nhầm đề : \(\sqrt[4]{x+8}+\sqrt{x+4}=\sqrt{2x+3}+\sqrt{3x}\)

\(\sqrt[4]{x+8}+\sqrt{x+4}=\sqrt{2x+3}+\sqrt{3x}\)

\(\Leftrightarrow\sqrt[4]{x+8}-\sqrt{3}+\sqrt{x+4}-\sqrt{5}=\sqrt{2x+3}-\sqrt{5}+\sqrt{3x}-\sqrt{3}\)

\(\Leftrightarrow\frac{x+8-9}{\sqrt[4]{x+8}^3+\sqrt[4]{x+8}^2\sqrt{3}+3\sqrt[4]{x+8}+\sqrt{3}^3}+\frac{x+4-5}{\sqrt{x+4}+\sqrt{5}}=\frac{2x+3-5}{\sqrt{2x+3}+\sqrt{5}}+\frac{3x-3}{\sqrt{3x}+\sqrt{3}}\)

\(\Leftrightarrow\frac{x-1}{\sqrt[4]{x+8}^3+\sqrt[4]{x+8}^2\sqrt{3}+3\sqrt[4]{x+8}+\sqrt{3}^3}+\frac{x-1}{\sqrt{x+4}+\sqrt{5}}-\frac{2\left(x-1\right)}{\sqrt{2x+3}+\sqrt{5}}-\frac{3\left(x-1\right)}{\sqrt{3x}+\sqrt{3}}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt[4]{x+8}^3+\sqrt[4]{x+8}^2\sqrt{3}+3\sqrt[4]{x+8}+\sqrt{3}^3}+\frac{1}{\sqrt{x+4}+\sqrt{5}}-\frac{2}{\sqrt{2x+3}+\sqrt{5}}-\frac{31}{\sqrt{3x}+\sqrt{3}}\right)=0\)

Dễ thấy : pt trong ngoặc vô nghiệm

\(\Rightarrow x-1=0\Rightarrow x=1\)

bình 2 vế \(\frac{x-2}{2018}=\frac{x-2018}{2}\)

\(\Leftrightarrow\frac{x-2}{2018}-1=\frac{x-2018}{2}-1\)

\(\Leftrightarrow\frac{x-2020}{2018}-\frac{x-2020}{2}=0\)

\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2018}-\frac{1}{2}\right)=0\)

Thấy: \(\frac{1}{2018}-\frac{1}{2}\ne0\Rightarrow x=2020\)

Bình phương 2 vế

(X-2)/2018 = (X-2018)/2

   (X-2)×2=(X-2018)×2018

X= 2020

Căn bậc 2 của (x-1)^2 = x-1 hoặc 1-x

Trường hợp 1: x-1 >=0 => x>=1

 Thì    x-1+x =2x-1 >=1 (1)

Trường hợp 2 :1-x >=0  => x <=1

   Thì    1-x +x = 1 (2)

   (1) VÀ (2) => ĐPCM

\(\left(\sqrt{x+1}+1\right)\left(5-x\right)=2x\)

ĐK: \(x\ge-1\)

\(pt\Leftrightarrow\sqrt{x+1}=\frac{2x}{5-x}-1\)

\(\Leftrightarrow\sqrt{x+1}-2=\frac{2x}{5-x}-3\)

\(\Leftrightarrow\frac{x+1-4}{\sqrt{x+1}+2}-\frac{5x-15}{5-x}=0\)

\(\Leftrightarrow\frac{x-3}{\sqrt{x+1}+2}-\frac{5\left(x-3\right)}{5-x}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{x+1}+2}-\frac{5}{5-x}\right)=0\)

Dễ thấy: \(\frac{1}{\sqrt{x+1}+2}-\frac{5}{5-x}=0\) vô nghiệm

\(\Rightarrow x-3=0\Rightarrow x=3\)

Ta có:

\(\left(\sqrt{x+1}+1\right)\left(5-x\right)=2x\)

\(\Leftrightarrow\left(5-x\right)\sqrt{x+1}+5-3x=0\)

Đặt: \(\hept{\begin{cases}5-x=a\\\sqrt{x+1}=b\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}ab+3a=10\\a+b^2=6\end{cases}}\)​​

\(\Leftrightarrow\hept{\begin{cases}\left(6-b^2\right)b+3\left(6-b^2\right)=10\\a=6-b^2\end{cases}}\)

​​\(\Leftrightarrow\hept{\begin{cases}\left(2-b\right)\left(b+1\right)\left(b+4\right)=0\\a=6-b^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=2\\a=2\end{cases}}\)

\(\Rightarrow x=3\)

a/ BN và CN cắt nhau tại I => \(NI=\frac{BI}{2}\) và \(MI=\frac{CI}{2}\)

+ Ta có \(AC=2CN\Rightarrow AC^2=4CN^2\)và \(AB=2BM\Rightarrow AB^2=4BM^2\)

+ Xét tg vuông BIM có \(BM^2=BI^2+MI^2\Rightarrow4BM^2=AB^2=4\left(BI^2+MI^2\right)=4\left(BI^2+\frac{CI^2}{4}\right)\)

+ Xét tg vuông CIN có \(CN^2=CI^2+NI^2\Rightarrow4CN^2=AC^2=4\left(CI^2+NI^2\right)=4\left(CI^2+\frac{BI^2}{4}\right)\)

\(\Rightarrow AB^2+AC^2=4\left[\left(BI^2+CI^2\right)+\frac{BI^2+CI^2}{4}\right]\)

Mà trong tg vuông BIC có \(BC^2=BI^2+CI^2\)

\(\Rightarrow AB^2+AC^2=4\left(BC^2+\frac{BC^2}{4}\right)=5BC^2\)

b/