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Câu 3:
\(A=cos\frac{\pi}{7}.cos\frac{5\pi}{7}.cos\frac{4\pi}{7}=cos\frac{\pi}{7}.cos\left(\pi-\frac{2\pi}{7}\right).cos\frac{4\pi}{7}\)
\(A=-cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.2sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}\)
\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{8}sin\frac{8\pi}{7}=-\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=\frac{1}{8}sin\frac{\pi}{7}\)
\(\Rightarrow A=\frac{1}{8}\)
Câu 4:
Đầu tiên ta chứng minh công thức:
\(tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}\)
Áp dụng để biến đổi tử số:
\(tan30+tan60+tan40+tan50=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{cos30.cos60}+\frac{1}{cos40.cos50}\)
\(=\frac{2}{cos90+cos30}+\frac{2}{cos90+cos10}=\frac{2}{cos30}+\frac{2}{cos10}=2\left(\frac{cos30+cos10}{cos30.cos10}\right)\)
\(=2\left(\frac{2cos20.cos10}{cos30.cos10}\right)=\frac{4.cos20}{cos30}=\frac{8\sqrt{3}}{3}.cos20\)
\(\Rightarrow A=\frac{\frac{8\sqrt{3}}{3}cos20}{cos20}=\frac{8\sqrt{3}}{3}\)
Câu 5:
\(cos54.cos4-cos36.cos86=cos54.cos4-cos\left(90-54\right).cos\left(90-4\right)\)
\(=cos54.cos4-sin54.sin4=cos\left(54+4\right)=cos58\)
Câu 1:
\(A=\frac{1}{2sin10}-2sin70=\frac{1-4sin10.sin70}{2sin10}=\frac{1+2\left(cos80-cos60\right)}{2sin10}\)
\(=\frac{1+2cos80-1}{2sin10}=\frac{2cos80}{2sin10}=\frac{sin10}{sin10}=1\)
Câu 2:
\(cos10.cos30.cos50.cos70=cos10.cos30.\frac{1}{2}\left(cos120+cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+cos10.cos20\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}\left(cos30+cos10\right)\right)\)
\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}cos30+\frac{1}{2}cos10\right)\)
\(=\frac{1}{2}.\frac{\sqrt{3}}{2}\left(-\frac{1}{2}cos10+\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}cos10\right)\)
\(=\frac{3}{16}\)
4. đặt \(\sqrt[3]{x+24}=a\) và \(\sqrt{12-x}=b\)(b>=0)
==>ta có hệ pt
\(\int_{a^3+b^2=36}^{a+b=6}\)<=> \(\int_{a^3+\left(6-a\right)^2=36}^{b=6-a}\)<=> \(\int_{b=6-a}^{a^3+a^2-12a=0}\)<=> \(\int_{b=6-a}^{a\left(a^2+a-12\right)=0}\)<=>\(\int_{b=6-a}^{a\left(a+4\right)\left(a-3\right)=0}\)
đến đây bạn tự tìm a;b rufit hay vào tìm x là ok
3. \(\Leftrightarrow\sqrt[3]{2x^2}-\sqrt[3]{x+1}+\sqrt[3]{2x^2+1}-\sqrt[3]{x+2}=0\)
\(\Leftrightarrow\frac{2x^2-x-1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{2x^2-x-1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}=0\)
\(\Leftrightarrow2x^2-x-1=0\)
( do \(\frac{1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}>0\forall xTMĐK\))
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2=\frac{9}{8}\Leftrightarrow\left(x-\frac{1}{4}\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}=\frac{3}{4}\\x-\frac{1}{4}=-\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\) ( TM )
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x>-2\\x\ne3\end{matrix}\right.\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ne1\\x\ne\frac{1}{2}\end{matrix}\right.\)
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\) => bpt vô nghiệm
