4. We are worried that he will fail the exam.

5. The man who lives next door is my grandfather.

4. Chúng tôi lo lắng rằng anh ấy sẽ trượt kỳ thi.

5. Người đàn ông sống bên cạnh là ông nội của tôi.

ĐKXĐ: \(x\ge-1\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=x^2+2\)

Pt trở thành:

\(5ab=2\left(a^2+b^2\right)\Leftrightarrow2a^2-5ab+2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)

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

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

\(\Leftrightarrow...\)

\(Q=ac+bc-2022ab\le ac+bc=c\left(a+b\right)\le\dfrac{1}{4}\left(c+a+b\right)^2=\dfrac{1}{4}\)

\(Q_{max}=\dfrac{1}{4}\) khi \(\left\{{}\begin{matrix}a+b+c=1\\ab=0\\c=a+b\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;\dfrac{1}{2};\dfrac{1}{2}\right);\left(\dfrac{1}{2};0;\dfrac{1}{2}\right)\)

\(Q=c\left(a+b\right)-2022ab\ge c\left(a+b\right)-\dfrac{1011}{2}\left(a+b\right)^2\)

\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}\left(1-c\right)^2\)

\(Q\ge c\left(1-c\right)-\dfrac{1011}{2}c\left(c-2\right)-\dfrac{1011}{2}\)

\(Q\ge\dfrac{c\left(1011+1013\left(1-c\right)\right)}{2}-\dfrac{1011}{2}\ge-\dfrac{1011}{2}\)

\(Q_{min}=-\dfrac{1011}{2}\) khi \(\left(a;b;c\right)=\left(\dfrac{1}{2};\dfrac{1}{2};0\right)\)

Bill hasn't made up his mind what he should bring with him on the trip to America.

Have a great day.

Bill hasn't made up his mind what he should bring with him on the trip to America (make up sb's mind = decide: quyết định)

ko bé ơi

có

Help mik trg hôm nay vs ạ mik đag cần gấp

ai yêu mik kb nha

a) Ta có \(\widehat{AKB}=90^0\) (góc nội tiếp chắn nửa đường tròn)

\(\widehat{BEC}=90^0\) (Do \(CD\) là trung trực của \(OA\))

\(\Rightarrow\widehat{BKC}+\widehat{BEC}=90^0+90^0=180^0\)

\(\Rightarrow BEHK\) là tứ giác nội tiếp.

b) Ta có \(OC=OD=R\) nên tam giác \(OCD\) cân tại O

Mà \(OE\perp CD\Rightarrow OE\) là phân giác \(\widehat{COD}\Rightarrow\widehat{COA}=\widehat{DOA}\)

\(\Rightarrow sđ\stackrel\frown{AC}=sđ\stackrel\frown{AD}\)

Do \(\left\{{}\begin{matrix}\widehat{ACH}=\dfrac{1}{2}sđ\stackrel\frown{AD}\\\widehat{AKC}=\dfrac{1}{2}sđ\stackrel\frown{AC}\end{matrix}\right.\Rightarrow\widehat{ACH}=\widehat{AKC}\)

Xét \(\Delta ACH\) và \(\Delta AKC\) có 

\(\widehat{CAK}\) chung

\(\widehat{ACH}=\widehat{AKC}\) (cmt)

\(\Rightarrow\Delta ACH\sim\Delta AKC\) (g.g) \(\Rightarrow\dfrac{AC}{AH}=\dfrac{AK}{AC}\Rightarrow AC^2=AH.AK\)

Ta có: Tam giác \(AOC\) cân tại \(O\) (do \(OC=OA=R\))

Mặt khác: \(\Delta OEC\) vuông tại \(E\), có \(OE=\dfrac{1}{2}OA=\dfrac{1}{2}OC\)

\(\Rightarrow\widehat{OCE}=30^0\Rightarrow\widehat{AOC}=60^0\)

\(\Rightarrow\Delta OAC\) đều hay \(AC=OA=OC=R\)