x-8x+16=0\)
\(\Leftrightarrow x\left(x-2\right)-8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)
Bài 2:
a)(x-2)²-(x-3)(x-1)\(=x^2-4x+4-\left(x^2-x-3x+3\right)\)
=\(x^2-4x+4-x^2+4x-3\)
\(=1\).Vậy biểu thức a không phụ thuộc vào biến x
b)(x-1)³-(x+1)³+6(x+1)(x-1)
\(=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)
\(=-8\).Vậy biểu thức b không phụ thuộc vào biến x

4x(x-1)-3(x²-5)-x²=(x-3)-(x+4)
\(\Leftrightarrow4x^2-4x-3x^2-15-x^2=x-3-x-4\)
⇔\(-4x-15=-7\)
\(\Leftrightarrow-4x=8\)
\(\Leftrightarrow x=-2\)

a) (4x-5)^2 -4 (x-2)^2 =0
⇔(4x-5)²-[2(x-2)]²=0
​⇔(4x-5)²-(2x-4)²=0
​⇔(4x-5-2x+4)(4x-5+2x-4)=0
​⇔(2x-1)(6x-9)=0
​⇔\(\left[{}\begin{matrix}2x-1=0\\6x-9=0\end{matrix}\right.\)
​⇔\(\left[{}\begin{matrix}2x=1\\6x=9\end{matrix}\right.\text{​⇔}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

(x - 2)³ + 6(x + 1)² - (x² + 3x + 9)(x - 3) = 15
⇔x³-6x²+12x-8+6(x²+2x+1)-(x³-27)=15
⇔x³-6x²+12x-8+6x²+12x+6-x³+27=15
⇔24x+25=15
⇔24x=-10
⇔x=\(\dfrac{-5}{12}\)

(x² - 2x + 4)(x + 2) - x(x² + 2) = 17
⇔x³+8-x³-2x=17
⇔8-2x=17
⇔-2x=9
⇔2x=-9
⇔x=\(\dfrac{-9}{2}\)