a. Xét (O) , có

CD \(\perp\)AB = {I}

=> \(\widehat{CIB}=90^o\Rightarrow\widehat{FIB}=90^o\) 

Có: \(\widehat{AEB}\)là góc nội tiếp chắn nửa đường tròn đường kính AB

\(\Rightarrow\widehat{AEB}=90^o\Rightarrow\widehat{IEB}=90^o\)

Xét tứ giác EFIB, có:

\(\widehat{FEB}+\widehat{FIB}=90^o+90^o=180^o\)

2 góc \(\widehat{FEB}\)và \(\widehat{FIB}\)là 2 góc đối nhau




=> EFIB là tứ giác nội tiếp (dhnb) (đpcm)

Để dễ hình dung thì ta đặt: \(\hept{\begin{cases}\sqrt[3]{x}=a\\\sqrt[3]{y}=b\\\sqrt[3]{z}=c\end{cases}}\) khi đó BT cần chứng minh trở thành:

\(a^3+b^3+c^3-3abc=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

\(VP=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=a^3+b^3+c^3-3abc\) (HĐT sau vở hoặc nhân ra)

=> đpcm

Đặt \(\sqrt[3]{x}=a;\sqrt[3]{y}=b;\sqrt[3]{z}=c\)

Ta có :(+) \(x+y+z-3\sqrt[3]{xyz}=a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\left(a+b+c\right)\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=\frac{1}{2}.\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2^{ }\right]\)

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

\(\frac{1}{2}.\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

Suy ra điều phải chứng minh 

Với \(x\ge0;x\ne1\)

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

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

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

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

\(=\frac{x\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(x-1\right)}=\frac{\sqrt{x}\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=1\)

Ta có: \(\left(x+2\right)\left(x+4\right)\left(x^2-1\right)=27\)

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x-1\right)\left(x+1\right)=27\)

\(\Leftrightarrow\left[\left(x+2\right)\left(x+1\right)\right]\left[\left(x+4\right)\left(x-1\right)\right]=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x^2+3x-4\right)=27\)

Đặt \(x^2+3x-1=a\)

\(PT\Leftrightarrow\left(a-3\right)\left(a+3\right)=27\)

\(\Leftrightarrow a^2-9=27\Leftrightarrow a^2=36\Leftrightarrow\orbr{\begin{cases}a=6\\a=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+3x-1=6\\x^2+3x-1=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+3x-7=0\\x^2+3x+5=0\left(kcn^0\right)\end{cases}}\)

\(\Rightarrow x^2+3x-7=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-3+\sqrt{37}}{2}\\x=\frac{-3-\sqrt{37}}{2}\end{cases}}\)

Vậy tập nghiệm của PT S = ...

Bạn tham khỏa link này nha 

@Câu hỏi của Vân knth - Toán lớp 9 - Học trực tuyến OLM

#chuccauhoctot

Cậut k giúp mk nha

Ta có:

\(A=x-\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{1}{\sqrt{x}+\sqrt{x-1}}\right)\)

\(A=x-\frac{\sqrt{x}+\sqrt{x-1}-\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}\)

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

\(A=x-2\sqrt{x-1}\)

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

\(A=\left(\sqrt{x-1}-1\right)^2\ge0\left(\forall x\ge1\right)\)

=> đpcm