\(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\left(ĐKXĐ:x>0;x\ne1\right)\\ P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ P=\left(\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{4\sqrt{x}}\\ P=\dfrac{x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{4\sqrt{x}}\\ P=\dfrac{x+1}{4\sqrt{x}}\)

\(P=\left(1+\dfrac{\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\\ P=\dfrac{a+\sqrt{a}+1}{a+1}:\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)}\right)\\ P=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{a+1+2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\\ P=\dfrac{a+\sqrt{a}+1}{a+1}:\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\\ P=\dfrac{a+\sqrt{a}+1}{a+1}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(a+1\right)}{\left(\sqrt{a}+1\right)^2}\\ P=\dfrac{\sqrt{a^3}-1}{\left(\sqrt{a}+1\right)^2}\)

bạn bổ sung \(ĐKXĐ:a\ge0;a\ne1\) giúp mình nhé, lúc làm mình quên mất T_T

a) \(\sqrt[]{x^2-2x+4}=2x-2\)

\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)

Giải pt \(3x^2-6x=0\)

\(\Leftrightarrow3x\left(x-2\right)=0\)

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

\(\left(1\right)\Leftrightarrow x=2\)

c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)

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

\(\sqrt{x^2-6x+5}=2\)

ĐK: \(\left(x-5\right)\left(x-1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le1\end{matrix}\right.\)

\(\Leftrightarrow x^2-6x+5=2^2\)

\(\Leftrightarrow x^2-6x+1=0\)

\(\Rightarrow\Delta=\left(-6\right)^2-4\cdot1\cdot1=32>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{6+\sqrt{32}}{2}=3+2\sqrt{2}\left(tm\right)\\x_2=\dfrac{6-\sqrt{32}}{2}=3-2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)

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

\(\sqrt[]{x^2-6x+5}=2\left(2>0\right)\)

\(\Leftrightarrow x^2-6x+5=4\)

\(\Leftrightarrow x^2-6x+1=0\left(1\right)\)

\(\Delta'=9-1=8\)

\(\Rightarrow\sqrt[]{\Delta'}=2\sqrt[]{2}\)

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

a) Ta có:

\(sinB=\dfrac{AC}{AB}=\dfrac{15}{16}\)

\(\Leftrightarrow\widehat{B}\approx70^o\)

b), c) Xem lại đề  

Ta có

\(\dfrac{BD}{AB}=\dfrac{CD}{AC}\) (Trong tam giác, đường phân giác của một góc chia cạnh đối diện thành hai đoạn thẳng tỉ lệ với hai cạnh kề hai đoạn ấy)

\(\Rightarrow\dfrac{BD}{CD}=\dfrac{AB}{AC}=\dfrac{15}{20}=\dfrac{3}{4}\Rightarrow AB=\dfrac{3AC}{4}\)

\(BC=BD+CD=15+20=35cm\)

Ta có

\(BC^2=AB^2+AC^2\) (Pitago)

\(\Rightarrow35^2=\left(\dfrac{3AC}{4}\right)^2+AC^2\Rightarrow AC^2=784\Rightarrow AC=28cm\)

Ta có

\(AC^2=CH.BC\) (Trong tg vuông bình phương 1 cạnh góc vuông bằng tích giữa hình chiếu cạnh góc vuông đó trên cạnh huyền với cạnh huyền)

\(\Rightarrow CH=\dfrac{AC^2}{BC}=\dfrac{784}{35}=22,4cm\)

\(\Rightarrow BH=BC-CH=35-22,4=12,6cm\)

Ta có

\(AH^2=BH.CH\) (trong tg vuông bình phương đường cao hạ từ đỉnh góc vuông bằng tích giữa các hình chiếu của 2 cạnh góc vuông trên cạnh huyền)

\(\Rightarrow AH^2=12,6^2+22,4^2=660,52\Rightarrow AH=\sqrt{660,52}\)

Ta có

\(HD=BD-BH=15-12,6=2,4cm\)

Xét tg vuông AHD có

\(AD^2=AH^2+HD^2\) (Pitago)

Bạn tự tính nốt nhé

\(x+y+z+8=2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\left(1\right)\)

Áp dụng Bđt Bunhiacopxki :

\(\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le\left(2^2+4^2+6^2\right)\left(x-1+y-2+z-3\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z-6\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z+8\right)-784\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=\dfrac{x+y+z-6}{14}\left(2\right)\)

Đặt \(t=x+y+z+8\)

\(\left(1\right)\Leftrightarrow t^2=56t-784\)

\(\Leftrightarrow t^2-56t+784=0\)

\(\Leftrightarrow\left(t-28\right)^2=0\)

\(\Leftrightarrow t=28\)

\(\Leftrightarrow x+y+z+8=28\)

\(\Leftrightarrow x+y+z-6=14\)

\(\left(2\right)\Leftrightarrow\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.4=4\\z-2=1.8=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=6\\z=10\end{matrix}\right.\) thỏa mãn đề bài