a) \(\Leftrightarrow\sqrt{\left(x-\sqrt{3}\right)^2}=6\Leftrightarrow\left|x-\sqrt{3}\right|=6\Leftrightarrow\orbr{\begin{cases}x-\sqrt{3}=6\\x-\sqrt{3}=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6+\sqrt{3}\\x=3-\sqrt{6}\end{cases}}\)

b) \(\Leftrightarrow\sqrt{\left(x-3\right)^2}=4-x\Leftrightarrow\left|x-3\right|=4-x\Leftrightarrow\orbr{\begin{cases}x-3=4-x\left(x\ge3\right)\\3-x=4-x\left(x< 3\right)\end{cases}\Leftrightarrow}x=\frac{7}{2}\)

c) \(\Leftrightarrow\hept{\begin{cases}x\ge\frac{\sqrt{6}}{2}\\2x^2-3=4x-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{\sqrt{6}}{2}\\2x\left(x-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{\sqrt{6}}{2}\\x=0\left(loai\right);x=2\left(nhan\right)\end{cases}}\)

d) \(\Leftrightarrow\left|3-2x\right|=4\Leftrightarrow\orbr{\begin{cases}3-2x=4\\3-2x=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

e) \(\Leftrightarrow\sqrt{\left(x^2\right)^2}=5\Leftrightarrow x^2=5\Leftrightarrow x=\pm\sqrt{5}\)

f) \(\Leftrightarrow\hept{\begin{cases}x\ge-\frac{1}{2}\\\sqrt{2x+1}=-3\left(voli\right)\end{cases}}\)=> pt vô nghiệm

ta có 

\(sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-cos^2x=1-0.6^2=0.64\)

TH1.\(sinx=\sqrt{0.64}=0.8\Rightarrow\hept{\begin{cases}tanx=\frac{sinx}{cosx}=\frac{0.8}{0.6}=\frac{4}{3}\\cotx=\frac{1}{tanx}=\frac{3}{4}\end{cases}}\)

TH2.\(sinx=-\sqrt{0.64}=-0.8\Rightarrow\hept{\begin{cases}tanx=\frac{sinx}{cosx}=\frac{-0.8}{0.6}=-\frac{4}{3}\\cotx=\frac{1}{tanx}=-\frac{3}{4}\end{cases}}\)

A B C K H

ta có \(AH=\sqrt{AB^2-BH^2}=\sqrt{10^2-6^2}=8cm\)

khi đó \(sinABC=\frac{AH}{AB}=\frac{8}{10}=\frac{4}{5}\)

ta có \(BK.AC=AH.BC=2S_{ABC}\Rightarrow BK=\frac{AH.BC}{AC}=\frac{36}{5}cm\)

nên \(sinBAC=\frac{BK}{BA}=\frac{18}{25}\)

a) \(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}11\left(x+1\right)=22\\3\left(x+1\right)+2\left(x+2y\right)=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\4y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)

b) ĐK : y khác 0

\(\hept{\begin{cases}x+\frac{1}{y}=-\frac{1}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}3x+\frac{3}{y}=-\frac{3}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=-5\\3x+\frac{3}{y}=-\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-1\\-3+\frac{3}{y}=-\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\\frac{3}{y}=\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\left(tm\right)\end{cases}}\)

Bạn tham khảo bài tại link :

https://olm.vn/hoi-dap/detail/244883081409.html

hoặc :

Câu hỏi của Vũ Nguyễn Phương Thảo - Toán lớp 8 - Học trực tuyến OLM

Hok tốt

Trả lời :

Bạn vào hoc 24 có bài đấy

a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

b) Áp dụng đẳng thức ở câu a: \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\ge\left(ac+bd\right)^2\)

Dấu "=" xảy ra khi \(\left(ad-bc\right)^2=0\Leftrightarrow ad=bc\)

Link tham khảo : https://hoidap247.com/cau-hoi/165024

Nguồn : hoidap247.com

Bài làm : 

a.

(ac + bd)² + (ad – bc)²

= a² c² + 2acbd + b² d² + a² d² - 2adbc + b² c²

= a² c² + b² d² + a² d² + b² c²

= ( a² c² + a² d² ) + ( b² d² + b² c² )

= a² ( c² + d² ) + b² ( c² + d² )

= ( a² + b² ) . ( c² + d² )

Vậy (ac + bd)² + (ad – bc)² = (a² + b²)(c² + d²)

b.

( a² + b² ) . ( c² + d² ) - ( ac + bd )²

= a² c² + a² d² + b² c² + b² d² - a² c²  - 2acbd - b² d²

= a² d² + b² c² - 2acbd

= ( ad )² - 2ad . bc + ( bc )²

= ( ad - bc )² \(\ge\)0

\(\Rightarrow\) (ac + bd)² ≤ (a² + b²)(c² + d²)

Vậy (ac + bd)² ≤ (a² + b²)(c² + d²)

\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\left(2\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)

\(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{4+2\sqrt{3}}\)

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

\(\sqrt{15-\sqrt{216}}+\sqrt{33+12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3+2\sqrt{6}\right)^2}\)

\(=3-\sqrt{6}+3+2\sqrt{6}=6+\sqrt{6}\)

a)\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(1+2\sqrt{2}\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}=\sqrt{13+30\sqrt{\left(1+\sqrt{2}\right)^2}}=\sqrt{13+30\left(1+\sqrt{2}\right)}\)

\(=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)

b) \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{6+2\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}\)

\(=\sqrt{6+2\sqrt{5-\left(1+2\sqrt{3}\right)}}=\sqrt{6+2\sqrt{\left(1-\sqrt{3}\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

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

c)\(\sqrt{15-\sqrt{216}}+\sqrt{33+12\sqrt{6}}=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3+2\sqrt{6}\right)^2}=3-\sqrt{6}+3+2\sqrt{6}=6+\sqrt{6}\)

d) Sửa đề nha, đề bị sai 1 chỗ:

\(\sqrt{24-3\sqrt{15}}-\sqrt{36-9\sqrt{15}}=\sqrt{\frac{48-6\sqrt{15}}{2}}-\sqrt{\frac{72-18\sqrt{15}}{2}}\)

\(=\sqrt{\frac{\left(\sqrt{3}-3\sqrt{5}\right)^2}{2}}-\sqrt{\frac{\left(3\sqrt{3}-3\sqrt{5}\right)^2}{2}}=\frac{3\sqrt{5}-\sqrt{3}}{\sqrt{2}}-\frac{3\sqrt{5}-3\sqrt{3}}{\sqrt{2}}\)

\(=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

ta có điều kiện của căn thức là : \(\left|4x-9\right|\ge0\)luôn đúng

Vậy điều kiện xác định là \(x\in R\)