b/ ĐKXĐ: \(x>1\)
\(bpt\Leftrightarrow x-2< 2\Leftrightarrow x< 4\)
\(\Rightarrow1< x< 4\)
c/ \(\frac{x+2}{3}-2x-2>0\)
\(\Leftrightarrow\frac{x+2-6x-6}{3}>0\Leftrightarrow x+2-6x-6>0\Leftrightarrow x< -\frac{4}{5}\)
d/ \(bpt\Leftrightarrow\frac{3x+5}{2}-\frac{x+2}{3}-x-1\le0\)
\(\Leftrightarrow\frac{9x+15-2x-4-6x-6}{6}\le0\)
\(\Leftrightarrow x\le-5\)
a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)
hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)
\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)
\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)
\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)
b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)
Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)
Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được
\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)
(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)
\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)
(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)
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1. ĐK x >1
pt \(\Leftrightarrow\frac{1}{\sqrt{x}-\sqrt{x-1}}\left(m\sqrt{x}+\frac{1}{\sqrt{x-1}}-16\sqrt[4]{\frac{x^3}{x-1}}\right)=1\)
\(\Leftrightarrow m\sqrt{x}+\frac{1}{\sqrt{x-1}}-16\sqrt[4]{\frac{x^3}{x-1}}=\sqrt{x}-\sqrt{x-1}\)
\(\Leftrightarrow m\sqrt{x\left(x-1\right)}+1-16\sqrt[4]{x^3\left(x-1\right)}=\sqrt{x\left(x-1\right)}-x+1\)
\(\Leftrightarrow\left(m-1\right)\sqrt{x\left(x-1\right)}-16\sqrt[4]{x^3\left(x-1\right)}+x=0\)
\(\Leftrightarrow\left(m-1\right)\sqrt{\frac{x-1}{x}}-16\sqrt[4]{\frac{x-1}{x}}+1=0\)
Đặt rồi đưa về phương trình bậc 2: \(\left(m-1\right)t^2-16t+1=0\)
2. ĐK:...
\(\sqrt{x-4-2\sqrt{x-4}+1}+\sqrt{x-4-2.\sqrt{x-4}.3+9}=m\)
\(\Leftrightarrow\left|\sqrt{x-4}-1\right|+\left|\sqrt{x-4}-3\right|=m\)Tìm m để pt có đúng 2 nghiệm. Tự làm nhé!
\(3.\) ĐK:...
Đặt: \(\left(x^2-3x-4\right)=a\)
\(\sqrt{x+7}=b\)
Ta có: \(ab-m\left(a-b\right)-m^2=0\Leftrightarrow m^2+m\left(a-b\right)-ab=0\)
\(\Delta=\left(a-b\right)^2+4ab=\left(a+b\right)^2\)
pt có 2 nghiệm : \(\orbr{\begin{cases}m=\frac{b-a-\left(a+b\right)}{2}=-a\\m=\frac{b-a+\left(a+b\right)}{2}=b\end{cases}}\)
Khi đó: \(\orbr{\begin{cases}m=-\left(x^2-3x-4\right)\\m=\sqrt{x+7}\end{cases}}\)
pt <=> \(\left(m+x^2-3x-4\right)\left(m-\sqrt{x+7}\right)=0\)Tìm m để pt có nhiều nghiệm nhất .
Nhìn BĐT 4 số ngán quá
\(1\ge4\sqrt[4]{\frac{1}{a^2b^2c^2d^2}}\Rightarrow abcd\ge16\)
\(\Rightarrow VT=\frac{abcd}{8}+2\ge4\) (1)
Mà \(VP=\frac{a+c}{\sqrt{ac}}+\frac{b+d}{\sqrt{bd}}\le\frac{2\left(a+c\right)}{a+c}+\frac{2\left(b+d\right)}{b+d}=4\) (2)
(1);(2) \(\Rightarrow\) đpcm
Dấu "=" xảy ra khi \(a=b=c=d=2\)
\(S=\frac{\sqrt{3}sin70-cos70}{\sqrt{3}sin70.cos70}=\frac{\frac{\sqrt{3}}{2}sin70-\frac{1}{2}cos70}{\frac{\sqrt{3}}{4}.2sin70.cos70}\)
\(=\frac{sin70.cos30-sin30.cos70}{\frac{\sqrt{3}}{4}sin140}=\frac{sin\left(70-30\right)}{\frac{\sqrt{3}}{4}sin\left(180-40\right)}\)
\(=\frac{sin40}{\frac{\sqrt{3}}{4}sin40}=\frac{4}{\sqrt{3}}\)