Tam giác ABC cân ở A, có góc A = 120^o nên góc ACB  = 30^o

  \(\widehat{AOB}=2\widehat{ACB}\)= 2.30^o = 60^o

Tam giác AOB đều nên OA = AB = 4 (cm)

S_(O) = \(\pi.OA^2=\pi.4^2=16\pi\)

          \(\approx3,14.16\approx50,24\)(cm²)

\(x^2+9\ge9\forall x\)

\(\Rightarrow pt:x^2+9=0\)Vô nghiệm

    \(x^2+9=0\)

\(\Leftrightarrow x^2=-9\)(vô lí)

Vậy \(x=\varnothing\)

\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)

\(\tan^2\alpha\left(2.\cos^2\alpha+\sin^2\alpha-1\right)=\tan^2\alpha\left(\cos^2\alpha+\left(\sin^2\alpha+\cos^2\alpha\right)-1\right)\)\(=\tan^2\alpha.\cos^2\alpha=\left(\frac{1}{\cos^2\alpha}-1\right)\cos^2\alpha=1-\cos^2\alpha=\sin^2\alpha\)

gọi d=( n+1, 2n+1)

=> n+1 chia hết cho d=> 2n+2 chia hết cho d

=>2n+1 chia hết cho d=> 2n+1 chia hết cho d

=> ( 2n+2)-( 2n+1) chia hết cho d

=> 1 chia hết cho d

=> d= -1 hoặc +1

=> phân số n+1/2n+1 là phân số tối giản

b, giải 

  Gọi d là \(UCLN\left(n+1,n+2\right)\)

\(\Rightarrow\orbr{\begin{cases}n+1⋮d\\n+2⋮d\end{cases}}\)

\(\Rightarrow\left(n+1\right)-\left(n+2\right)⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d=1\)

\(\Rightarrow UCLN\left(n+1,n+2\right)=1\)

\(\Rightarrow\frac{n+1}{n+2}\) là phân số tối giản (ĐPCM)

\(ab+b\sqrt{a}+\sqrt{a}+1\)

(đk: \(a\ge0\))

\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}+1=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

ĐK: \(x,y\ge0\)

\(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}=x\left(\sqrt{x}+\sqrt{y}\right)-y\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)\)

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

\(\left(\sqrt{\frac{1}{4}\cdot\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\cdot10\sqrt{2}\right):\frac{1}{8}\)

\(=\left(\sqrt{\frac{1}{4}\cdot\frac{1\cdot2}{2\cdot2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\cdot10\sqrt{2}\right):\frac{1}{8}\)

\(=\left(\sqrt{\frac{1}{4}\cdot\frac{2}{4}}-\frac{3}{2}\sqrt{2}+8\sqrt{2}\right):\frac{1}{8}\)

\(=\left(\sqrt{\frac{2}{16}}-\frac{3}{2}\sqrt{2}+8\sqrt{2}\right):\frac{1}{8}\)

\(=\left(\frac{1}{4}\sqrt{2}-\frac{3}{2}\sqrt{2}+8\sqrt{2}\right):\frac{1}{8}\)

\(=\sqrt{2}\left(\frac{1}{4}-\frac{3}{2}+8\right):\frac{1}{8}\)

\(=\sqrt{2}\left(\frac{1}{4}-\frac{6}{4}+\frac{32}{4}\right):\frac{1}{8}\)

\(=\sqrt{2}\cdot\frac{27}{4}:\frac{1}{8}\)

\(=\frac{27\sqrt{2}}{4}\cdot\frac{8}{1}\)

\(=2\cdot27\sqrt{2}=54\sqrt{2}\)

\(\Rightarrow\left(\sqrt{\frac{1}{4}\cdot\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\cdot10\sqrt{2}\right):\frac{1}{8}=54\sqrt{2}\)

\(\left(\frac{1}{4}\sqrt{2}-\frac{3}{2}\sqrt{2}+\frac{4}{5}.10\sqrt{2}\right):\frac{1}{8}\)

=\(\sqrt{2}\)(1/4-3/2+8):1/8

=\(\sqrt{2}\).27/4.8

=54\(\sqrt{2}\)

Sửa đề lại cho đúng nhé : 

\(\sqrt{12x^2-17x+5}=\sqrt{12x^2-12x-5x+5}\)

\(=\)\(\sqrt{12x\left(x-1\right)-5\left(x-1\right)}=\sqrt{\left(x-1\right)\left(12x-5\right)}\)

\(btxđ\Leftrightarrow\left(x-1\right)\left(12x-5\right)\ge0\)

\(\Rightarrow\orbr{\begin{cases}x-1\ge0;12x-5\ge0\\x-1< 0;12x-5< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x\ge1;x\le\frac{12}{5}\left(tm\right)\\x< 1;x>\frac{12}{5}\left(ktm\right)\end{cases}}\)

\(\Rightarrow1\le x\le\frac{12}{5}\)

hình như ko phải \(\frac{12}{5}\)mà là \(\frac{5}{12}\)

Anh ơi mik mấy bài toán khó như thế này mik tham khảo trên H.vn nhé

Toán lớp 9 của bn hơi khó , có gì bn lên lazi,vn hoặc hoc.24.vn để hỏi nha 
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